SCHEME OF WORK (STATISTICS & ALGEBRA)
WEEK TOPICS
1. Tabulation of data; Presentation of data using Histogram and cumulative frequency curve.
2. Measures of Central Tendency Mean , Median, Mode of Ungrouped data
3. Mean, Mode and Median of grouped data
4. Cumulative Frequency & Median
5 Percentile, Quartile & Inter- Quartile
6. Heights and Distances; Elevation and Depression ( continuation )
7. Theory of Logarithms and Laws of Logarithms; Solving problems without log tables and miscellaneous problems on Logarithms
8. Solution of Quadratic Equation; Word Problems Leading to Quadratic equations. Solving Simultaneous Linear and Quadratic Equations
9. Trigonometry: Trigonometric Ratios of Angles between 00 and 3600
10. Graphs of Trigonometric Functions; Harder Problems on Trigonometric Graphs.
11. REVISION
1ST TERM
WEEK 1
Topic: Grouped Data
Tabulation of group data
Presentation of group data
CONTENT.
When data has a large number of values, it is cumbersome to prepare its frequency table, hence the data are organize into classes or groups to overcome this problem. E.g. 0-4, 5-9, 10-14 etc.
The range of the classes are first considered before we group the data.
When data is divided into groups it is called a grouped frequency distribution
Grouped Frequency Distribution ; The groups into which the data are arranged are called class intervals
e.g 15-19.
Class Limit: The end number of each class intervals are called class limits of that intervals, Consider the class interval 10-14,
10 = lower class limit
14 = upper class limit.
Class Boundaries
When data is given to the nearest unit, the class interval 34-37, has a lower class boundary of 33.5 and upper class boundary 37.5.
Hence, 33.5 — 37.5 are called the boundaries.
e.g Consider the interval below: 10-14, 15-19 etc. To obtain the class boundaries of 15-19
14 + 15 ,
2
= 14.5
19+ 20
2
= 19.5
Class Width: This is the difference between the upper class boundary and the lower class boundary. Class Mark: This is the centre or mid-point of any class interval. It is obtained by adding the lower and upper limits together and dividing the result by 2. e.g. Find the class of the following class intervals 40-44, 45 —49, 50- 54 etc.
Solution:
Cumulative Frequency Table: This is the table that shows the cumulative frequency of each of the classes and it is the running total of the frequencies class by class, giving the total frequency.
Example: In a mock examination for the final year chemistry class,, the following were obtained by 50 students.
71 63 70 45 59 82 61 79 37 89
33 56 39 42 64 73 59 67 72 60
46 36 61 87 91 67 54 72 39 43
57 65 45 52 35 46 64 37 95 86
76 73 67 71 74 82 61 59 58 43.
Using class interval 31 — 40, 41-50 etc.
Construct a table showing the following columns: class interval, class boundary, class mark, frequency and cumulative frequency.
Solution:
https://youtu.be/QI5OZN9t-D0
Evaluation: The following figures show how many people visited an art gallery each day for 50 days,
30 60 53 54 35 51 13 36 43 44
44 38 39 52 45 39 25 27 31 44
29 46 49 42 47 43 34 52 50 39
53 25 28 51 54 33 35 45 51 59
19 28 34 42 48 51 20 25 37 38.
Arrange the figures in classes, beginning with 11-20, 21-30 etc and make a table showing class interval, class boundary, class mark, frequency and cumulative frequency.
https://youtu.be/7CcmBxf7lu0
https://youtu.be/tcU_hApd-j0
ASSIGNMENT.
1 . The thickness of 20 samples of steel plate are measured and the results (in mm) to two significant figures are as follows:
7.3, 7.1, 6.6, 7.0, 7.8, 7.3, 7.5, 6.2, 6.9,
6.7, 6.5, 6.8, 7.2, 7.4, 6.5, 6.9, 7.2, 7.6,
7.0, 6.8
Compile a table showing the frequency distribution, cumulative frequency, class boundary, class mark and class intervals using class 6.2 — 6.4, 6.5 — 6.7 etc.
TOPIC: STATISTICS I
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Present data in a frequency table
ii. Represent statistical data graphically using pie charts and bar chart
iii. Read graphs and draw sample references from them
iv. Classify large data into groups and represent the grouped data graphically using histograms and frequency polygons
v. Estimate the mode of a grouped data from the histogram
REFERENCE: NEW GEN. MATHS AND COMPRE. MATHS
CONTENT: DATA:
They are any numerical facts or information which can be measured or given a numerical qualification.
TABULAR REPRESENTATION OF DATA
When data are collected they may be presented in the raw form in which they are gathered. It may also be represented in a tabular form.
FREQUENCY TABLE:
This shows the number of times a particular event occurs in a given data or information. It shows at a glance the number of time each event appears.
EXAMPLE I:
A teacher gives a spelling test to 40 students. The number of errors made by the students is shown in table below. Represent the data in a frequency table.
Number of errors in a spelling test
Thus the frequency table is shown below
Number of errors - 0 1 2 3 4 5
Frequency - 8 7 10 7 5 3
BAR CHART
This is a statistical graph in which bars are drawn such that their lengths or height are proportional to the quantities they represent.
PIE CHART
This is a circular graph in which numerical data are represented by sectors of a circle. The angles of the sectors are proportional to the frequencies of the item they represent.
EXAMPLE 2:
The table below shows the numbers of babies born in 12 villages on a certain day
No. of babies | 4| 5| 6| 7| 8|
.................................................
No. of villages| 1| 4| 2| 3| 2|
a. Draw a bar chart to illustrate this information
b. State the mode and median of the distribution
c. Calculate the mean of the distribution to the nearest number
The mode is 5 babies
The median is 6 i.e
4 5 5 5 5 6 6 7 7 7 8 8
Mean number of babies per village
= total number of babies
Total number of villages
= 1 x 4 + 4 x 5 + 2 x 6 + 3 x 7 + 2 x 8
12
= 4 + 20 + 12 + 2 + 21 + 16
12
= 73
12
= 6 ½
= 6 to the nearest whole number
https://youtu.be/gVr2eYfc4vk
EXAMPLE 3
In a certain school, the lesson periods for each week are itemized below English 10, Mathematics 7, Biology 3, Statistics 4, Igbo 3, others 9. Draw a pie chart to illustrate this information.
GROUPED DATA
(Grouped frequency distribution)
When the number of entries in a given data is large, the construction of a frequency distribution may be difficult, hence there will be need to group the data.
In grouping any numerical data, the following statistical parameters must be examined.
CLASS INTERVAL
This is the number of groups that is classified. Consider the weights of a number of students given below measured in Kg
18 34 20 39 36 32 29 19 25 38
25 22 25 28 37 40 39 27 23 19
30 19 21 25 40 27 18 32 19 40
26 18 39 31 19 29 40 30 21 26
40 20 37 33 26 35 35 29 25 18
Suppose we want to classify the data into 8 equal intervals, the process is as follows:
The least weight is = 18
The greatest weight is = 40
The range (40-18) = 22
Therefore the width of each class which is the number of units in each grouping shall be 22/8 = 2.75.
This figure can be approximated to 3.
Therefore in order to get 8 equal intervals each grouping must contain 3 units. The following will then be the required class intervals:
Class intervals
CLASS LIMITS
The end numbers in the class intervals are called the class limits.
CLASS BOUNDARY
In theory, the class 18-20 will include all weights between 17.5 and 20.5. hence the umbers 17.5 and 20.5 are referred to as class boundaries.
CLASS MARK:
The centre or mid point of any class intervals is called the class mark. This is got by adding the lower and upper limits together and dividing the result by 2. Class marks are usually used for easy calculations most especially in finding the arithmetic mean of a grouped.
HISTOGRAM
This consists of a number of rectangles. The horizontal width of each rectangle is given by the class width. The height is such that the area of the rectangle is proportional to the frequency in that class interval.
FREQUENCY POLYGON
The frequency distribution can be shown on a line graph called a frequency polygon. Here the frequencies are plotted at the midpoints of each class interval. The points are joined by straight lines. The two end points are joined to the midpoints of the adjacent intervals on the horizontal axis.
https://youtu.be/PDpBPKf8z9E
EXAMPLE 4
Draw a histogram and frequency polygon for the frequency distribution given in table below
Class
Interval 0-4 5-7 10-14 15-19 20-24
Frequency 4 6 8 7 5
Calculate the mean of the distribution
EVALUATION/ ASSIGNMENT:
1. A company’s sales manager spent his monthly salary as itemized below
Food ----- 30%
Rent ----- 18%
Car maintenance ----- 25%
Savings ----- 12%
Taxes ----- 5%
Others ----- 10%
a. Represent this information on a pie chart
b. Find his savings at the end of the month, if his monthly salary was N60,000.
2. The distribution of workers at Dank Ventures is
DRIVERS - 25
MESSENGERS - 15
CLERKS - 30
CLEANERS - 60
TYPISTS - 20
Represent the information above using
(i) Pie chart
(ii) Bar chart
(iii) Histogram
(iv) Frequency polygon
3. Draw a histogram and frequency polygon of the frequency distribution in the table below:
CLASS| 8-14 |15-21 |22-28 |29-35 |36-42 |43-49
FREQUENCY| 3 |5 |8 |18 |9 |7
4a. Draw a histogram of the frequency distribution below:
CLASS| 1-10 |11-20 |21-30 |31-40 |41-50
FREQUENCY| 5 |12 |17 |10 |6
b. Estimate the mode of the distribution
5. Draw a histogram and a frequency polygon for the frequency distribution in table below:
CLASS| 1-5 |6-10 |11-15 |16-20 |21-25
FREQUENCY| 2 |4 |6 |5 |3
https://youtu.be/g08P2pk1_Es
Tabulation of group data
Presentation of group data
CONTENT.
When data has a large number of values, it is cumbersome to prepare its frequency table, hence the data are organize into classes or groups to overcome this problem. E.g. 0-4, 5-9, 10-14 etc.
The range of the classes are first considered before we group the data.
When data is divided into groups it is called a grouped frequency distribution
Grouped Frequency Distribution ; The groups into which the data are arranged are called class intervals
e.g 15-19.
Class Limit: The end number of each class intervals are called class limits of that intervals, Consider the class interval 10-14,
10 = lower class limit
14 = upper class limit.
Class Boundaries
When data is given to the nearest unit, the class interval 34-37, has a lower class boundary of 33.5 and upper class boundary 37.5.
Hence, 33.5 — 37.5 are called the boundaries.
e.g Consider the interval below: 10-14, 15-19 etc. To obtain the class boundaries of 15-19
14 + 15 ,
2
= 14.5
19+ 20
2
= 19.5
Class Width: This is the difference between the upper class boundary and the lower class boundary. Class Mark: This is the centre or mid-point of any class interval. It is obtained by adding the lower and upper limits together and dividing the result by 2. e.g. Find the class of the following class intervals 40-44, 45 —49, 50- 54 etc.
Solution:
Cumulative Frequency Table: This is the table that shows the cumulative frequency of each of the classes and it is the running total of the frequencies class by class, giving the total frequency.
Example: In a mock examination for the final year chemistry class,, the following were obtained by 50 students.
71 63 70 45 59 82 61 79 37 89
33 56 39 42 64 73 59 67 72 60
46 36 61 87 91 67 54 72 39 43
57 65 45 52 35 46 64 37 95 86
76 73 67 71 74 82 61 59 58 43.
Using class interval 31 — 40, 41-50 etc.
Construct a table showing the following columns: class interval, class boundary, class mark, frequency and cumulative frequency.
Solution:
https://youtu.be/QI5OZN9t-D0
Evaluation: The following figures show how many people visited an art gallery each day for 50 days,
30 60 53 54 35 51 13 36 43 44
44 38 39 52 45 39 25 27 31 44
29 46 49 42 47 43 34 52 50 39
53 25 28 51 54 33 35 45 51 59
19 28 34 42 48 51 20 25 37 38.
Arrange the figures in classes, beginning with 11-20, 21-30 etc and make a table showing class interval, class boundary, class mark, frequency and cumulative frequency.
https://youtu.be/7CcmBxf7lu0
https://youtu.be/tcU_hApd-j0
ASSIGNMENT.
1 . The thickness of 20 samples of steel plate are measured and the results (in mm) to two significant figures are as follows:
7.3, 7.1, 6.6, 7.0, 7.8, 7.3, 7.5, 6.2, 6.9,
6.7, 6.5, 6.8, 7.2, 7.4, 6.5, 6.9, 7.2, 7.6,
7.0, 6.8
Compile a table showing the frequency distribution, cumulative frequency, class boundary, class mark and class intervals using class 6.2 — 6.4, 6.5 — 6.7 etc.
TOPIC: STATISTICS I
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Present data in a frequency table
ii. Represent statistical data graphically using pie charts and bar chart
iii. Read graphs and draw sample references from them
iv. Classify large data into groups and represent the grouped data graphically using histograms and frequency polygons
v. Estimate the mode of a grouped data from the histogram
REFERENCE: NEW GEN. MATHS AND COMPRE. MATHS
CONTENT: DATA:
They are any numerical facts or information which can be measured or given a numerical qualification.
TABULAR REPRESENTATION OF DATA
When data are collected they may be presented in the raw form in which they are gathered. It may also be represented in a tabular form.
FREQUENCY TABLE:
This shows the number of times a particular event occurs in a given data or information. It shows at a glance the number of time each event appears.
EXAMPLE I:
A teacher gives a spelling test to 40 students. The number of errors made by the students is shown in table below. Represent the data in a frequency table.
Number of errors in a spelling test
Thus the frequency table is shown below
Number of errors - 0 1 2 3 4 5
Frequency - 8 7 10 7 5 3
BAR CHART
This is a statistical graph in which bars are drawn such that their lengths or height are proportional to the quantities they represent.
PIE CHART
This is a circular graph in which numerical data are represented by sectors of a circle. The angles of the sectors are proportional to the frequencies of the item they represent.
EXAMPLE 2:
The table below shows the numbers of babies born in 12 villages on a certain day
No. of babies | 4| 5| 6| 7| 8|
.................................................
No. of villages| 1| 4| 2| 3| 2|
a. Draw a bar chart to illustrate this information
b. State the mode and median of the distribution
c. Calculate the mean of the distribution to the nearest number
The mode is 5 babies
The median is 6 i.e
4 5 5 5 5 6 6 7 7 7 8 8
Mean number of babies per village
= total number of babies
Total number of villages
= 1 x 4 + 4 x 5 + 2 x 6 + 3 x 7 + 2 x 8
12
= 4 + 20 + 12 + 2 + 21 + 16
12
= 73
12
= 6 ½
= 6 to the nearest whole number
https://youtu.be/gVr2eYfc4vk
EXAMPLE 3
In a certain school, the lesson periods for each week are itemized below English 10, Mathematics 7, Biology 3, Statistics 4, Igbo 3, others 9. Draw a pie chart to illustrate this information.
GROUPED DATA
(Grouped frequency distribution)
When the number of entries in a given data is large, the construction of a frequency distribution may be difficult, hence there will be need to group the data.
In grouping any numerical data, the following statistical parameters must be examined.
CLASS INTERVAL
This is the number of groups that is classified. Consider the weights of a number of students given below measured in Kg
18 34 20 39 36 32 29 19 25 38
25 22 25 28 37 40 39 27 23 19
30 19 21 25 40 27 18 32 19 40
26 18 39 31 19 29 40 30 21 26
40 20 37 33 26 35 35 29 25 18
Suppose we want to classify the data into 8 equal intervals, the process is as follows:
The least weight is = 18
The greatest weight is = 40
The range (40-18) = 22
Therefore the width of each class which is the number of units in each grouping shall be 22/8 = 2.75.
This figure can be approximated to 3.
Therefore in order to get 8 equal intervals each grouping must contain 3 units. The following will then be the required class intervals:
Class intervals
CLASS LIMITS
The end numbers in the class intervals are called the class limits.
CLASS BOUNDARY
In theory, the class 18-20 will include all weights between 17.5 and 20.5. hence the umbers 17.5 and 20.5 are referred to as class boundaries.
CLASS MARK:
The centre or mid point of any class intervals is called the class mark. This is got by adding the lower and upper limits together and dividing the result by 2. Class marks are usually used for easy calculations most especially in finding the arithmetic mean of a grouped.
HISTOGRAM
This consists of a number of rectangles. The horizontal width of each rectangle is given by the class width. The height is such that the area of the rectangle is proportional to the frequency in that class interval.
FREQUENCY POLYGON
The frequency distribution can be shown on a line graph called a frequency polygon. Here the frequencies are plotted at the midpoints of each class interval. The points are joined by straight lines. The two end points are joined to the midpoints of the adjacent intervals on the horizontal axis.
https://youtu.be/PDpBPKf8z9E
EXAMPLE 4
Draw a histogram and frequency polygon for the frequency distribution given in table below
Class
Interval 0-4 5-7 10-14 15-19 20-24
Frequency 4 6 8 7 5
Calculate the mean of the distribution
EVALUATION/ ASSIGNMENT:
1. A company’s sales manager spent his monthly salary as itemized below
Food ----- 30%
Rent ----- 18%
Car maintenance ----- 25%
Savings ----- 12%
Taxes ----- 5%
Others ----- 10%
a. Represent this information on a pie chart
b. Find his savings at the end of the month, if his monthly salary was N60,000.
2. The distribution of workers at Dank Ventures is
DRIVERS - 25
MESSENGERS - 15
CLERKS - 30
CLEANERS - 60
TYPISTS - 20
Represent the information above using
(i) Pie chart
(ii) Bar chart
(iii) Histogram
(iv) Frequency polygon
3. Draw a histogram and frequency polygon of the frequency distribution in the table below:
CLASS| 8-14 |15-21 |22-28 |29-35 |36-42 |43-49
FREQUENCY| 3 |5 |8 |18 |9 |7
4a. Draw a histogram of the frequency distribution below:
CLASS| 1-10 |11-20 |21-30 |31-40 |41-50
FREQUENCY| 5 |12 |17 |10 |6
b. Estimate the mode of the distribution
5. Draw a histogram and a frequency polygon for the frequency distribution in table below:
CLASS| 1-5 |6-10 |11-15 |16-20 |21-25
FREQUENCY| 2 |4 |6 |5 |3
https://youtu.be/g08P2pk1_Es
WEEK 2
MEASUREMENT OF CENTRAL TENDENCY.
Mean, Median and Mode of ungrouped data (revision )
Mean, Median and Mode of grouped data .
Measures of Central tendency: This is a measure of how the data are centrally placed. The three commonest measures of position, depending on the information required are the arithmetic mean, median and the mode.
https://youtu.be/kn83BA7cRNM
https://youtu.be/Cl7cnOyMvHs
MEAN: it is most widely used measure and sometimes called the arithmetical averages.
The mean of the number x1, x2, x3, x4 .........xn is given by
x = x1, x2, x3, x4 .........xn / n
or x = x/n
where x = sum of all items
n = number of items and if the data involves frequency ; x = Fx/F
Example: Calculate the mean of the numbers 37.5, 25.5, 30.5, 41.5, 52.5, 28.5
Solution.
MODE: The mode of a distribution is the value of the variable which occurs most often in the distribution. It is possible for a distribution to have more than one mode, if there were more than one item having the highest frequency.
Example.
1. Find the mode of the data
Solution the mode is 8 ( it appears three (3) times).
MEDIAN: This is the middle value of a set of data, when arranged in ascending or descending order.
Example. Find the median of these numbers
1. 35, 28, 42, 28, 56,70, 35
2. 25, 28. Solution:
1. Re-arranging the numbers : The median is 35.
2. (22, 25), 25, 28, 20
Median = 22 + 25/2 = 47/2 = 23.5/ 2
Example 3.
The table below is the distribution of test scored in a class:
a. if the mean score of the class is 6, find
i. the value of x
ii. the median score
iii. the modal score.
Solution ,
Mode = 7
EVALUATION
The table give the frequency distribution of marks obtained by a group of students in a test.
MEAN, MODE AND MEDIAN OF GROUPED DATA.
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Estimate the mode of a grouped data from the histogram
b. Estimate the median of a grouped data from the histogram
REFERENCE:
1. New General mathematics for West Africa SS3 by M.F. Macrae et al.
2. Excellence in mathematics for Senior secondary Schools 3
MEAN : The arithmetic mean of grouped frequency distribution can be obtained using class mark.
x = Fx/F Where x is the mid point of the class interval.
Assumed Mean Method: The mean of a frequency distribution can also be obtained by using a working mean called assumed mean.
X =A+Fd/F
Where d = x —A
x = class mark.
Example: The number of matches in 100 boxes are counted, and the results are shown in the table below:
MODE: The mode of the grouped frequency distribution can be determined geometrically and by interpolation method.
Mode from Histogram: The highest bar is the modal class. The mode can be determined by drawing a straight line from the right top corner of this rectangle to the right top corner of the adjacent rectangle on the left. Draw another line from the left top corner of the rectangle of the modal class to the left top corner of the adjacent rectangle on the right.
Example: The table below gives the distribution of ages in an institution.
Modal Class = 22 - 24
Mode = 21.5 + 1 = 22.5 Approx. 23yrs
https://youtu.be/lp2nTFdYGec
MODE FROM INTERPOLATION FORMULA.
The mode can be obtained using the formula
Median of Grouped Data
The median of a grouped data can be determined from a cumulative frequency curve and from interpolation formulae.
Median from cumulative frequency curve: The cumulative frequency curve can be used to determine the median.
Example: the table below shows the masses of 50 students in a secondary school
So to get the median, find (N/2) and check the table on the curve.
Median = 50/2 = 25th
Check 25th on the cumulative frequency and trace to the upper class boundary.
Median -29.5 +0.5 =31.5
Median = 30kg approximately.
Median from interpolation formula.
https://youtu.be/Urz4NGkXwis
EVALUATION :
The table below gives the distribution of masses (kg) of 40 people.
1. State the modal class of the distribution and obtain the mode.
2. Draw a cumulative frequency curve to illustrate the distribution.
3. Use your curve to estimate the median of the distribution
4, Calculate the mean of the distribution.
Assignment
New General Maths for SSS 3. Page 69, Ex. 8b, No 5h
https://youtu.be/wrGYheq0whk
TOPIC: ESTIMATING THE MODE AND MEDIAN
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Estimate the mode of a grouped data from the histogram
b. Estimate the median of a grouped data from the histogram
REFERENCE: NEW GEN. MATHS AND COMPRE. MATHS
CONTENT: THE MEDIAN
This is one of the measures of location (central tendency). The medium for an ungrouped data is the value of the middle item when the items are arranged in order of magnitude either ascending or descending when the number of items is odd. If the number of item is even, the arithmetic mean of the two middle items give the median score or value.
https://youtu.be/zjHfAhcU6kE
EXAMPLE 1:
Find the median of the following numbers
16, 13, 10, 23, 36, 9, 8, 48, 24
Solution:
Arrange the items in order
8, 9,10, 13, 16, 24, 36, 48
The middle term (median) is 16
EXAMPLE 2:
The table below shows the distinction of marks scored by some students in a physics test
Marks |22 24 36 42 45 48 56 60
Frequency |11 2 7 13 10 3 9 5
Solution:
A cumulative frequency table is prepared for the distribution
Marks |22 24 36 42 45 48 56
Frequency |11 2 7 13 10 3 9
Cumulative frequency| 11 13 20 33 43 46 55
Median = 60/2 = 30th of 31st/2
= 42 + 42/2 = 42%
MEDIAN FROM THE HISTOGRAM
Consider the item which divides the whole histogram into two equal areas.
Note: the total area is equal to the total frequency
EXAMPLE 3:
The following table shows the distribution of marks scored by a class of 80 students
Marks | number of students
--------------------------------------------------
10-14 | 18
15-19 | 9
20-24 | 11
25-29 | 25
30-34 | 14
35-39 | 3
Draw a histogram of the distribution and use it to estimate the median mark.
Solution:
Frequency density = frequency/ Mark
Marks Class marks Frequency Frequency density
10-14 12 18 1.5
15-19 17 9 0.5
20-24 22 11 0.5
25-29 27 25 0.9
30-34 32 14 0.4
35-39 37 3 0.1
------------------------------------------80
The total area here is equal to the total frequency and this is 80. Therefore we require half of 80 that is the area of 40.
From the histogram, the first three rectangles give area 37.5 i.e. (15 X12) + (0.5 X17) + (0.5 X22).
We therefore need 2.5 square units more. The area of the fourth rectangle is 0.9 X 25= 22.5. hence we need 1/9th of the whole area.
The median is 23.7 because this is the point at which the dotted line meets with the marked axis.
https://youtu.be/Urz4NGkXwis
MODE FROM HISTOGRAM
The mode can be read off from the histogram. Here are the steps to follow:
a. Identify the modal class by choosing the tallest rectangle from the histogram
b. Draw a straight line from the right top corner of the rectangle to the right top corner of the rectangle to its left.
c. Draw another line from the left top corner of the modal rectangle to the left top corner of the rectangle to the right.
d. Identify the point of intersection of the two lines just drawn.
e. Draw perpendicular line from the this point of intersection to the horizontal axes and read off the modal item from there.
https://youtu.be/vU1O_Gg0VJ8
EXAMPLE 4:
The table below gives the age distribution of 50 men in a village
Age (years) 20-29 30-39 40-49 50-59 60-69
No. of men 6 3 11 19 6
a. Draw the histogram of the distribution
b. Use your diagram to estimate the modal age correct to three significant figures.
Solution: the mode from the histogram is approximately 53.5
https://youtu.be/u9o-VYNU_l4
The table below shows the mark distribution of 100 students in a chemistry test
Mark 21-30 31-40 41-50 51-60 61-70 71-80 81-90
No. of students 22 16 18 9 10 13 12
Prepare a histogram of the distribution.
Find the median from the histogram
Find the mode from the histogram
https://youtu.be/2S3nVDY6Uso
ASSIGNMENT:
1. The table below shows the monthly profit in 100,000 of naira of a supermarket
Monthly profit 11-20 21-30 31-40 41-50 51-60 61-70
frequency 5 11 9 10 7 8
a. What is the modal monthly profit
b. Estimate the mean and median profit
TOPIC: USE OF ASSUMED MEAN IN CALCULATING A.M
PERIOD: 1
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Use assumed mean method in calculating the arithmetic mean.
b. Use the coded factor method in calculating the arithmetic mean.
REFERENCE: NEW GEN. MATHS AND COMPRE. MATHS
CONTENT:
When the data that is being considered consists of large items a short-cut to make the calculation quick can be adopted. This short cut method involves the use of an arbitrary number from the list of items being considered. The arbitrary number calculating the arithmetic mean is the assumed mean (A.M)
The actual arithmetic mean of the distribution is the addition of the assumed mean and the mean deviation
X = A + ∑fd/∑f
A = Assumed mean
d = Deviation from the assumed
f = Frequency
x= Arithmetic mean
https://youtu.be/P2MfpMO6sY8
EXAMPLE :
Using an assumed mean of 14 years calculate the mean age of the following ages of nine members of a household
43 12 19 14 35 22 13 16 26
Solution:
Age x deviation d= x –A
12 12-14 = -2
13 13-14 = -1
14 14-14 = 0
16 16-14 = 2
19 19-14 = 5
22 22-14 = 8
26 26-14 = 12
35 35-14 = 21
40 43-14 = 29
∑d = 74
Mean = 14 + 8.2 = 22.20
EVALUATION:
The masses of 35 girls in a class are
Masses (Kg) 30 35 40 45 50 55
Frequency 5 9 7 6 4 4
a. Use assumed mean of 40Kg. calculate the mean mark
EXAMPLE
The table below shows the shoe size worn by some employees, using an assumed of 72mm, calculate the mean sizes of the distribution
Shoe sizes 58 61 64 72 75 78 82
Frequency 5 7 4 3 4 6 1
Solution:
X = A + ∑fd/∑f
= 72 + -121/30
= 72 – 4.03
= 67.97mm
Shoe size dev = x-A f fd
56 58-72 = 14 5 -70
61 61-72 = -11 7 -77
64 64-72 = -8 4 -32
72 72-72 = 0 3 0
75 75-72 = +3 4 12
78 78-72 = 6 6 36
82 82-72 = 10 1 10
30 121
Mean = 72 – 4.03
= 67.97mm
https://youtu.be/6WUPyiFwdcA
EVALUATION:
The frequency distribution below shows the weights distribution of 100 applicants looking for job at XYZ industries
Weight (Kg) | No. of applicants
50-59 | 30
60-69 | 17
70-79 | 16
80-89 | 22
90-99 | 9
100-109 | 4
110-119 | 2
Use an assumed mean of 85Kg to calculate correct to two significant figures the mean of the distribution.
Use of Coded Factor
Example
The table below shows the distribution of the masses of a number of in mates in a zoological garden
Masses (Kg) | No. of inmates
40-49 | 7
50-59 | 5
60-69 | 10
70-79 | 18
80-89 | 8
90-99 | 12
100-109 | 10
Using an assumed mean of 84.5, calculate the mean mass of the distribution
Solution:
https://youtu.be/cHzLupR_uWY
EVALUATION:
1. The table below shows the monthly profit in 100,000 naira of a supermarket
Monthly profit N100,000 11-20 21-30 31-40 41-50 51-60 61-70
frequency 5 11 9 10 7 8
Use the coded factor to estimate the arithmetic mean.
https://youtu.be/Q35OitaKdQw
Topic: REVISION
Sub-topic:- Assumed mean method of calculating mean of a distribution.
Behavioral objective:- At the end of the lesson, the should be able to; calculate the mean of grouped distribution using assumed mean method.
QUESTION 1
The marks of 30 students in a mathematical test are as follows;
9 1 5 6 4 7 4 5 2 4
7 3 4 3 2 10 4 1 10 8
7 6 5 3 2 1 4 4 5 3
(a) construct a frequency table using class intervals of 1-2, 3-4 etc.
(b) from the frequency table calculate the mean of the distribution using the assumed mean method.
QUESTION 2
The table below represent the heights of 50 SS II students in a certain secondary school
Heights (cm) 156-160 161-165 166-170 171-175 176-180
No of Std. 9 11 10 5 5
Calculate the mean height using the class mark of the third class as the assumed mean.
Assignment: Ngm, page 132, Ex. 14b, No 2b
QUESTION 3
The number of days workers in an establishment were found absent on duty in a certain month are given in the table below.
No of absent (x) 1-5 6-10 11-15 16-20 21-25
No of worker (f) 10 7 7 5 1
Calculate the mean by assumed mean method
QUESTION 4 Calculate the mean of the distribution below by assumed mean method.
Length (cm) 131-140 141-150 151-160 161-170 171-180
No of planks 11 2 14 9 2
QUESTION 5 The table below shoes the distribution of market obtained by 100 students in a test.
Mark 10-19 20-29 30-39 40-49 50-59 60-69 70-79 80-89 90-99
F 1 3 7 20 30 22 10 5 2
Using 54.5 as an assumed mean, calculate the mean of the distribution.
QUESTION 6
The weight in kilograms of 100 men are shown in the table below.
Weights (kg) 100-104 105-109 110-114 115-119 120-124 125-129
Freq. 2 10 38 35 14 1
Find the mean of the distribution by assumed mean method
QUESTION 7
The following table shows the frequency table of weights of some items in a warehouse.
Weights (kg) 40-41 42-43 44-45 46-47 48-19 50-51 52-53
Freq. 1 8 26 38 19 7 1
Calculate the mean of the distribution using assumed mean method.
Mean, Median and Mode of ungrouped data (revision )
Mean, Median and Mode of grouped data .
Measures of Central tendency: This is a measure of how the data are centrally placed. The three commonest measures of position, depending on the information required are the arithmetic mean, median and the mode.
https://youtu.be/kn83BA7cRNM
https://youtu.be/Cl7cnOyMvHs
MEAN: it is most widely used measure and sometimes called the arithmetical averages.
The mean of the number x1, x2, x3, x4 .........xn is given by
x = x1, x2, x3, x4 .........xn / n
or x = x/n
where x = sum of all items
n = number of items and if the data involves frequency ; x = Fx/F
Example: Calculate the mean of the numbers 37.5, 25.5, 30.5, 41.5, 52.5, 28.5
Solution.
MODE: The mode of a distribution is the value of the variable which occurs most often in the distribution. It is possible for a distribution to have more than one mode, if there were more than one item having the highest frequency.
Example.
1. Find the mode of the data
Solution the mode is 8 ( it appears three (3) times).
MEDIAN: This is the middle value of a set of data, when arranged in ascending or descending order.
Example. Find the median of these numbers
1. 35, 28, 42, 28, 56,70, 35
2. 25, 28. Solution:
1. Re-arranging the numbers : The median is 35.
2. (22, 25), 25, 28, 20
Median = 22 + 25/2 = 47/2 = 23.5/ 2
Example 3.
The table below is the distribution of test scored in a class:
a. if the mean score of the class is 6, find
i. the value of x
ii. the median score
iii. the modal score.
Solution ,
Mode = 7
EVALUATION
The table give the frequency distribution of marks obtained by a group of students in a test.
MEAN, MODE AND MEDIAN OF GROUPED DATA.
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Estimate the mode of a grouped data from the histogram
b. Estimate the median of a grouped data from the histogram
REFERENCE:
1. New General mathematics for West Africa SS3 by M.F. Macrae et al.
2. Excellence in mathematics for Senior secondary Schools 3
MEAN : The arithmetic mean of grouped frequency distribution can be obtained using class mark.
x = Fx/F Where x is the mid point of the class interval.
Assumed Mean Method: The mean of a frequency distribution can also be obtained by using a working mean called assumed mean.
X =A+Fd/F
Where d = x —A
x = class mark.
Example: The number of matches in 100 boxes are counted, and the results are shown in the table below:
MODE: The mode of the grouped frequency distribution can be determined geometrically and by interpolation method.
Mode from Histogram: The highest bar is the modal class. The mode can be determined by drawing a straight line from the right top corner of this rectangle to the right top corner of the adjacent rectangle on the left. Draw another line from the left top corner of the rectangle of the modal class to the left top corner of the adjacent rectangle on the right.
Example: The table below gives the distribution of ages in an institution.
Modal Class = 22 - 24
Mode = 21.5 + 1 = 22.5 Approx. 23yrs
https://youtu.be/lp2nTFdYGec
MODE FROM INTERPOLATION FORMULA.
The mode can be obtained using the formula
Median of Grouped Data
The median of a grouped data can be determined from a cumulative frequency curve and from interpolation formulae.
Median from cumulative frequency curve: The cumulative frequency curve can be used to determine the median.
Example: the table below shows the masses of 50 students in a secondary school
So to get the median, find (N/2) and check the table on the curve.
Median = 50/2 = 25th
Check 25th on the cumulative frequency and trace to the upper class boundary.
Median -29.5 +0.5 =31.5
Median = 30kg approximately.
Median from interpolation formula.
https://youtu.be/Urz4NGkXwis
EVALUATION :
The table below gives the distribution of masses (kg) of 40 people.
1. State the modal class of the distribution and obtain the mode.
2. Draw a cumulative frequency curve to illustrate the distribution.
3. Use your curve to estimate the median of the distribution
4, Calculate the mean of the distribution.
Assignment
New General Maths for SSS 3. Page 69, Ex. 8b, No 5h
https://youtu.be/wrGYheq0whk
TOPIC: ESTIMATING THE MODE AND MEDIAN
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Estimate the mode of a grouped data from the histogram
b. Estimate the median of a grouped data from the histogram
REFERENCE: NEW GEN. MATHS AND COMPRE. MATHS
CONTENT: THE MEDIAN
This is one of the measures of location (central tendency). The medium for an ungrouped data is the value of the middle item when the items are arranged in order of magnitude either ascending or descending when the number of items is odd. If the number of item is even, the arithmetic mean of the two middle items give the median score or value.
https://youtu.be/zjHfAhcU6kE
EXAMPLE 1:
Find the median of the following numbers
16, 13, 10, 23, 36, 9, 8, 48, 24
Solution:
Arrange the items in order
8, 9,10, 13, 16, 24, 36, 48
The middle term (median) is 16
EXAMPLE 2:
The table below shows the distinction of marks scored by some students in a physics test
Marks |22 24 36 42 45 48 56 60
Frequency |11 2 7 13 10 3 9 5
Solution:
A cumulative frequency table is prepared for the distribution
Marks |22 24 36 42 45 48 56
Frequency |11 2 7 13 10 3 9
Cumulative frequency| 11 13 20 33 43 46 55
Median = 60/2 = 30th of 31st/2
= 42 + 42/2 = 42%
MEDIAN FROM THE HISTOGRAM
Consider the item which divides the whole histogram into two equal areas.
Note: the total area is equal to the total frequency
EXAMPLE 3:
The following table shows the distribution of marks scored by a class of 80 students
Marks | number of students
--------------------------------------------------
10-14 | 18
15-19 | 9
20-24 | 11
25-29 | 25
30-34 | 14
35-39 | 3
Draw a histogram of the distribution and use it to estimate the median mark.
Solution:
Frequency density = frequency/ Mark
Marks Class marks Frequency Frequency density
10-14 12 18 1.5
15-19 17 9 0.5
20-24 22 11 0.5
25-29 27 25 0.9
30-34 32 14 0.4
35-39 37 3 0.1
------------------------------------------80
The total area here is equal to the total frequency and this is 80. Therefore we require half of 80 that is the area of 40.
From the histogram, the first three rectangles give area 37.5 i.e. (15 X12) + (0.5 X17) + (0.5 X22).
We therefore need 2.5 square units more. The area of the fourth rectangle is 0.9 X 25= 22.5. hence we need 1/9th of the whole area.
The median is 23.7 because this is the point at which the dotted line meets with the marked axis.
https://youtu.be/Urz4NGkXwis
MODE FROM HISTOGRAM
The mode can be read off from the histogram. Here are the steps to follow:
a. Identify the modal class by choosing the tallest rectangle from the histogram
b. Draw a straight line from the right top corner of the rectangle to the right top corner of the rectangle to its left.
c. Draw another line from the left top corner of the modal rectangle to the left top corner of the rectangle to the right.
d. Identify the point of intersection of the two lines just drawn.
e. Draw perpendicular line from the this point of intersection to the horizontal axes and read off the modal item from there.
https://youtu.be/vU1O_Gg0VJ8
EXAMPLE 4:
The table below gives the age distribution of 50 men in a village
Age (years) 20-29 30-39 40-49 50-59 60-69
No. of men 6 3 11 19 6
a. Draw the histogram of the distribution
b. Use your diagram to estimate the modal age correct to three significant figures.
Solution: the mode from the histogram is approximately 53.5
https://youtu.be/u9o-VYNU_l4
The table below shows the mark distribution of 100 students in a chemistry test
Mark 21-30 31-40 41-50 51-60 61-70 71-80 81-90
No. of students 22 16 18 9 10 13 12
Prepare a histogram of the distribution.
Find the median from the histogram
Find the mode from the histogram
https://youtu.be/2S3nVDY6Uso
ASSIGNMENT:
1. The table below shows the monthly profit in 100,000 of naira of a supermarket
Monthly profit 11-20 21-30 31-40 41-50 51-60 61-70
frequency 5 11 9 10 7 8
a. What is the modal monthly profit
b. Estimate the mean and median profit
TOPIC: USE OF ASSUMED MEAN IN CALCULATING A.M
PERIOD: 1
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Use assumed mean method in calculating the arithmetic mean.
b. Use the coded factor method in calculating the arithmetic mean.
REFERENCE: NEW GEN. MATHS AND COMPRE. MATHS
CONTENT:
When the data that is being considered consists of large items a short-cut to make the calculation quick can be adopted. This short cut method involves the use of an arbitrary number from the list of items being considered. The arbitrary number calculating the arithmetic mean is the assumed mean (A.M)
The actual arithmetic mean of the distribution is the addition of the assumed mean and the mean deviation
X = A + ∑fd/∑f
A = Assumed mean
d = Deviation from the assumed
f = Frequency
x= Arithmetic mean
https://youtu.be/P2MfpMO6sY8
EXAMPLE :
Using an assumed mean of 14 years calculate the mean age of the following ages of nine members of a household
43 12 19 14 35 22 13 16 26
Solution:
Age x deviation d= x –A
12 12-14 = -2
13 13-14 = -1
14 14-14 = 0
16 16-14 = 2
19 19-14 = 5
22 22-14 = 8
26 26-14 = 12
35 35-14 = 21
40 43-14 = 29
∑d = 74
Mean = 14 + 8.2 = 22.20
EVALUATION:
The masses of 35 girls in a class are
Masses (Kg) 30 35 40 45 50 55
Frequency 5 9 7 6 4 4
a. Use assumed mean of 40Kg. calculate the mean mark
EXAMPLE
The table below shows the shoe size worn by some employees, using an assumed of 72mm, calculate the mean sizes of the distribution
Shoe sizes 58 61 64 72 75 78 82
Frequency 5 7 4 3 4 6 1
Solution:
X = A + ∑fd/∑f
= 72 + -121/30
= 72 – 4.03
= 67.97mm
Shoe size dev = x-A f fd
56 58-72 = 14 5 -70
61 61-72 = -11 7 -77
64 64-72 = -8 4 -32
72 72-72 = 0 3 0
75 75-72 = +3 4 12
78 78-72 = 6 6 36
82 82-72 = 10 1 10
30 121
Mean = 72 – 4.03
= 67.97mm
https://youtu.be/6WUPyiFwdcA
EVALUATION:
The frequency distribution below shows the weights distribution of 100 applicants looking for job at XYZ industries
Weight (Kg) | No. of applicants
50-59 | 30
60-69 | 17
70-79 | 16
80-89 | 22
90-99 | 9
100-109 | 4
110-119 | 2
Use an assumed mean of 85Kg to calculate correct to two significant figures the mean of the distribution.
Use of Coded Factor
Example
The table below shows the distribution of the masses of a number of in mates in a zoological garden
Masses (Kg) | No. of inmates
40-49 | 7
50-59 | 5
60-69 | 10
70-79 | 18
80-89 | 8
90-99 | 12
100-109 | 10
Using an assumed mean of 84.5, calculate the mean mass of the distribution
Solution:
https://youtu.be/cHzLupR_uWY
EVALUATION:
1. The table below shows the monthly profit in 100,000 naira of a supermarket
Monthly profit N100,000 11-20 21-30 31-40 41-50 51-60 61-70
frequency 5 11 9 10 7 8
Use the coded factor to estimate the arithmetic mean.
https://youtu.be/Q35OitaKdQw
Topic: REVISION
Sub-topic:- Assumed mean method of calculating mean of a distribution.
Behavioral objective:- At the end of the lesson, the should be able to; calculate the mean of grouped distribution using assumed mean method.
QUESTION 1
The marks of 30 students in a mathematical test are as follows;
9 1 5 6 4 7 4 5 2 4
7 3 4 3 2 10 4 1 10 8
7 6 5 3 2 1 4 4 5 3
(a) construct a frequency table using class intervals of 1-2, 3-4 etc.
(b) from the frequency table calculate the mean of the distribution using the assumed mean method.
QUESTION 2
The table below represent the heights of 50 SS II students in a certain secondary school
Heights (cm) 156-160 161-165 166-170 171-175 176-180
No of Std. 9 11 10 5 5
Calculate the mean height using the class mark of the third class as the assumed mean.
Assignment: Ngm, page 132, Ex. 14b, No 2b
QUESTION 3
The number of days workers in an establishment were found absent on duty in a certain month are given in the table below.
No of absent (x) 1-5 6-10 11-15 16-20 21-25
No of worker (f) 10 7 7 5 1
Calculate the mean by assumed mean method
QUESTION 4 Calculate the mean of the distribution below by assumed mean method.
Length (cm) 131-140 141-150 151-160 161-170 171-180
No of planks 11 2 14 9 2
QUESTION 5 The table below shoes the distribution of market obtained by 100 students in a test.
Mark 10-19 20-29 30-39 40-49 50-59 60-69 70-79 80-89 90-99
F 1 3 7 20 30 22 10 5 2
Using 54.5 as an assumed mean, calculate the mean of the distribution.
QUESTION 6
The weight in kilograms of 100 men are shown in the table below.
Weights (kg) 100-104 105-109 110-114 115-119 120-124 125-129
Freq. 2 10 38 35 14 1
Find the mean of the distribution by assumed mean method
QUESTION 7
The following table shows the frequency table of weights of some items in a warehouse.
Weights (kg) 40-41 42-43 44-45 46-47 48-19 50-51 52-53
Freq. 1 8 26 38 19 7 1
Calculate the mean of the distribution using assumed mean method.
WEEK 3
TOPIC: CUMMULATIVE FREQUENCY & MEDIAN
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Represent data using a cumulative frequency curve
b. Read the quartiles and percentiles of a distribution from a cumulative frequency curve.
REFERENCE: NEW GEN. MATHS AND COMPRE. MATHS
CONTENT: CUMULATIVE FREQUENCY
The data below can be illustrated by plotting the cumulative frequencies against the corresponding upper class boundary of the class intervals. The points are joined by a smooth curve called an OGIVE.
Figure below shows the ogive or cumulative frequency curve for the data
Class interval Frequency Cumulative frequency
21-30 2 2
31-40 5 5 + 2 = 7
41-50 7 7 + 7 = 14
51-60 9 9 + 14 = 23
61-70 11 11 + 23 = 34
71-80 8 8 + 34 = 42
81-90 5 5 + 42 = 47
91-100 3 3 + 47 = 50
EVALUATION:
Make a cumulative frequency table for the data below:
marks 1-10 11-20 21-30 31-40 41-50
No. of students 2 7 9 11 13
Hence draw the cumulative frequency curve
ASSIGNMENT:
CONTENT: MEDIAN OF GROUPED DATA
The median of a grouped data can be estimated form the cumulative frequency curve. The median is the mark that corresponds to the middle.
QUARTILES
Just as the median is the half –way up the distribution, the lower quartile is one quartile of the way up and the upper quartile is three-quartile of the way up. If the total is 7, then the lower quartile is the value of the ¼(n+1)th item and the upper quartile is the value of the ¾ (n+1)th item.
The lower quartile is usually called Q1
The second quartile is usually called Q2
The third quartile is usually called Q3
The semi-inter quartile range Q is defined as Q = Q3-Q1/2
Inter quartile range is (Q3-Q4)
https://youtu.be/K3wsOqIqA6k
Example
The table below shows the frequency distribution of the marks scored by 50 students in an examination
Marks % 0-9 10-19 20-29 30-39 40-49 50-59 60-69 70-79 80-89 90-99
Frequency 2 3 4 6 13 10 5 3 2 2
a. Make a cumulative frequency table for the distribution
b. Draw a cumulative frequency curve for the distribution.
Use your curve to estimate
- the median
- semi quartile range
- pass mark if 60% of students passed
Solution:
Mark % frequency Upper class boundary Cumulative frequency
0-9 2 9.5 2
10-19 3 19.5 5
20-29 4 29.5 9
30-39 6 39.5 15
40-49 13 49.5 28
50-59 10 59.5 38
60-69 5 69.5 43
70-79 3 79.5 46
80-89 2 89.5 48
90-99 2 99.5 50
50
c. from the graph
(i) the median = ½ (n + 1)th mark
= ½ (50 + 1)th mark
= 25.5th mark
Q2 =47.5
(ii) lower quartile, Q1 = ¼ (50 +1)th
= 12.75th mark
Q1 = 35.5
Upper quartile, Q3 = ¾ (30 +1) th
= 38.25th
Q3= 59.5
Semi quartile range
Q = Q3 – Q1/2= 59.5 – 35.5/2
= 24/2 = 12
If 60% of the students passed then 40% failed. Using the percentile axis the 40% corresponds to a mark of 43.5%. Thus, the pass mark is 43.5%.
https://youtu.be/wNamjO-JzUg
ASSIGNMENT:
Table below shows the number of eggs laid by chickens in a poultry farm in a year
a. draw a cumulative frequency curve of the distribution.
b. Use your graph to find the inter quartile range.
c. If a woman buys a chicken from the farm, what is the probability that the chicken lays at least 60 eggs in a year?
No. of eggs per year No. of chickens
45-49 10
50-54 36
55-59 64
60-64 52
65-69 28
70-74 10
EXERCISES
A grouped of 50 students were asked to estimate the length of a line to the nearest mm. their results arranged in order of size, are given below.
67 69 69 71 71 72 73 74 75 75
76 76 76 77 77 77 77 78 78 79
79 79 79 80 80 80 80 81 81 81 81
81 83 83 83 83 84 84 85 85
85 86 86 86 87 89 90 91 91 92 94
(a) make a frequency distribution table, taking five equal interval 65-69 etc. (b) what is the modal class of the distribution (c) draw a histogram of the distribution.
Assignment: Ngm, page 32, Ex. 4b, No 7
Content: the table below gives the masses in kg of 50 international athletics
67 75 79 56 59 60 64 76 58 50
54 65 78 66 65 65 70 62 70 62
70 61 83 51 74 69 59 73 71 74
73 81 69 82 71 53 67 72 66 74
85 63 58 69 75 61 62 68 52 68
Draw a frequency polygon of the distribution.
Assignment: Ngm. Page 31, Ex 4b, No 7
Content: the following gives the height in an of 30 students.
145 163 149 152 166 156 159
139 145 141 150 158 150 149 143
150 154 167 146 147 152 162
144 169 162 150 173 160 167 171
Draw (a) histogram (b) frequency polygon.
Evaluation: Ngm, page 31, Ex 4b, No 8
Assignment: Ngm, page 32, Ex., 4b, No 11
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Represent data using a cumulative frequency curve
b. Read the quartiles and percentiles of a distribution from a cumulative frequency curve.
REFERENCE: NEW GEN. MATHS AND COMPRE. MATHS
CONTENT: CUMULATIVE FREQUENCY
The data below can be illustrated by plotting the cumulative frequencies against the corresponding upper class boundary of the class intervals. The points are joined by a smooth curve called an OGIVE.
Figure below shows the ogive or cumulative frequency curve for the data
Class interval Frequency Cumulative frequency
21-30 2 2
31-40 5 5 + 2 = 7
41-50 7 7 + 7 = 14
51-60 9 9 + 14 = 23
61-70 11 11 + 23 = 34
71-80 8 8 + 34 = 42
81-90 5 5 + 42 = 47
91-100 3 3 + 47 = 50
EVALUATION:
Make a cumulative frequency table for the data below:
marks 1-10 11-20 21-30 31-40 41-50
No. of students 2 7 9 11 13
Hence draw the cumulative frequency curve
ASSIGNMENT:
CONTENT: MEDIAN OF GROUPED DATA
The median of a grouped data can be estimated form the cumulative frequency curve. The median is the mark that corresponds to the middle.
QUARTILES
Just as the median is the half –way up the distribution, the lower quartile is one quartile of the way up and the upper quartile is three-quartile of the way up. If the total is 7, then the lower quartile is the value of the ¼(n+1)th item and the upper quartile is the value of the ¾ (n+1)th item.
The lower quartile is usually called Q1
The second quartile is usually called Q2
The third quartile is usually called Q3
The semi-inter quartile range Q is defined as Q = Q3-Q1/2
Inter quartile range is (Q3-Q4)
https://youtu.be/K3wsOqIqA6k
Example
The table below shows the frequency distribution of the marks scored by 50 students in an examination
Marks % 0-9 10-19 20-29 30-39 40-49 50-59 60-69 70-79 80-89 90-99
Frequency 2 3 4 6 13 10 5 3 2 2
a. Make a cumulative frequency table for the distribution
b. Draw a cumulative frequency curve for the distribution.
Use your curve to estimate
- the median
- semi quartile range
- pass mark if 60% of students passed
Solution:
Mark % frequency Upper class boundary Cumulative frequency
0-9 2 9.5 2
10-19 3 19.5 5
20-29 4 29.5 9
30-39 6 39.5 15
40-49 13 49.5 28
50-59 10 59.5 38
60-69 5 69.5 43
70-79 3 79.5 46
80-89 2 89.5 48
90-99 2 99.5 50
50
c. from the graph
(i) the median = ½ (n + 1)th mark
= ½ (50 + 1)th mark
= 25.5th mark
Q2 =47.5
(ii) lower quartile, Q1 = ¼ (50 +1)th
= 12.75th mark
Q1 = 35.5
Upper quartile, Q3 = ¾ (30 +1) th
= 38.25th
Q3= 59.5
Semi quartile range
Q = Q3 – Q1/2= 59.5 – 35.5/2
= 24/2 = 12
If 60% of the students passed then 40% failed. Using the percentile axis the 40% corresponds to a mark of 43.5%. Thus, the pass mark is 43.5%.
https://youtu.be/wNamjO-JzUg
ASSIGNMENT:
Table below shows the number of eggs laid by chickens in a poultry farm in a year
a. draw a cumulative frequency curve of the distribution.
b. Use your graph to find the inter quartile range.
c. If a woman buys a chicken from the farm, what is the probability that the chicken lays at least 60 eggs in a year?
No. of eggs per year No. of chickens
45-49 10
50-54 36
55-59 64
60-64 52
65-69 28
70-74 10
EXERCISES
A grouped of 50 students were asked to estimate the length of a line to the nearest mm. their results arranged in order of size, are given below.
67 69 69 71 71 72 73 74 75 75
76 76 76 77 77 77 77 78 78 79
79 79 79 80 80 80 80 81 81 81 81
81 83 83 83 83 84 84 85 85
85 86 86 86 87 89 90 91 91 92 94
(a) make a frequency distribution table, taking five equal interval 65-69 etc. (b) what is the modal class of the distribution (c) draw a histogram of the distribution.
Assignment: Ngm, page 32, Ex. 4b, No 7
Content: the table below gives the masses in kg of 50 international athletics
67 75 79 56 59 60 64 76 58 50
54 65 78 66 65 65 70 62 70 62
70 61 83 51 74 69 59 73 71 74
73 81 69 82 71 53 67 72 66 74
85 63 58 69 75 61 62 68 52 68
Draw a frequency polygon of the distribution.
Assignment: Ngm. Page 31, Ex 4b, No 7
Content: the following gives the height in an of 30 students.
145 163 149 152 166 156 159
139 145 141 150 158 150 149 143
150 154 167 146 147 152 162
144 169 162 150 173 160 167 171
Draw (a) histogram (b) frequency polygon.
Evaluation: Ngm, page 31, Ex 4b, No 8
Assignment: Ngm, page 32, Ex., 4b, No 11
WEEK 4
TOPIC: PERCENTILE, QUARTILE & INTER QUARTILE
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Calculate the percentiles of a given distribution.
b. Calculate the quartiles of a given distribution
c. Calculate the inter quartile rage of a given distribution
REFERENCE: NEW GEN. MATHS AND COMPRE. MATHS
https://youtu.be/9ACtsUvafK0
CONTENT: QUARTILES
The median has earlier been described as the value which divides the whole area into two equal parts.
Quartiles are the values which divide the distribution into four equal parts namely:
First quartile Q1
Second quartile Q2
Third quartile Q3
Qi = Li + c/fi (N/4 – fb)
Where i= 1,2 or 3
And Li = lower class boundary
N = total frequency
Fb = cumulative frequency before the first quartile
F= frequency of the quartile
C= class size of the quartile
https://youtu.be/Snf6Wpn-L4c
Example 1:
The data below shows the age distribution of dwellers in a village.
Age Frequency
20-29 12
30-39 13
40-49 39
50-59 28
60-69 11
70-79 7
80-89 10
Calculate the first and the third quartiles for the distribution
Solution:
Age frequency cumulative frequency
19.5 - 29.5 12 12
29.5 – 39.5 13 25
39.5 – 49.5 39 64
49.5 – 59.5 28 92
59.5 – 69.5 11 103
69.5 – 79.5 7 110
79.5 – 89.5 10 120
From the table, the first quartile lies in the third row
i.e. Qi = N/4 =120/4 = 30th cumulative frequency
hence
Li = 39.5, N = 120, F = 25, f =39, c= 10
Qi = Li + c/fi (N/4 –f)
= 39.5 + 10/39 (30-25)
= 39.5 + 1.28 = 40.75
L2 = 49.5 N = 120 Fb = 64 f2 = 28 c=10
Q2 = 49.5 + 10/28 (90-64)
= 49.5 + 9.29
= 58.79
CONTENT: PERCENTILES AND DECILES
Percentiles are averages that divide the whole distribution into 100 equal parts.
On the other hand, deciles divide the whole distribution into 10 equal parts. We can calculate any percentile by the formula.
Pi = Li + c/fi (Ni/100 – Fb)
Where i = 1,2,3,…, or 99
Li = lower class boundary of the ith percentile
N = total frequency
Fb = cumulative frequency before the ith percentile
f = frequency of the ith percentile
c = class size of the ith percentile
Similarly, the formula for calculating any decile is
Di = Li + c/fi (N x I – Fb/100)
Students should note the following
100th percentile = 10th Decile
50th percentile = 5th Decile
80th percentile = 8th Decile
https://youtu.be/9QhU2grGU_E
EXAMPLE 2:
The table shows the mark distributions of 50 students
Marks | Frequency
16-20 | 2
21-25 | 4
26-20 | 15
31-35 | 9
36-40 | 7
41-45 | 13
For the distribution, calculate the following
i. 15th percentile
ii. 70th percentile
iii. 9th decile
iv. 6th decile
v. 60th percentile
Solution:
Marks Frequency Cumulative frequency
15.5 – 20.5 2 2
20.5 – 25.5 4 6
25.5. – 30.5 15 21
30.5 – 35.5 9 30
35.5 – 40.5 7 37
40.5 – 45.5 13 50
i. 15th which is 15/100 of 50
= 71/2th item
P15 = 25.5 + 5/15 (71/2 -6)
= 25.5 + 0.5 = 30
ii. 70th percentile which is 70/100 X 50
= 35th item
P70 = 35.5 + 5/7 (35-30)
= 35.5 + 3.6
= 39.1
iii. 9th decile is the 9/10 of 50 = 45th item
D9 = 40.5 +5/13 (45 -37)
= 40.5 + 3.08
+ 43.6
iv. 6th decile is the 6/10 of 50 = 30th item
D6 = 30.5 + 5/9 (30-21)
= 30.5 + 5 = 35.5
v. 60th percentile is the 60/100 X 50 = 30th item
vi. P60 = 30.5 + 5/9 (30-21) = 35.5
Note that the 6th decile and the 60th percentile give the same value. This is to confirm the earlier statement and analysis.
https://youtu.be/FFYvNrRGVOo
EVALUATION:
The table below shows the distribution of 100 men
Weight Kg 21-30 31-40 41-50 51-60 61-70 71-80 81-90
Frequency 7 13 20 15 22 15 8
a. Calculate the mode of the distribution
b. Find the lower and the upper quartiles
c. Calculate the 70th percentile
ASSIGNMENT:
The table below shows the monthly profit in 100,000 of naira of a supermarket.
Monthly profit 11-20 21-30 31-40 51-60 61-70 71-80
Frequency 5 11 9 10 7 8
a. What is the modal monthly profit?
b. Estimate the mean and median profit
c. Find the upper quartile profit
d. Find the probability of making a profit of above N3,500,000 in a particular month
e. Find the 70th percentile of the distribution
Reference books:-
1. New General mathematics for West Africa SS3 by M.F. Macrae et al.
2. MAN mathematics for Senior secondary Schools 3
REVISION
QUESTION 1
The following are the ages of the people that were interrogated by the police over the following charge.
Class (yrs) 6-10 1-15 16-20 21-25 26-30 31-35
Frequency 1 5 7 9 6 2
Construct a cumulative frequency table, hence draw the cumulative frequency curve (ogive)
Evaluation: Ngm, page 36, Ex. 4c, No 4, Ngm. Page 37, Ex. 4c, No 6
QUESTION 2
The table below shows the weight of 100 people in the community.
Weights (kg)| 1-20 21-40 41-60 61-80 81-100 101-120 121-140
Frequency | 7 12 10 10 12 26 23
Draw the cumulative frequency curve and estimate the median, quartile, decile, percentile from the ogive.
Evaluation: Ngm. Page 36, Ex. 4c, No 5. Ngm
QUESTION 3
The distribution of scores obtained by 20 students in a literature test is as follows
Scores| 0-2 3-5 6-8 9-11 12-14 15-17
F| 5 6 4 3 1 1
Draw the cumulative frequency curve and estimate the interquartile range of the distribution
Evaluation: Ngm, page 36, Ex, 4c, No 3 & 4
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Calculate the percentiles of a given distribution.
b. Calculate the quartiles of a given distribution
c. Calculate the inter quartile rage of a given distribution
REFERENCE: NEW GEN. MATHS AND COMPRE. MATHS
https://youtu.be/9ACtsUvafK0
CONTENT: QUARTILES
The median has earlier been described as the value which divides the whole area into two equal parts.
Quartiles are the values which divide the distribution into four equal parts namely:
First quartile Q1
Second quartile Q2
Third quartile Q3
Qi = Li + c/fi (N/4 – fb)
Where i= 1,2 or 3
And Li = lower class boundary
N = total frequency
Fb = cumulative frequency before the first quartile
F= frequency of the quartile
C= class size of the quartile
https://youtu.be/Snf6Wpn-L4c
Example 1:
The data below shows the age distribution of dwellers in a village.
Age Frequency
20-29 12
30-39 13
40-49 39
50-59 28
60-69 11
70-79 7
80-89 10
Calculate the first and the third quartiles for the distribution
Solution:
Age frequency cumulative frequency
19.5 - 29.5 12 12
29.5 – 39.5 13 25
39.5 – 49.5 39 64
49.5 – 59.5 28 92
59.5 – 69.5 11 103
69.5 – 79.5 7 110
79.5 – 89.5 10 120
From the table, the first quartile lies in the third row
i.e. Qi = N/4 =120/4 = 30th cumulative frequency
hence
Li = 39.5, N = 120, F = 25, f =39, c= 10
Qi = Li + c/fi (N/4 –f)
= 39.5 + 10/39 (30-25)
= 39.5 + 1.28 = 40.75
L2 = 49.5 N = 120 Fb = 64 f2 = 28 c=10
Q2 = 49.5 + 10/28 (90-64)
= 49.5 + 9.29
= 58.79
CONTENT: PERCENTILES AND DECILES
Percentiles are averages that divide the whole distribution into 100 equal parts.
On the other hand, deciles divide the whole distribution into 10 equal parts. We can calculate any percentile by the formula.
Pi = Li + c/fi (Ni/100 – Fb)
Where i = 1,2,3,…, or 99
Li = lower class boundary of the ith percentile
N = total frequency
Fb = cumulative frequency before the ith percentile
f = frequency of the ith percentile
c = class size of the ith percentile
Similarly, the formula for calculating any decile is
Di = Li + c/fi (N x I – Fb/100)
Students should note the following
100th percentile = 10th Decile
50th percentile = 5th Decile
80th percentile = 8th Decile
https://youtu.be/9QhU2grGU_E
EXAMPLE 2:
The table shows the mark distributions of 50 students
Marks | Frequency
16-20 | 2
21-25 | 4
26-20 | 15
31-35 | 9
36-40 | 7
41-45 | 13
For the distribution, calculate the following
i. 15th percentile
ii. 70th percentile
iii. 9th decile
iv. 6th decile
v. 60th percentile
Solution:
Marks Frequency Cumulative frequency
15.5 – 20.5 2 2
20.5 – 25.5 4 6
25.5. – 30.5 15 21
30.5 – 35.5 9 30
35.5 – 40.5 7 37
40.5 – 45.5 13 50
i. 15th which is 15/100 of 50
= 71/2th item
P15 = 25.5 + 5/15 (71/2 -6)
= 25.5 + 0.5 = 30
ii. 70th percentile which is 70/100 X 50
= 35th item
P70 = 35.5 + 5/7 (35-30)
= 35.5 + 3.6
= 39.1
iii. 9th decile is the 9/10 of 50 = 45th item
D9 = 40.5 +5/13 (45 -37)
= 40.5 + 3.08
+ 43.6
iv. 6th decile is the 6/10 of 50 = 30th item
D6 = 30.5 + 5/9 (30-21)
= 30.5 + 5 = 35.5
v. 60th percentile is the 60/100 X 50 = 30th item
vi. P60 = 30.5 + 5/9 (30-21) = 35.5
Note that the 6th decile and the 60th percentile give the same value. This is to confirm the earlier statement and analysis.
https://youtu.be/FFYvNrRGVOo
EVALUATION:
The table below shows the distribution of 100 men
Weight Kg 21-30 31-40 41-50 51-60 61-70 71-80 81-90
Frequency 7 13 20 15 22 15 8
a. Calculate the mode of the distribution
b. Find the lower and the upper quartiles
c. Calculate the 70th percentile
ASSIGNMENT:
The table below shows the monthly profit in 100,000 of naira of a supermarket.
Monthly profit 11-20 21-30 31-40 51-60 61-70 71-80
Frequency 5 11 9 10 7 8
a. What is the modal monthly profit?
b. Estimate the mean and median profit
c. Find the upper quartile profit
d. Find the probability of making a profit of above N3,500,000 in a particular month
e. Find the 70th percentile of the distribution
Reference books:-
1. New General mathematics for West Africa SS3 by M.F. Macrae et al.
2. MAN mathematics for Senior secondary Schools 3
REVISION
QUESTION 1
The following are the ages of the people that were interrogated by the police over the following charge.
Class (yrs) 6-10 1-15 16-20 21-25 26-30 31-35
Frequency 1 5 7 9 6 2
Construct a cumulative frequency table, hence draw the cumulative frequency curve (ogive)
Evaluation: Ngm, page 36, Ex. 4c, No 4, Ngm. Page 37, Ex. 4c, No 6
QUESTION 2
The table below shows the weight of 100 people in the community.
Weights (kg)| 1-20 21-40 41-60 61-80 81-100 101-120 121-140
Frequency | 7 12 10 10 12 26 23
Draw the cumulative frequency curve and estimate the median, quartile, decile, percentile from the ogive.
Evaluation: Ngm. Page 36, Ex. 4c, No 5. Ngm
QUESTION 3
The distribution of scores obtained by 20 students in a literature test is as follows
Scores| 0-2 3-5 6-8 9-11 12-14 15-17
F| 5 6 4 3 1 1
Draw the cumulative frequency curve and estimate the interquartile range of the distribution
Evaluation: Ngm, page 36, Ex, 4c, No 3 & 4
WEEK 5
Topic: The mean deviation of discrete and ungrouped data.
Behavioral objective:- At the end of the lesson, the should be able to; calculate the mean deviation of discrete and ungrouped data.
Reference books:-
1. New General mathematics for West Africa SS3 by M.F. Macrae et al.
2. MAN mathematics for Senior secondary Schools 3
https://youtu.be/aFpAgJ709nw
Content Element:-
The students estimated the mass in kilograms, of a piece of yam with the following result.
3.3, 2.4, 3.8, 3.0, 1.8,
3.1, 4.0, 2.7, 3.2, 2.5
Calculate
(a) the mean
(b) the mean deviation
Solution
Evaluation: Ngm. Page 69, Ex. 8b, No 4b, 4e & 5c, 5d
Topic: The mean deviation of grouped data.
Content: Calculate (i) the mean (ii) the mean deviation of the following distribution.
Mark interval 40-44 45-49 50-54 55-59 60-64 65-69 70-74
Frequency 2 5 5 8 10 5 3
Solution:
Mean deviation = ∑ |f| x – ‾x
∑f
Evaluation: Ngm, page 69, Ex. 8b, No 5e,f & h
https://youtu.be/ZSaXU7mzUso
Topic: Statistics
Sub-topic: Mean deviation
Behavioral Objective: At the end of lesson, the students should be able to estimate should be able to calculate mean deviation of distribution.
Reference books:-
1. New General mathematics for West Africa SS3 by M.F. Macrae et al.
The table below shows the distribution of weekly profit in naira recorded form certain mini-markets.
Weekly profit| 1-10 |11-20 |21-30 |31-40 |41-50
Frequency |6 |3 |12 |9 |10
Calculate (i) mean (ii) mean deviation of the distribution
Solution:
Evaluation: The table below shows the distribution of the weights of packets of biscuits per gram.
Weights of packets 10-19 20-29 30-39 40-49 50-59
Frequency 5 9 12 11 8
Find the mean and mean deviation of the distribution
Assignment: Ngm Page 69, Ex. 8b, No 5h
https://youtu.be/qqOyy_NjflU
REVISION EXERCISE1
(a) calculate the standard deviation of the numbers below 3, 5,6,7,8,1,2
(b) calculate the standard deviation and variance of the distribution below.
Score 1 2 3 4 5 6 7 8
Frequency 2 4 7 14 12 6 4 1
Solution:
Evaluation: The teacher gives class work to the students. Ngm. Page 69, Ex. 8b, No 5c & d
Assignment: The teacher gives assignment to the students. Ngm, page 69, Ex. 8b, No 5e & g
REVISION EXERCISE 2
Calculate the standard deviation and variance of the distribution below.
No of tomatoes 21-25 26-30 31-35 36-40 41-45
No of baskets 10 12 5 4 2
Solution:-
Evaluation: Ngm, page 71, Ex. 8c, No 2b&c
REVISION EXERCISE 3
The distance to the nearest of 30 students is given in the table below.
Distances (km) 0-4 5-9 10-14 15-19 20-24 25-29
No of students 2 10 8 6 3 1
Calculate the standard deviation and variance of the distribution.
Solution:
Evaluation: Ngm, page 72, Ex, 8c, No2d & 2e
Behavioral objective:- At the end of the lesson, the should be able to; calculate the mean deviation of discrete and ungrouped data.
Reference books:-
1. New General mathematics for West Africa SS3 by M.F. Macrae et al.
2. MAN mathematics for Senior secondary Schools 3
https://youtu.be/aFpAgJ709nw
Content Element:-
The students estimated the mass in kilograms, of a piece of yam with the following result.
3.3, 2.4, 3.8, 3.0, 1.8,
3.1, 4.0, 2.7, 3.2, 2.5
Calculate
(a) the mean
(b) the mean deviation
Solution
Evaluation: Ngm. Page 69, Ex. 8b, No 4b, 4e & 5c, 5d
Topic: The mean deviation of grouped data.
Content: Calculate (i) the mean (ii) the mean deviation of the following distribution.
Mark interval 40-44 45-49 50-54 55-59 60-64 65-69 70-74
Frequency 2 5 5 8 10 5 3
Solution:
Mean deviation = ∑ |f| x – ‾x
∑f
Evaluation: Ngm, page 69, Ex. 8b, No 5e,f & h
https://youtu.be/ZSaXU7mzUso
Topic: Statistics
Sub-topic: Mean deviation
Behavioral Objective: At the end of lesson, the students should be able to estimate should be able to calculate mean deviation of distribution.
Reference books:-
1. New General mathematics for West Africa SS3 by M.F. Macrae et al.
The table below shows the distribution of weekly profit in naira recorded form certain mini-markets.
Weekly profit| 1-10 |11-20 |21-30 |31-40 |41-50
Frequency |6 |3 |12 |9 |10
Calculate (i) mean (ii) mean deviation of the distribution
Solution:
Evaluation: The table below shows the distribution of the weights of packets of biscuits per gram.
Weights of packets 10-19 20-29 30-39 40-49 50-59
Frequency 5 9 12 11 8
Find the mean and mean deviation of the distribution
Assignment: Ngm Page 69, Ex. 8b, No 5h
https://youtu.be/qqOyy_NjflU
REVISION EXERCISE1
(a) calculate the standard deviation of the numbers below 3, 5,6,7,8,1,2
(b) calculate the standard deviation and variance of the distribution below.
Score 1 2 3 4 5 6 7 8
Frequency 2 4 7 14 12 6 4 1
Solution:
Evaluation: The teacher gives class work to the students. Ngm. Page 69, Ex. 8b, No 5c & d
Assignment: The teacher gives assignment to the students. Ngm, page 69, Ex. 8b, No 5e & g
REVISION EXERCISE 2
Calculate the standard deviation and variance of the distribution below.
No of tomatoes 21-25 26-30 31-35 36-40 41-45
No of baskets 10 12 5 4 2
Solution:-
Evaluation: Ngm, page 71, Ex. 8c, No 2b&c
REVISION EXERCISE 3
The distance to the nearest of 30 students is given in the table below.
Distances (km) 0-4 5-9 10-14 15-19 20-24 25-29
No of students 2 10 8 6 3 1
Calculate the standard deviation and variance of the distribution.
Solution:
Evaluation: Ngm, page 72, Ex, 8c, No2d & 2e
WEEK 6
Topic:
TRIGONOMETRY.
Trigonometry Ratios,
The basic trigonometric ratios can be defined in terms of the sides and angles of a right angled triangle.
Example 1: Given that sin 9 = 8/17, 0 < Ɵ<900
Find the values of (a) Cos Ɵ (b) tan Ɵ (c) Cos Ɵ (d) Cos Ɵ - Sin Ɵ/Sin2 Ɵ
Solution.
sin e = 8/17, Sin O = Qpp hyp Opp = 8, hyp using a right-angled triangle
https://youtu.be/KQhQSd7Wigo
COMPLEMENTARY ANGLE: Complementary angles are two angles whose sum is 900. Consider a right angled triangle XYZ.
Sin (Θ) = cos ( 90-Θ)
cos Θ = Sin (90- Θ)
Tan Θ = Cos (90-Θ)
Example: Solve the following equations:
1. Sin P = cos 26º
2. Cos x = sin ( x + 28º)
Sin s Θ = cos7 Θ
Solution.
Sin P = cos 26º= P+ 26º = 90º
= P=90-26º
P = 64º
2. Cos x = sin (x + 28º)
x + (x +28 ) 90
2x = 90 - 28,
2x = 62
x = 31
3. Sin 2 Θ = Cos 7Θ
2Θ + 7Θ = 90
9Θ = 90
Θ = 10.
https://youtu.be/o891OmYwemY
https://youtu.be/5tp74g4N8EY
EVALUATION: Solve the following equations.
1.Sin= Cos ( -20º)
2. Sin 2 Θ - Cos3Θ = 0
ASSIGNMENT :
Solve the equation:
1a. Sin = Cos 3
b Cos x = Sin (x+ 22º)
2. Given that cos x = 5/13, find the value of
i. tan x ii. Cos x iii. Sin2x+ cos2x
iv. Sinx— Cos x /Tanx
Topic: HEIGHTS AND DISTANCES
- Angles of Elevation and Depression.
Angle of Elevation: This is the angle between the normal eye level and the line though which the observer view an object above
Angle ACB = Θ is the angle of elevation.
Angle of Depression.
The angle which BD makes with the horizontal AB is angle of depression of the angle formed between the eye level of the observer and the object below:
Angle of elevation and depression problem(s) can be solved using the basic trigonometric ratios,
Sine and Cosine Rules
NB. Sine Rule for ABC= a/Sin A = b/Sin B = c/Sin C
Cosine rule :
a2 =B2 + C2 — 2bc Cos A
B2 = a2+C2 - 2ac Cos B
C2 = a2 + c2 - 2ab Cos C.
Example 1. A ladder 30m long rests against a vertical wall. If the ladder makes an angle of 63º with the wall, find the distance between the foot of the ladder and the wall.
Solution
AC = ladder, CB = wall
AB = distance between the foot of the ladder and the wall.
:.Sin 63º = AB
AB = x sin 63º
AB 30 x 0.8910
AB 26.7m
2. The angle of elevation of the top of a tower from a point on the horizontal ground, 40m away from the foot of the tower is 30º, find the height of the tower.
AB = height of the tower
Tan 30º = AB/40
AB = 40 x tan 30º
AB = 23.1m
https://youtu.be/DvpUgVnrYxs
https://youtu.be/K-LQNIAOEJU
https://youtu.be/NOv89Jc_YwQ
EVALUATION:
From the top of a building 10m high, the angle of elevation of a stone lying on the horizontal ground is 69º.
Calculate correct to I decimal places, the distance of the stone from the foot of the building.
ASSIGNMENT.
Find the angle of elevation of the nearest degree of the top of a church tower 180m high, from a point on the ground 75m from its foot.
Heights and Distances ( continuation).
Example 1:
The angle of elevation of the top of a vertical pole from a height I. 54m above on a horizontal ground is 40º. The foot of the pole is on the same horizontal ground and the point of observation is 20m from the pole. Calculate correct to 3.s.f.
i. the height of the pole
ii. The angle of depression of the foot of the pole from the point of observation
AB = x tan 43º ................... 1
From ADB, tan 32º = AB/8+x
AB = (8 + x) tan 32º . . . . . . . . .2
Equating eqn 1 and 2
x tan 43º = (8 + x ) tan 32º
x tan 43º = 8 tan 32º = x tan 32º
x tan 43º -x tan 32º = 8 tan 32º
x (tan 43º — tan 32º ) = 8 tan 32º
x = 8 tan 32º/Tan43º - tan 32º
8 x 0.6249/0.9325 - 0.6249
x = 4.9992/0.3076
x = 16.25m
Substitute x into equation ...1 or equation ...2 Using equation 1; AB = 16.25 x tan 43º
= 16.25 x 0.9325.
AB 15.15m Height of the tower — 15.15m.
https://youtu.be/QmDhai2fKzk
https://youtu.be/qXxTu0fg4_8
ASSIGNMENT.
The feet of two vertical pole of height 3m and 7m are in line with a point P on the ground, the smaller pole being between the taller pole and P and at a distance of 20m from P. the angle of elevation of the top (T) of the taller pole from the top ( R ) of the smaller pole is 30º Calculate :
1. distance RT
2. Distance of the foot of the taller pole from P, correct to 3 significant figures.
3. Angles of elevation of T from P, correct to one decimal place.
TRIGONOMETRY.
Trigonometry Ratios,
The basic trigonometric ratios can be defined in terms of the sides and angles of a right angled triangle.
Example 1: Given that sin 9 = 8/17, 0 < Ɵ<900
Find the values of (a) Cos Ɵ (b) tan Ɵ (c) Cos Ɵ (d) Cos Ɵ - Sin Ɵ/Sin2 Ɵ
Solution.
sin e = 8/17, Sin O = Qpp hyp Opp = 8, hyp using a right-angled triangle
https://youtu.be/KQhQSd7Wigo
COMPLEMENTARY ANGLE: Complementary angles are two angles whose sum is 900. Consider a right angled triangle XYZ.
Sin (Θ) = cos ( 90-Θ)
cos Θ = Sin (90- Θ)
Tan Θ = Cos (90-Θ)
Example: Solve the following equations:
1. Sin P = cos 26º
2. Cos x = sin ( x + 28º)
Sin s Θ = cos7 Θ
Solution.
Sin P = cos 26º= P+ 26º = 90º
= P=90-26º
P = 64º
2. Cos x = sin (x + 28º)
x + (x +28 ) 90
2x = 90 - 28,
2x = 62
x = 31
3. Sin 2 Θ = Cos 7Θ
2Θ + 7Θ = 90
9Θ = 90
Θ = 10.
https://youtu.be/o891OmYwemY
https://youtu.be/5tp74g4N8EY
EVALUATION: Solve the following equations.
1.Sin= Cos ( -20º)
2. Sin 2 Θ - Cos3Θ = 0
ASSIGNMENT :
Solve the equation:
1a. Sin = Cos 3
b Cos x = Sin (x+ 22º)
2. Given that cos x = 5/13, find the value of
i. tan x ii. Cos x iii. Sin2x+ cos2x
iv. Sinx— Cos x /Tanx
Topic: HEIGHTS AND DISTANCES
- Angles of Elevation and Depression.
Angle of Elevation: This is the angle between the normal eye level and the line though which the observer view an object above
Angle ACB = Θ is the angle of elevation.
Angle of Depression.
The angle which BD makes with the horizontal AB is angle of depression of the angle formed between the eye level of the observer and the object below:
Angle of elevation and depression problem(s) can be solved using the basic trigonometric ratios,
Sine and Cosine Rules
NB. Sine Rule for ABC= a/Sin A = b/Sin B = c/Sin C
Cosine rule :
a2 =B2 + C2 — 2bc Cos A
B2 = a2+C2 - 2ac Cos B
C2 = a2 + c2 - 2ab Cos C.
Example 1. A ladder 30m long rests against a vertical wall. If the ladder makes an angle of 63º with the wall, find the distance between the foot of the ladder and the wall.
Solution
AC = ladder, CB = wall
AB = distance between the foot of the ladder and the wall.
:.Sin 63º = AB
AB = x sin 63º
AB 30 x 0.8910
AB 26.7m
2. The angle of elevation of the top of a tower from a point on the horizontal ground, 40m away from the foot of the tower is 30º, find the height of the tower.
AB = height of the tower
Tan 30º = AB/40
AB = 40 x tan 30º
AB = 23.1m
https://youtu.be/DvpUgVnrYxs
https://youtu.be/K-LQNIAOEJU
https://youtu.be/NOv89Jc_YwQ
EVALUATION:
From the top of a building 10m high, the angle of elevation of a stone lying on the horizontal ground is 69º.
Calculate correct to I decimal places, the distance of the stone from the foot of the building.
ASSIGNMENT.
Find the angle of elevation of the nearest degree of the top of a church tower 180m high, from a point on the ground 75m from its foot.
Heights and Distances ( continuation).
Example 1:
The angle of elevation of the top of a vertical pole from a height I. 54m above on a horizontal ground is 40º. The foot of the pole is on the same horizontal ground and the point of observation is 20m from the pole. Calculate correct to 3.s.f.
i. the height of the pole
ii. The angle of depression of the foot of the pole from the point of observation
AB = x tan 43º ................... 1
From ADB, tan 32º = AB/8+x
AB = (8 + x) tan 32º . . . . . . . . .2
Equating eqn 1 and 2
x tan 43º = (8 + x ) tan 32º
x tan 43º = 8 tan 32º = x tan 32º
x tan 43º -x tan 32º = 8 tan 32º
x (tan 43º — tan 32º ) = 8 tan 32º
x = 8 tan 32º/Tan43º - tan 32º
8 x 0.6249/0.9325 - 0.6249
x = 4.9992/0.3076
x = 16.25m
Substitute x into equation ...1 or equation ...2 Using equation 1; AB = 16.25 x tan 43º
= 16.25 x 0.9325.
AB 15.15m Height of the tower — 15.15m.
https://youtu.be/QmDhai2fKzk
https://youtu.be/qXxTu0fg4_8
ASSIGNMENT.
The feet of two vertical pole of height 3m and 7m are in line with a point P on the ground, the smaller pole being between the taller pole and P and at a distance of 20m from P. the angle of elevation of the top (T) of the taller pole from the top ( R ) of the smaller pole is 30º Calculate :
1. distance RT
2. Distance of the foot of the taller pole from P, correct to 3 significant figures.
3. Angles of elevation of T from P, correct to one decimal place.
WEEK 7
Topic: Theory of Logarithm and Laws of Logarithms
The logarithm of a number N to a given base 'a' is the index to which the base must be raised in order to get the given number.
Thus if N = ax, it implies that loga N = x. The form N = ax is known as the index from while Loga N = x is known as the logarithmic form.
https://youtu.be/ntBWrcbAhaY
Examples:
Evaluate the following logarithms:
(1) Log4 8 (2) Log 6216. (3) log0.2 25 (4) Loga 0.0625.
Solutions
1. Let log4 8 = y then 4y = 8 change both sides to base 2 (22)y=23)
i.e 22y =23 equating the indices:
2y = 3
y = 3/2
log4 8 = 1½
(2) Let 4 log 6216 = n
then 6n = 216
6n = 63 by changing both sides to the same base 6.
Then equating the indices: N = 3
. Log6216 = 3
(3) Let Log0.225 = t then 0.2t= 25
i.e (2/10)t =25
(1/5)t =25
(5-1)t = 25
Change both sides to the same base 5:
5 —t = 52
Equating the
-t = 2
-1 x -t = -1 x2
t = -2
thus: Log0.2 25 = -2
(4) Let Log8 0.0625 = y then 8 = 0.0625
Evaluate the following logarithms:
(1) 0.001
(2) Log 3 81
(3) Log 1. 728
https://youtu.be/kqVpPSzkTYA
Laws of Logarithms
The sixth laws of indices can be stated in their equivalent logarithms form:
Law of indices
https://youtu.be/PLME8ZRKMWU
Laws of Logarithms
Calculations based on the applications of the basic rules.
https://youtu.be/2Bu5NG1KIL4
Examples;
I. Express the following as logarithms of single numbers:
(a) 2 Log 3 + Log 6
(b) I - log 5
(c) log 15 — log 3 log 16.
Solutions
(a) 2 log3 +Log 6
= Log 32 + log 6
= Log (32 x 6)
= Log (9 x 6)
= Log 54
(b) I - log 5
= log 10 — log 5
log 10/5
= log 2
( c) Log 15 - Log 3
= Log 15/3
= Log 5
(d) ¾ Log 16
= Log (16) ¾
=Log (24) ¾
= Log 23
Log 8.
2. Simplify the following :
https://youtu.be/Flbx-yjfzVQ
EVALUATION
Express the following as logarithms of single numbers:
(a) I + log 3
(b) ½ log 25
(c) log 81 - log 3
2. Simplify the following :
(a ) Log 0.2/Log 25
(b) Log 8 + Log 4
ASSIGNMENT.
If Log10x = 2, What is x?
(a) 100 (b) 10 (c) 1/10 (d) 1/100 (e) 20.
2. Evaluate Log1050 + Log1064 log1032.
(a) -2 (b)20 (c) 1 (d) 2 (e) 5
3. Express the following as a single logarithm. 2 -2 log 5
(a) Log 5 (b) Log 8 (c ) Log4 (d) Log 16 (e) Log 40.
4. If Logax = p, express x in terms of a and p.
(a) px = a (b) pa = x ( c) x = ap (d) x — ap (e) x = a/p
5. Solve the equation Log5 2x = 3
(a) 125 (b ) 2/125 (c ) 64 (d) 62 %
THEORY
1.Simplify: 3 log4 6 + log4 8 + log4 36 - Log4 48.
2(a) Find the value of x if Log10 5 + Log10 (x + 2) — Log 10 (x -l) = 2
(b) Given that 1/3 log P = 1, Find the value of P.
READING ASSIGNMENT.
Exam Focus pg 22-24 by S.A Ilori et al Ex 1.5 Nos 5 and 24 pg 24
SOLVING PROBLEMS WITHOUT LOG TABLES AND MISCELLANEOUS PROBLEMS ON LOGARITHMS Examples.
1. Evaluate the following without any tables:
(a) 3 log 2 + log 20 — log 1.6
(b) Log 45 — log 9 + log 20
2, Given that log 2=0.3010, log3 = 0.4771 and log 0.8451, evaluate the following ;
a) log 42 (b) log 35
Solutions.
EVALUATION
Evaluate without using mathematical tables Log1025 + log1032 — log108
2. Given that log 2 = 0.69, log 3 1.20 and log 7 = 1.80, all to a fixed base, find log 10.5 to the same base without using tables.
Solving of Miscellaneous problems on logarithms.
Examples:
1. Express the following in index form:
(a ) Log x — log y — log z
(b ) Log x + 2 log y = 3
Solutions.
(a) Log x — log y = log z
= Log x/y = log z
= x/y = z
i.e. x = yz
(b) Log x +2 log y = 3
Log x + logy2 = 3
Log (X x Y2 ) = 3
Log xy2 3
Since the logs are given to base 10, then xy2 = 103
https://youtu.be/fnhFneOz6n8
https://youtu.be/NhDcGOmGa-w
2. Use logarithms tables to evaluate the following, giving answers to two decimal places:
(a) Log5 12.64
(b) Log12 8.621
Solutions. Let Log5 12.64 = p
Then 5p = 12.64
Take log of both sides to base 10
= Log10 5P = log10 12,64
= P log10 5 = Log10 12.64
= P = Log10 12.64
= Log 10
= 1.1018 / 0.6990
2. Find the value of x if log105 + log10(x+2) — log10(x-1) = 2
Solution
Log10 5 + log10 (x+2) — log10(x-1) =2
Taking logs of both sides to base 10 i.e log105 + log10 (x+2) — log10 (x-1)= log10 100
(Since the logs are to base 10)
4. If log x = 2.3675 and log y = 0.9750, what is the value of x + y, correct to 3 significant figures.
= 242.541 = 243 to 3 s.f
https://youtu.be/0-kZdwxFjY4
EVALUATION
1. Solve the equation
Log2 (x2 —x) = Log2 (x-1) + I
ASSIGNMENT.
THEORY
1. if 3 log a + 5 log a —6 loga = log 64
What is the value of a?
2a. Evaluate without using tables log10 1.44 — log1090 + log10 0.0625
b Solve the equation
Log2(x2-2) = Log2 (x -1) +1
c. If logs 2 = 0.431 and logs 3 = 0.682, find the value of log 1.5.
Reading Assignment
( l) Exam Focus, Maths for WASSCE and SSCE by SA Ilori et al pages 22-25.
(2) NGM for Senior Secondary School Bk 3 by MF Macrae et al pages 1-3
The logarithm of a number N to a given base 'a' is the index to which the base must be raised in order to get the given number.
Thus if N = ax, it implies that loga N = x. The form N = ax is known as the index from while Loga N = x is known as the logarithmic form.
https://youtu.be/ntBWrcbAhaY
Examples:
Evaluate the following logarithms:
(1) Log4 8 (2) Log 6216. (3) log0.2 25 (4) Loga 0.0625.
Solutions
1. Let log4 8 = y then 4y = 8 change both sides to base 2 (22)y=23)
i.e 22y =23 equating the indices:
2y = 3
y = 3/2
log4 8 = 1½
(2) Let 4 log 6216 = n
then 6n = 216
6n = 63 by changing both sides to the same base 6.
Then equating the indices: N = 3
. Log6216 = 3
(3) Let Log0.225 = t then 0.2t= 25
i.e (2/10)t =25
(1/5)t =25
(5-1)t = 25
Change both sides to the same base 5:
5 —t = 52
Equating the
-t = 2
-1 x -t = -1 x2
t = -2
thus: Log0.2 25 = -2
(4) Let Log8 0.0625 = y then 8 = 0.0625
Evaluate the following logarithms:
(1) 0.001
(2) Log 3 81
(3) Log 1. 728
https://youtu.be/kqVpPSzkTYA
Laws of Logarithms
The sixth laws of indices can be stated in their equivalent logarithms form:
Law of indices
https://youtu.be/PLME8ZRKMWU
Laws of Logarithms
Calculations based on the applications of the basic rules.
https://youtu.be/2Bu5NG1KIL4
Examples;
I. Express the following as logarithms of single numbers:
(a) 2 Log 3 + Log 6
(b) I - log 5
(c) log 15 — log 3 log 16.
Solutions
(a) 2 log3 +Log 6
= Log 32 + log 6
= Log (32 x 6)
= Log (9 x 6)
= Log 54
(b) I - log 5
= log 10 — log 5
log 10/5
= log 2
( c) Log 15 - Log 3
= Log 15/3
= Log 5
(d) ¾ Log 16
= Log (16) ¾
=Log (24) ¾
= Log 23
Log 8.
2. Simplify the following :
https://youtu.be/Flbx-yjfzVQ
EVALUATION
Express the following as logarithms of single numbers:
(a) I + log 3
(b) ½ log 25
(c) log 81 - log 3
2. Simplify the following :
(a ) Log 0.2/Log 25
(b) Log 8 + Log 4
ASSIGNMENT.
If Log10x = 2, What is x?
(a) 100 (b) 10 (c) 1/10 (d) 1/100 (e) 20.
2. Evaluate Log1050 + Log1064 log1032.
(a) -2 (b)20 (c) 1 (d) 2 (e) 5
3. Express the following as a single logarithm. 2 -2 log 5
(a) Log 5 (b) Log 8 (c ) Log4 (d) Log 16 (e) Log 40.
4. If Logax = p, express x in terms of a and p.
(a) px = a (b) pa = x ( c) x = ap (d) x — ap (e) x = a/p
5. Solve the equation Log5 2x = 3
(a) 125 (b ) 2/125 (c ) 64 (d) 62 %
THEORY
1.Simplify: 3 log4 6 + log4 8 + log4 36 - Log4 48.
2(a) Find the value of x if Log10 5 + Log10 (x + 2) — Log 10 (x -l) = 2
(b) Given that 1/3 log P = 1, Find the value of P.
READING ASSIGNMENT.
Exam Focus pg 22-24 by S.A Ilori et al Ex 1.5 Nos 5 and 24 pg 24
SOLVING PROBLEMS WITHOUT LOG TABLES AND MISCELLANEOUS PROBLEMS ON LOGARITHMS Examples.
1. Evaluate the following without any tables:
(a) 3 log 2 + log 20 — log 1.6
(b) Log 45 — log 9 + log 20
2, Given that log 2=0.3010, log3 = 0.4771 and log 0.8451, evaluate the following ;
a) log 42 (b) log 35
Solutions.
EVALUATION
Evaluate without using mathematical tables Log1025 + log1032 — log108
2. Given that log 2 = 0.69, log 3 1.20 and log 7 = 1.80, all to a fixed base, find log 10.5 to the same base without using tables.
Solving of Miscellaneous problems on logarithms.
Examples:
1. Express the following in index form:
(a ) Log x — log y — log z
(b ) Log x + 2 log y = 3
Solutions.
(a) Log x — log y = log z
= Log x/y = log z
= x/y = z
i.e. x = yz
(b) Log x +2 log y = 3
Log x + logy2 = 3
Log (X x Y2 ) = 3
Log xy2 3
Since the logs are given to base 10, then xy2 = 103
https://youtu.be/fnhFneOz6n8
https://youtu.be/NhDcGOmGa-w
2. Use logarithms tables to evaluate the following, giving answers to two decimal places:
(a) Log5 12.64
(b) Log12 8.621
Solutions. Let Log5 12.64 = p
Then 5p = 12.64
Take log of both sides to base 10
= Log10 5P = log10 12,64
= P log10 5 = Log10 12.64
= P = Log10 12.64
= Log 10
= 1.1018 / 0.6990
2. Find the value of x if log105 + log10(x+2) — log10(x-1) = 2
Solution
Log10 5 + log10 (x+2) — log10(x-1) =2
Taking logs of both sides to base 10 i.e log105 + log10 (x+2) — log10 (x-1)= log10 100
(Since the logs are to base 10)
4. If log x = 2.3675 and log y = 0.9750, what is the value of x + y, correct to 3 significant figures.
= 242.541 = 243 to 3 s.f
https://youtu.be/0-kZdwxFjY4
EVALUATION
1. Solve the equation
Log2 (x2 —x) = Log2 (x-1) + I
ASSIGNMENT.
THEORY
1. if 3 log a + 5 log a —6 loga = log 64
What is the value of a?
2a. Evaluate without using tables log10 1.44 — log1090 + log10 0.0625
b Solve the equation
Log2(x2-2) = Log2 (x -1) +1
c. If logs 2 = 0.431 and logs 3 = 0.682, find the value of log 1.5.
Reading Assignment
( l) Exam Focus, Maths for WASSCE and SSCE by SA Ilori et al pages 22-25.
(2) NGM for Senior Secondary School Bk 3 by MF Macrae et al pages 1-3
WEEK 8
SOLUTION OF QUADRATIC EQUATIONS.
The following methods can be used for solving quadratic equations.
1. Factorization
2. Completing the square
3. use of the formula.
4. Drawing quadratic graph.
The examples below demonstrate these methods.
Example I.
Solve the following equation by first factorizing the quadratic expression:
2(x2 + 1) -5x = 0.
Solution .
2(x2 + 1) -5x = 0.
2x2 + 2 - 5x = 0
2x2 -5x + 2 = 0
2x2 -4x -x + 2 = 0
2x2(x-2) - 1 (x-2) = 0
(x-2) (2x - 1) = 0
Solution .
Graphical method may also be used to solve quadratic equations as shown in example 4 below.
https://youtu.be/YtN9_tCaRQc
https://youtu.be/qeByhTF8WEw
Example 4.
The table below is an incomplete table for the relation y =px2 + 2x + 4, where p is a constant.
|x |-2 |-1 |0 |1 |2 |3 |4
|y |-4 |1 |4 | | | |
(a) Use the table to find the value of the constant p
(b) Copy and complete the table
( c) Using a scale of 2cm to represent 1 unit on both axes, draw a graph of the relation Y = px2 +2x + 4.
For -2 ≤ x ≤ 4
(d) Use your graph to obtain.
(i) the roots of the equation px2 + 2x + 4 = 0
(ii) the equation of the line of symmetry of the curve .
Solution.
(a) From the first column of the table, when x = -2, -4.
Substitute these values into
(b) Substituting p = - 1, the relation is y -x2 + 2x + 4.
Substituting the values x = 1, 2, 3, and 4 in turn into the relation; the complete table is obtained as follows:
|x|-2|-1|0|1|2|3|4|
|........................|
|y|-4|1|4|5|4|1|-4|
The figure below shows the graph of the relation y = px2 + 2x + 4 (where p = -1).
d.(i) the roots of the equation px2 +2x + 4 = 0 are found where the curve cuts the x axis i.e at x = -1.2 and x = 3.2.
The roots are -1.2 and 3.2,
(ii) The line of symmetry is the broken line in Fig A above.
In general, as in the example above, the graph of y = ax2 + bx + C gives the solution of ax2 + bx + C = 0 at the points where the curve cuts the x axis.
https://youtu.be/reRSfNfmcsk
https://youtu.be/cKZVcmxlOwY
EVALUATION,
1 .Solve the following equation by the method of completing the square. Give roots correct to 2 decimal places:
t2 - 3t = 9
2.Solve the equation 2x 3x + 7 using the general formula., Give roots correct to 2 decimal places.
ASSIGNMENT.
1. Find the sum of the roots of the equation 2x2 — 7x + 2 = 0
(a)-7 (b)-31/2 (c)2 (d) 31/2 (e) 7
2. Solve the equation C2 —7c - 18 = 0
(a) (-2,9) (b) (-6,3) (c) (-9,2 ) (d) (-3,6) (e) (18,1)
3. Find the value of k such that x2 — 5x + k is a perfect square
(a)2½ (b)4¼ (c)6¼ (d) 25 (e) 100.
4. Solve the equation .3d2 = 2d + 5
(a)-5,3 (b), -5) ( c) (15,-1) (d) (-15,1) (e) (-1, 5/3)
5. The greater of the two roots of the equation (2x -5) ( 3x + 10) = 0 is .....................
THEORY.
1. Solve the equation 2x2 + 6x + I = 0
(a) by completing the square
(b) by formula.
2. Solve the following equations, giving each answer correct to 2 decimal places
3r2 - r - 7 = 0
3 = 85 - 2s2
Reading Assignment,
NGM SS Bk 2 pages 40-42. Ex 3d No 17 pg 41 and Ex NO IS pg 42.
Topic: WORD PROBLEMS LEADING TO QUADRATIC EQUATIONS.
Take the following steps when solving such problems.
Use a letter to represent the unknown
2. Translate all the information given in the question to a quadratic equation.
3. Solve, the resulting quadratic equation.
Examples.
The sum of the digits of a two-digit number is 14. The tens digit is the square of a number which is 2 less than the units digit.
Find the two digits number o
Solution
Let the required number xy
From the information given in the question, then
x + y = 14 ..................Eqn(1)
Also
x = (y - 2)2
x = y2 - 4y + 4 .........Eqn (2)
From Eqn 1
x + y = 14
x = 14 - y
Substitute ( 14 —y ) for x in eqn (2):
14 — y = y2 - 4y + 4
0 = y2 — 4y + y + 4 -14
0 = y2 -3y - 10
or
y2 -3y - 10 = 0
y2 -5y + 2y - 10 = 0
y(y-5) + 2(y-5) = 0
(y-5) (y+2) = 0
Either y — 5 = 0 or y + 2 = 0
Y = 0 + 5 or y = 0—2
Y = 5 or y -2
Units digit cannot be a negative number. Hence the possible unit digit of the required two digit number is 5,
Substituting 5 for y in eqn (1) x + 5 = 14
Thus the required two-digits number is 95 (since x = 9 and y =5).
https://youtu.be/dNcAb9zvRhY
2. A cyclist travels from A to B, a distance of 8km, at x km/hr. he travels from B to A, the same distance at ( x- 4) km/hr. If the total time taken is I hour 10 minutes, find x.
Solution.
Time = Distance/Speed
Time taken for outward journey from A to B is : 8/x
Also, time taken for return journey from B to A is : ( 8/x )hr.
Thus, from the information, the total time taken for the whole journey is
8/x + 8/x - 4 =101/60
8/x + 8/x - 4 =11/6
8/x + 8/x - 4 =7/6
L.C. M of the denominators
x(x-4) x 6 = 6x (x-4)
multiply each term by the L.C.M.
48 (x-4) + 48x = '7x (x - 4)
48x - 192 + 48x = 7x2 - 28x
0 = 7x2 - 28x -48x - 48x + 192
0 = 7x2 - 124x + 192
or
7x2 - 124x + 192 = 0
7x2 (x - 16) - 12 (x - 16) = 0
(x - 16) (7x - 12) = 0
Either
(x - 16) = 0 or (7x - 12) = 0
x = 0 + 16 or 7x = 0+ 12
x = 16 or 7x = 12
x = 16 or x = 12/7
X = 16 or x = 15/12
From these two solutions, the speed cannot be 15/12km /hr because
X —4 will be a negative speed. The possible solution is
X = 16km/hr,
Check.
8/16 + 8/16-4
½ + 8/12
½ + 2/3
= 3 + 4/6
= 7/6 = RHS
https://youtu.be/ZRfTRVZaCSU
EVALUATION.
1, The length of a rectangle is twice its width. If the width is reduced by 1cm and the length is also reduced by 2cm,
the area will be 15cm2 . Find the dimensions of the original rectangle.
2. If 13 times a number n is added to twice its square, the result is 84.
(a) Obtain a quadratic expression in x.
(b) Find x.
ASSIGNMENT.
1. Two numbers differ by 2. The sum of their squares is 244. Find the numbers
(a) (3, 18) (b) (10,12) (c 9d) (-12, 10) (e) (2,12)
2. A trader spent NS50 on x articles, each costing N (2x — 28) Find x
hint : Use formula method to solve the resulting quadratic equation.
(a) 50 (b) 75 (c) 40 (d) 25 (e) 30.
3. The product of two consecutive odd numbers is 323. Find the numbers.
(a) (17,18) (b) (12,14) (c) (17,19 ) (d) (15,17 ) (e) 19,21).
4. A rectangular farm is one hectare in area. Its perimeter is 500 meters.. Find the dimensions of the farm in metes.
(Hint: I hectare = 10,000m2).
(a) (300m, 25m) (b) (100m, 75m) (c) (200m, 50m ) (d) (150m, 80m) (e) (90m, 80m)
5. If a number is increased by 30, it is less than its square by 12. Find the number.
(a) 7 or -6 (c) 5 or -6 (d) 3 or 6 (e) 10, or 6. THEORY.
1. The numerator of a fraction is 5 less than its denominator. If 6 is added to the numerator and 4 added to the denominator, the fraction is doubled. What is the fraction?
2. The length of a room is 3 metres longer than its breadth,. The area of the room is 28 m2 find its dimensions
Reading Assignment.
NGM SS Bk 2, pg 42- 43, ex 3f Nos 18and 20 pg 43.
The following methods can be used for solving quadratic equations.
1. Factorization
2. Completing the square
3. use of the formula.
4. Drawing quadratic graph.
The examples below demonstrate these methods.
Example I.
Solve the following equation by first factorizing the quadratic expression:
2(x2 + 1) -5x = 0.
Solution .
2(x2 + 1) -5x = 0.
2x2 + 2 - 5x = 0
2x2 -5x + 2 = 0
2x2 -4x -x + 2 = 0
2x2(x-2) - 1 (x-2) = 0
(x-2) (2x - 1) = 0
Solution .
Graphical method may also be used to solve quadratic equations as shown in example 4 below.
https://youtu.be/YtN9_tCaRQc
https://youtu.be/qeByhTF8WEw
Example 4.
The table below is an incomplete table for the relation y =px2 + 2x + 4, where p is a constant.
|x |-2 |-1 |0 |1 |2 |3 |4
|y |-4 |1 |4 | | | |
(a) Use the table to find the value of the constant p
(b) Copy and complete the table
( c) Using a scale of 2cm to represent 1 unit on both axes, draw a graph of the relation Y = px2 +2x + 4.
For -2 ≤ x ≤ 4
(d) Use your graph to obtain.
(i) the roots of the equation px2 + 2x + 4 = 0
(ii) the equation of the line of symmetry of the curve .
Solution.
(a) From the first column of the table, when x = -2, -4.
Substitute these values into
(b) Substituting p = - 1, the relation is y -x2 + 2x + 4.
Substituting the values x = 1, 2, 3, and 4 in turn into the relation; the complete table is obtained as follows:
|x|-2|-1|0|1|2|3|4|
|........................|
|y|-4|1|4|5|4|1|-4|
The figure below shows the graph of the relation y = px2 + 2x + 4 (where p = -1).
d.(i) the roots of the equation px2 +2x + 4 = 0 are found where the curve cuts the x axis i.e at x = -1.2 and x = 3.2.
The roots are -1.2 and 3.2,
(ii) The line of symmetry is the broken line in Fig A above.
In general, as in the example above, the graph of y = ax2 + bx + C gives the solution of ax2 + bx + C = 0 at the points where the curve cuts the x axis.
https://youtu.be/reRSfNfmcsk
https://youtu.be/cKZVcmxlOwY
EVALUATION,
1 .Solve the following equation by the method of completing the square. Give roots correct to 2 decimal places:
t2 - 3t = 9
2.Solve the equation 2x 3x + 7 using the general formula., Give roots correct to 2 decimal places.
ASSIGNMENT.
1. Find the sum of the roots of the equation 2x2 — 7x + 2 = 0
(a)-7 (b)-31/2 (c)2 (d) 31/2 (e) 7
2. Solve the equation C2 —7c - 18 = 0
(a) (-2,9) (b) (-6,3) (c) (-9,2 ) (d) (-3,6) (e) (18,1)
3. Find the value of k such that x2 — 5x + k is a perfect square
(a)2½ (b)4¼ (c)6¼ (d) 25 (e) 100.
4. Solve the equation .3d2 = 2d + 5
(a)-5,3 (b), -5) ( c) (15,-1) (d) (-15,1) (e) (-1, 5/3)
5. The greater of the two roots of the equation (2x -5) ( 3x + 10) = 0 is .....................
THEORY.
1. Solve the equation 2x2 + 6x + I = 0
(a) by completing the square
(b) by formula.
2. Solve the following equations, giving each answer correct to 2 decimal places
3r2 - r - 7 = 0
3 = 85 - 2s2
Reading Assignment,
NGM SS Bk 2 pages 40-42. Ex 3d No 17 pg 41 and Ex NO IS pg 42.
Topic: WORD PROBLEMS LEADING TO QUADRATIC EQUATIONS.
Take the following steps when solving such problems.
Use a letter to represent the unknown
2. Translate all the information given in the question to a quadratic equation.
3. Solve, the resulting quadratic equation.
Examples.
The sum of the digits of a two-digit number is 14. The tens digit is the square of a number which is 2 less than the units digit.
Find the two digits number o
Solution
Let the required number xy
From the information given in the question, then
x + y = 14 ..................Eqn(1)
Also
x = (y - 2)2
x = y2 - 4y + 4 .........Eqn (2)
From Eqn 1
x + y = 14
x = 14 - y
Substitute ( 14 —y ) for x in eqn (2):
14 — y = y2 - 4y + 4
0 = y2 — 4y + y + 4 -14
0 = y2 -3y - 10
or
y2 -3y - 10 = 0
y2 -5y + 2y - 10 = 0
y(y-5) + 2(y-5) = 0
(y-5) (y+2) = 0
Either y — 5 = 0 or y + 2 = 0
Y = 0 + 5 or y = 0—2
Y = 5 or y -2
Units digit cannot be a negative number. Hence the possible unit digit of the required two digit number is 5,
Substituting 5 for y in eqn (1) x + 5 = 14
Thus the required two-digits number is 95 (since x = 9 and y =5).
https://youtu.be/dNcAb9zvRhY
2. A cyclist travels from A to B, a distance of 8km, at x km/hr. he travels from B to A, the same distance at ( x- 4) km/hr. If the total time taken is I hour 10 minutes, find x.
Solution.
Time = Distance/Speed
Time taken for outward journey from A to B is : 8/x
Also, time taken for return journey from B to A is : ( 8/x )hr.
Thus, from the information, the total time taken for the whole journey is
8/x + 8/x - 4 =101/60
8/x + 8/x - 4 =11/6
8/x + 8/x - 4 =7/6
L.C. M of the denominators
x(x-4) x 6 = 6x (x-4)
multiply each term by the L.C.M.
48 (x-4) + 48x = '7x (x - 4)
48x - 192 + 48x = 7x2 - 28x
0 = 7x2 - 28x -48x - 48x + 192
0 = 7x2 - 124x + 192
or
7x2 - 124x + 192 = 0
7x2 (x - 16) - 12 (x - 16) = 0
(x - 16) (7x - 12) = 0
Either
(x - 16) = 0 or (7x - 12) = 0
x = 0 + 16 or 7x = 0+ 12
x = 16 or 7x = 12
x = 16 or x = 12/7
X = 16 or x = 15/12
From these two solutions, the speed cannot be 15/12km /hr because
X —4 will be a negative speed. The possible solution is
X = 16km/hr,
Check.
8/16 + 8/16-4
½ + 8/12
½ + 2/3
= 3 + 4/6
= 7/6 = RHS
https://youtu.be/ZRfTRVZaCSU
EVALUATION.
1, The length of a rectangle is twice its width. If the width is reduced by 1cm and the length is also reduced by 2cm,
the area will be 15cm2 . Find the dimensions of the original rectangle.
2. If 13 times a number n is added to twice its square, the result is 84.
(a) Obtain a quadratic expression in x.
(b) Find x.
ASSIGNMENT.
1. Two numbers differ by 2. The sum of their squares is 244. Find the numbers
(a) (3, 18) (b) (10,12) (c 9d) (-12, 10) (e) (2,12)
2. A trader spent NS50 on x articles, each costing N (2x — 28) Find x
hint : Use formula method to solve the resulting quadratic equation.
(a) 50 (b) 75 (c) 40 (d) 25 (e) 30.
3. The product of two consecutive odd numbers is 323. Find the numbers.
(a) (17,18) (b) (12,14) (c) (17,19 ) (d) (15,17 ) (e) 19,21).
4. A rectangular farm is one hectare in area. Its perimeter is 500 meters.. Find the dimensions of the farm in metes.
(Hint: I hectare = 10,000m2).
(a) (300m, 25m) (b) (100m, 75m) (c) (200m, 50m ) (d) (150m, 80m) (e) (90m, 80m)
5. If a number is increased by 30, it is less than its square by 12. Find the number.
(a) 7 or -6 (c) 5 or -6 (d) 3 or 6 (e) 10, or 6. THEORY.
1. The numerator of a fraction is 5 less than its denominator. If 6 is added to the numerator and 4 added to the denominator, the fraction is doubled. What is the fraction?
2. The length of a room is 3 metres longer than its breadth,. The area of the room is 28 m2 find its dimensions
Reading Assignment.
NGM SS Bk 2, pg 42- 43, ex 3f Nos 18and 20 pg 43.
WEEK 9
Topic
SOLVING SIMULTANEOUS LINEAR AND QUADRATIC EQUATIONS
Examples:
(i) Solve equations
2x2 - y2 = -7
x + y = 2
Simultaneously
Solutions.
Let 2x2 - y2 = -7..............Eqn (1)
x + y = 2 ................Eqn (2)
From eqn (2)
x + y = 2
y = 2 - x
Substitute 2 — x for y in equation (i):
2x2 - (2 -x)2 = -7
2x2 — (4 -4x + x2) = -7
2x2- 4 + 4x - x = -7
2x2 — x2 + 4x —4 + 7 = 0
x2+ 4 + 3 = 0
x2+ x + 3x + 3 = 0
x(x + 1) + 3 (x + 1) = 0
(x + 1) (x + 3) = 0
either x + 1 = 0 or x + 3 = O
x = -1 or x = -3.
When x = -1 , substitute for x in eqn ( ii)
-1 + y = 2
y = 2 + 1
y = 3
Thus when x = -1, y = 3. Also, using eqn. 2
when x = -3
-3 + y = 2
y = 2 + 3
y = 5
Thus when x = -3, y = 5 then the solutions are x = -1, y = 3. and x = -3, y = 5
Or in ordered pairs ( -1,3) and (-3, 5).
https://youtu.be/NPzlCNDEJqA
2(a) Draw the graph of y = x2 + x— 5 for values of x ranging from —4 to + 3. Use a scale of 1cm to I unit on both axes.
(b)On the same graph draw the line y = 2 (x -1). From your graph.
(c ) Find the roots of the equation : x2 + x —5 = 0
(d) deduce the roots of the equation.: x2 + x — 7 = 0
( e) Deduce the roots of the equation : x2 + x— 5 = 2 (x-1)
(f) State the minimum value of x2 + x -5.
Solution
Given the eqn y = x2 + x — 5, the table below is the table of values for the given range.
|x|-4|-3|-2|-1|0|1|2|3|4
|-------------------------------|
|y|7|1|-3|-5|-5|-3| 1|7|
(b) To draw the line y = 2 (x - 1), it is sufficient to plot three points as given in the table below:
|x|-2|0|3|
|------------|
|y|-6|-2|4|
The figure on the next page shows the required graph.
The roots of the equation x2 + x— 5 = 0 are where the curve cuts the x- axis i.e where x = -2.8 and x = 1.8.
(d) Given : x2 + x— 7 = 0. Add 2 to both sides
x2 + x— 5 = 2
Thus the solution of x2 + x— 7 = 0 is where the curve y = x2 + x - 5 cuts the line y = 2 (shown as a line on the graph) i.e where x = -3.2 and x = 2.2
(e) the roots of x2 + x -5 = 2(x-1) are the values of x where the curve y = x + x— 5 cuts the line Y2(x-l) i.e where x = -1.3 and x = 2.3
(f) the minimum value of x2 + x — 5 is the value of y at the lowest point of the curve i.e -5,...25.
https://youtu.be/KvSs4MS8AwU
https://youtu.be/7C3f-sYMNCU
EVALUATION.
1. Given the simultaneous equations y = x (x -3) and y = 2 — 2x solve the equations algebraically.
2(a) Copy and complete the table below for the relation. Y = x2 — 2x - 5
(b) Draw the graph of the relation y = x2 — 2x— 5, using a scale of 2cm to 1 unit on the x axis and 2cm to 2 units on the y- axis.
(c ). Using the same axes, draw the graph of y = 2x — 3.
(d) Obtain, in the form ax2 + bx + C = 0, where a, b and C are integers, the equation which is satisfied by the x — coordinates of the points of intersection of the two graphs.
(e ) From your graphs determine the roots of the equation obtained in (d) above .
ASSIGNMENT.
1. Factorize the expression (a + b2— 4.
a. (a + b + 2) (a - b + 2)
b. (a - b - 2) (a + b + 2)
c. (a - b + 2) (a + b - 2)
d (a + b + 2)2
e. (a + b + 2) (a + b - 2)
2. The roots of a quadratic equation are ½ and -2/3. The quadratic equation is ?
(a) 3x2 - x - 1 = 0
(b) 3x2 + x - 1 = 0
(c) 6x2 + x —2 = 0
(d) 6x2 - x + 2 = 0
(e) 6x2 - x - 2 = 0
3. The perimeter of a floor of a rectangular length is xcm. Find the area of the floor in m2 in terms of x
(a) x (x-13) (b) x ( x -26) ( c) x ( 13 —x) (d) x (26 —x ) (e) x (x + 13)
4.1f x + 1/x = 3, evaluate x2 + 1/x2
5. Two numbers differ by 7. If their product is 120, find the number.
(a) 7 and 15 (b) 15 and 8 (c) -5 and 7 (d) -8 and 15 (e) 16 and 9
THEORY
In the triangle in the figure below, DHF = 90º,
|EF| = 9cm , |HE| xcm.
|DF| = 2xcm , |DE| (x + 3)cm
(a) obtain a quadratic equation in x.
(b) Using the method of completing the squares or otherwise, find x to two decimal places.
2(a) Copy and complete the table below for the relation Y = 3 + 3x - x2
(b) Taking 2cm as 1 unit on both axes, draw the graph of the relation for the given interval.
(c ) Draw on the same axes the graph of x — y = 0
(d) using your graphs
(i) solve the equation 3 + 2x — x2 = x.
(ii) Find the values of x for which 3 + —x2 = 2.
Reading Assignment.
NGM SS Bk 3 Pgs 4-11 by MF Macrea et al and Exam Focus by S, A Ilori et al pgs 49 — 56.
SOLVING SIMULTANEOUS LINEAR AND QUADRATIC EQUATIONS
Examples:
(i) Solve equations
2x2 - y2 = -7
x + y = 2
Simultaneously
Solutions.
Let 2x2 - y2 = -7..............Eqn (1)
x + y = 2 ................Eqn (2)
From eqn (2)
x + y = 2
y = 2 - x
Substitute 2 — x for y in equation (i):
2x2 - (2 -x)2 = -7
2x2 — (4 -4x + x2) = -7
2x2- 4 + 4x - x = -7
2x2 — x2 + 4x —4 + 7 = 0
x2+ 4 + 3 = 0
x2+ x + 3x + 3 = 0
x(x + 1) + 3 (x + 1) = 0
(x + 1) (x + 3) = 0
either x + 1 = 0 or x + 3 = O
x = -1 or x = -3.
When x = -1 , substitute for x in eqn ( ii)
-1 + y = 2
y = 2 + 1
y = 3
Thus when x = -1, y = 3. Also, using eqn. 2
when x = -3
-3 + y = 2
y = 2 + 3
y = 5
Thus when x = -3, y = 5 then the solutions are x = -1, y = 3. and x = -3, y = 5
Or in ordered pairs ( -1,3) and (-3, 5).
https://youtu.be/NPzlCNDEJqA
2(a) Draw the graph of y = x2 + x— 5 for values of x ranging from —4 to + 3. Use a scale of 1cm to I unit on both axes.
(b)On the same graph draw the line y = 2 (x -1). From your graph.
(c ) Find the roots of the equation : x2 + x —5 = 0
(d) deduce the roots of the equation.: x2 + x — 7 = 0
( e) Deduce the roots of the equation : x2 + x— 5 = 2 (x-1)
(f) State the minimum value of x2 + x -5.
Solution
Given the eqn y = x2 + x — 5, the table below is the table of values for the given range.
|x|-4|-3|-2|-1|0|1|2|3|4
|-------------------------------|
|y|7|1|-3|-5|-5|-3| 1|7|
(b) To draw the line y = 2 (x - 1), it is sufficient to plot three points as given in the table below:
|x|-2|0|3|
|------------|
|y|-6|-2|4|
The figure on the next page shows the required graph.
The roots of the equation x2 + x— 5 = 0 are where the curve cuts the x- axis i.e where x = -2.8 and x = 1.8.
(d) Given : x2 + x— 7 = 0. Add 2 to both sides
x2 + x— 5 = 2
Thus the solution of x2 + x— 7 = 0 is where the curve y = x2 + x - 5 cuts the line y = 2 (shown as a line on the graph) i.e where x = -3.2 and x = 2.2
(e) the roots of x2 + x -5 = 2(x-1) are the values of x where the curve y = x + x— 5 cuts the line Y2(x-l) i.e where x = -1.3 and x = 2.3
(f) the minimum value of x2 + x — 5 is the value of y at the lowest point of the curve i.e -5,...25.
https://youtu.be/KvSs4MS8AwU
https://youtu.be/7C3f-sYMNCU
EVALUATION.
1. Given the simultaneous equations y = x (x -3) and y = 2 — 2x solve the equations algebraically.
2(a) Copy and complete the table below for the relation. Y = x2 — 2x - 5
(b) Draw the graph of the relation y = x2 — 2x— 5, using a scale of 2cm to 1 unit on the x axis and 2cm to 2 units on the y- axis.
(c ). Using the same axes, draw the graph of y = 2x — 3.
(d) Obtain, in the form ax2 + bx + C = 0, where a, b and C are integers, the equation which is satisfied by the x — coordinates of the points of intersection of the two graphs.
(e ) From your graphs determine the roots of the equation obtained in (d) above .
ASSIGNMENT.
1. Factorize the expression (a + b2— 4.
a. (a + b + 2) (a - b + 2)
b. (a - b - 2) (a + b + 2)
c. (a - b + 2) (a + b - 2)
d (a + b + 2)2
e. (a + b + 2) (a + b - 2)
2. The roots of a quadratic equation are ½ and -2/3. The quadratic equation is ?
(a) 3x2 - x - 1 = 0
(b) 3x2 + x - 1 = 0
(c) 6x2 + x —2 = 0
(d) 6x2 - x + 2 = 0
(e) 6x2 - x - 2 = 0
3. The perimeter of a floor of a rectangular length is xcm. Find the area of the floor in m2 in terms of x
(a) x (x-13) (b) x ( x -26) ( c) x ( 13 —x) (d) x (26 —x ) (e) x (x + 13)
4.1f x + 1/x = 3, evaluate x2 + 1/x2
5. Two numbers differ by 7. If their product is 120, find the number.
(a) 7 and 15 (b) 15 and 8 (c) -5 and 7 (d) -8 and 15 (e) 16 and 9
THEORY
In the triangle in the figure below, DHF = 90º,
|EF| = 9cm , |HE| xcm.
|DF| = 2xcm , |DE| (x + 3)cm
(a) obtain a quadratic equation in x.
(b) Using the method of completing the squares or otherwise, find x to two decimal places.
2(a) Copy and complete the table below for the relation Y = 3 + 3x - x2
(b) Taking 2cm as 1 unit on both axes, draw the graph of the relation for the given interval.
(c ) Draw on the same axes the graph of x — y = 0
(d) using your graphs
(i) solve the equation 3 + 2x — x2 = x.
(ii) Find the values of x for which 3 + —x2 = 2.
Reading Assignment.
NGM SS Bk 3 Pgs 4-11 by MF Macrea et al and Exam Focus by S, A Ilori et al pgs 49 — 56.