John has 2456 pennies in his coin collection. He has the same number of pennies in each of 3 boxes. Estimate to the nearest hund
1 answer:
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The formula for the average value of a function is 
where b is the upper bound and a is the lower. For us, this formula will be filled in accordingly. 
. We will integrate that now: ![\frac{1}{2}[ \frac{2x^3}{3}+3x]](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%5B%20%5Cfrac%7B2x%5E3%7D%7B3%7D%2B3x%5D%20%20)
from 0 to 2. Filling in our upper and lower bounds we have ![\frac{1}{2}[( \frac{2(2^3)}{3}+3(2))-0]](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%5B%28%20%5Cfrac%7B2%282%5E3%29%7D%7B3%7D%2B3%282%29%29-0%5D%20%20)
which simplifies to 
and 
which is 17/3 or 5.667
Answer:
The number of individuals included in the summary is 146.
Step-by-step explanation:
The frequency distribution table provided is as follows:
Class Intervals Frequency
100 - 199 24
200 - 299 90
300 - 399 27
400 - 499 1
500 - 599 4
The lower class limit it the smallest value of each class interval.
Lower class limit = {100, 200, 300, 400, 500}
The upper class limit it the highest value of each class interval.
Upper class limit = {199, 299, 399, 499, 599}
The lower class boundaries are the lower class limits decreased by 0.5 and the upper class boundaries are the upper class limits increased by 0.5.
Class boundaries:
99.5 - 199.5
199.5 - 299.5
299.5 - 399.5
399.5 - 499.5
499.5 - 599.5
The class width is the difference between the class boundaries of each class.
Class width = 199.5 - 99.5 = 100
So, the class width is 100.
The midpoints of a class is the average value of the boundaries of a class.
The number of individuals included in the summary is the sum of all frequencies.
Thus, the number of individuals included in the summary is 146.