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!
If they together at the same rate for the same amount of time, they will solve 48 math problems,
From the question, we can see that two mathletes solve 32 math problems in a certain amount of time, this can be written as:
- 2 mathletes = 32 math problems
In order to determine the number of problems for 3 mathletes, this is expressed as:
DIvide both expressions:
Cross multiply
2x = 3* 32
x = 3 * 16
x = 48math problems
Hence if they together at the same rate for the same amount of time, they will solve 48 math problems,
Learn more on proportion here: brainly.com/question/1496357
