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BartSMP [9]
3 years ago
5

HELP!! ASAP

Mathematics
1 answer:
-Dominant- [34]3 years ago
8 0
The answer is two inches, C
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I need help on this question plz
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Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. The
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Answer:

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

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 z-score that has a p-value of 1 - \frac{\alpha}{2}.

They randomly survey 387 drivers and find that 298 claim to always buckle up.

This means that n = 387, \pi = \frac{298}{387} = 0.77

84% confidence level

So \alpha = 0.16, z is the value of Z that has a p-value of 1 - \frac{0.16}{2} = 0.92, so Z = 1.405.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 - 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.74

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 + 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.8

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

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I atttached some screenshots of something that explinaed better than I can. Another way to do it is rations. If you cna determine each of the angles, then you can set the sides equal to each other and find the missing sides otherwose this is proballu the best way to do it.

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