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sweet-ann [11.9K]
3 years ago
14

Each basketball comes in a cubic box with side lengths of 1 foot. The locker is 3 feet long, 2 feet wide, and 5 feet high. How m

any basketballs can Azul fit in the locker?
Mathematics
1 answer:
LuckyWell [14K]3 years ago
4 0

Answer:

30 basketballs

Step-by-step explanation:

one layer would be 3 x 2 = 6

  • to find how much you would need to cover the bottom of the locker, you would use the numbers 3 and 2, then multiply them and you get 6. You would need 6 basketballs to cover the bottom of the locker. That would be your first layer

5 layers would be 6 x 5 = 30

  • So the bottom would be filled with 6 basketballs. Now since the locker is 5 feet high multiply 6 x 5 and that would get you your total of basketballs
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Part A

Given info:

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Because n > 30 is not true, and we don't know the population standard deviation (sigma), this means we must use a T distribution.

The degrees of freedom here are n-1 = 23-1 = 22.

At 90% confidence and the degrees of freedom mentioned, the t critical value is roughly t = 1.717

Use a T distribution table or calculator to determine this. If you don't have a calculator for the task, then you can search out "inverse T calculator" and there are tons of free options to pick from.

The margin of error E is

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E = 1.717*1.12/sqrt(23)

E = 0.400982

This is approximate and accurate to 6 decimal places.

The confidence interval is going to be xbar plus or minus that E value

L = lower bound = xbar - E = 4.90 - 0.400982 = 4.499018 = 4.50

U = upper bound = xbar + E = 4.90 + 0.400982 = 5.300982 = 5.30

The confidence interval in the format of (L, U) is (4.50, 5.30)

You could also express it as the format L < mu < U and it would be 4.50 < mu < 5.30; however, I'll stick to the first method.

<h3>Answer:  (4.50, 5.30)</h3>

=====================================================

Part B

Since we know sigma = 2.6 is the population standard deviation, we can use a Z distribution now.

At 90% confidence, the z critical value is roughly 1.645; use a table or calculator to determine this.

n = \left(\frac{z*\sigma}{E}\right)^2\\\\n \approx \left(\frac{1.645*2.6}{0.03}\right)^2\\\\n \approx 20325.254444 \\\\

Round this up to the nearest integer to get 20326. For min sample size problems, <u>always</u> round up.

<h3>Answer: 20326</h3>
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