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4 years ago

I WILL MARK BRAINLIEST BUT PLEASE ANSWER FAST

Mathematics

1 answer:

Alinara [238K]4 years ago

8 0

Answer:

Step-by-step explanation:

It had this question on my test. I already answered it. I put c and got it wrong. The answer is B.

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Step-by-step explanation:

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3 years ago

A meatball recipe calls for 2 cups of breadcrumbs for every 5 cups of ground meat. How many cups of breadcrumbs are required for

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1/4 cup for every cup

Step-by-step explanation:

If you divide 2 by 5, you would get this answer. Sorry if it's wrong!

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3 years ago

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3 years ago

Question 4 of 5

belka [17]

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Angle B is a straight angel. Angels a and c are bot 45 degrees.

Step-by-step explanation:

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3 years ago

An automobile manufacturer would like to know what proportion of its customers are not satisfied with the service provided by th

butalik [34]

Answer:

a) A sample size of 5615 is needed.

b) 0.012

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

99.5% confidence level

So $\alpha = 0.005$ , z is the value of Z that has a pvalue of $1 - \frac{0.005}{2} = 0.9975$ , so $Z = 2.81$ .

(a) Past studies suggest that this proportion will be about 0.2. Find the sample size needed if the margin of the error of the confidence interval is to be about 0.015.

This is n for which M = 0.015.

We have that $\pi = 0.2$

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.015 = 2.81\sqrt{\frac{0.2*0.8}{n}}$

$0.015\sqrt{n} = 2.81\sqrt{0.2*0.8}$

$\sqrt{n} = \frac{2.81\sqrt{0.2*0.8}}{0.015}$

$(\sqrt{n})^{2} = (\frac{2.81\sqrt{0.2*0.8}}{0.015})^{2}$

$n = 5615$

A sample size of 5615 is needed.

(b) Using the sample size above, when the sample is actually contacted, 12% of the sample say they are not satisfied. What is the margin of the error of the confidence interval?

Now $\pi = 0.12, n = 5615$ .

We have to find M.

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$M = 2.81\sqrt{\frac{0.12*0.88}{5615}}$

$M = 0.012$

7 0

3 years ago