Hello!
First you have to list the data in both classes
Class A
41, 42, 45, 46, 47, 48, 52, 53, 54, 59, 61, 61, 64, 68, 71, 82, 85, 90
Class B
41, 42, 59, 62, 64, 69, 71, 75, 77, 78, 78, 80, 83, 84, 84, 86, 86, 87, 92, 92, 95
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First we are going to find the mode and mean of class A
The mode is the number that appears the most
The number that appears the most is 61
The mode is 61
To find the mean you add all the numbers together and divide the sum by the amount of numbers added
41 + 42 + 45 + 46 + 47 + 48 + 52 + 53 + 54 + 59 + 61 + 61 + 64 + 68 + 71 + 82 + 85 + 90 = 1069
Divide this by the amount of numbers added
1069 / 18 = 59.3888...
The mean is 59.3888
The mode is 61 and the mean is 59.39 for class A
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We are going to find the range and median for class B
To find the range you subtract the smallest number from the largest on
The smallest number is 41
The largest number is 95
Subtract these
95 - 41 = 54
The range is 54
To find the median you list the numbers from least to greatest and look for the number in the middle
41, 42, 59, 62, 64, 69, 71, 75, 77, 78, 78, 80, 83, 84, 84, 86, 86, 87, 92, 92, 95
The number in the middle is 78
The range is 54 and the median is 78
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Hope this helps!
Answer:
Algebrically the result can be written as -
Step-by-step explanation:
Let the number be 'x'.
- When multiplied by 3 , algebrically it is written as "3x".
- When 4 is added to 3x , algebrically it is written as "3x + 4".
- When (3x + 4) is squared completely , algebrically it is written as -
X^2 + y^2 = 8
X-y=0 so x = y
replace x = y into X^2 + y^2 = 8
y^2 + y^2 = 8
2y^2 = 8
y^2 = 8/2
y^2 = 4
y = - 2 and y = 2
because x = y
so x = - 2 and x = 2
solutions:
x= - 2 and x = + 2
y= - 2 and y = + 2
