Answer:
1. Five classes in the frequency distribution
2.The mid point of the third class is 24.5
3.The total number of frequency is 100
4.The relative frequency of the third class is (20/100=0.2)
5. We use n-1 because we are dealing with a sample data
6. The mean is 24.5
7. The value of the standard deviation is 11
Step-by-step explanation:
See attached picture for the solutions
Using proportions, it is found that it takes 886 more mini-bears than regular-bears to have the same weight as one super-bear.
<h3>What is a proportion?</h3>
A proportion is a fraction of a total amount, and the measures are related using a rule of three. Due to this, relations between variables, either direct(when both increase or both decrease) or inverse proportional(when one increases and the other decreases, or vice versa), can be built to find the desired measures in the problem, or equations to find these measures.
10 mini-bears weights to 12.1 grams, hence the weight of a mini-bear is of:
12.1/10 = 1.21 grams.
10 regular bears weights to 23.1 grams, hence the weight of a regular bear is of:
23.1/10 = 2.31 grams.
1 super bear weights to 2250 grams, hence the proportion between the <u>weight of a super bear and the weight of a mini-bear</u> is:
2250/1.21 = 1860.
The proportion between the <u>weight of a super bear and the weight of a regular bear</u> is:
2250/2.31 = 974.
The difference of proportions is given by:
1860 - 974 = 886.
It takes 886 more mini-bears than regular-bears to have the same weight as one super-bear.
More can be learned about proportions at brainly.com/question/24372153
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Answer:
C
Step-by-step explanation:
its 2/5 for the evens 3/5 for odds multiply and you get 6/25