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2 years ago

PLEASE HELP WHOEVER ANSWERS CORRECTLY GETS BRAINLIEST

Mathematics

1 answer:

Soloha48 [4]2 years ago

4 0

Answer:

3rd and 6th are correct

Step-by-step explanation:

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Express 7.4% as a fraction and a decimal.

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2 years ago

HELP ASAPPP 10 POINTS

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2 years ago

SAT scores are normally distributed with a mean of 1,500 and a standard deviation of 300. An administrator at a college is inter

Rama09 [41]

Answer:

The administrator should sample 968 students.

Step-by-step explanation:

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

$\alpha = \frac{1 - 0.88}{2} = 0.06$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a p-value of $1 - 0.06 = 0.94$ , so Z = 1.555.

Now, 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.

Standard deviation of 300.

This means that $n = 300$

If the administrator would like to limit the margin of error of the 88% confidence interval to 15 points, how many students should the administrator sample?

This is n for which M = 15. So

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

$15 = 1.555\frac{300}{\sqrt{n}}$

$15\sqrt{n} = 300*1.555$

Dividing both sides by 15

$\sqrt{n} = 20*1.555$

$(\sqrt{n})^2 = (20*1.555)^2$

$n = 967.2$

Rounding up:

The administrator should sample 968 students.

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2 years ago

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2 years ago

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