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Tatiana [17]
3 years ago
10

Steve is making breakfast. The recipes call for 7/8 cup of milk for grits and 3/4 cups .for biscuits he only has 2 cups of milk.

Does he have enough to make his breakfast? Explain
Mathematics
1 answer:
Zanzabum3 years ago
4 0
Answer: He does have enough.
Explanation: Simplifying 7/8 cups to fourths you get 1 3/4 cups of milk. If he has 2 cups of milk he still would have 1/4 cup left over of milk.
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86

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The time, t, required to complete a job
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19 hrs

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

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