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77julia77 [94]
3 years ago
14

Find the percent from 85 to 30​

Mathematics
1 answer:
Zanzabum3 years ago
6 0

Answer:

25.5

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. 500
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Answer:

The 95% confidence interval estimate for the true proportion of adults residents of this city who have cell phones is (0.81, 0.874).

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}.

For this problem, we have that:

n = 500, \pi = \frac{421}{500} = 0.842

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.842 - 1.96\sqrt{\frac{0.842*0.158}{500}} = 0.81

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.842 + 1.96\sqrt{\frac{0.842*0.158}{500}} = 0.874

The 95% confidence interval estimate for the true proportion of adults residents of this city who have cell phones is (0.81, 0.874).

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Find the product of 17892 and 36
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