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dimulka [17.4K]
3 years ago
10

Help please..I really need it​

Mathematics
1 answer:
jarptica [38.1K]3 years ago
4 0

Answer:

256

Step-by-step explanation:

V=1/3*π*4^2*7,46=124.993

32000/124.99=~256

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It is known that the population variance equals 484. With a .95 probability, the sample size that needs to be taken if the desir
Ksju [112]

Answer:

We need a sample size of at least 75.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

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Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

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Now, we find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

The standard deviation is the square root of the variance. So:

\sigma = \sqrt{484} = 22

With a .95 probability, the sample size that needs to be taken if the desired margin of error is 5 or less is

We need a sample size of at least n, in which n is found when M = 5. So

M = z*\frac{\sigma}{\sqrt{n}}

5 = 1.96*\frac{22}{\sqrt{n}}

5\sqrt{n} = 43.12

\sqrt{n} = \frac{43.12}{5}

\sqrt{n} = 8.624

(\sqrt{n})^{2} = (8.624)^{2}

n = 74.4

We need a sample size of at least 75.

6 0
3 years ago
Read 2 more answers
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Answer:

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