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Fynjy0 [20]
3 years ago
14

Tess ran 4 3 3 4 ​ start fraction, 4, divided by, 3, end fraction kilometers each day for 2 22 days. Cam ran 2 3 3 2 ​ start fra

ction, 2, divided by, 3, end fraction of a kilometer each day for 5 55 days. Who ran more kilometers in total? Choose 1 answer: Choose 1 answer: (Choice A) A Tess (Choice B) B Cam (Choice C) C They ran the same number of total kilometers. Stuck
Mathematics
1 answer:
mestny [16]3 years ago
7 0

We have been given that Tess ran \frac{4}{3} kilo-meters each day for 2 days.

Let us find total distance traveled by Tess in 2 days by multiplying \frac{4}{3} by 2 as:

\text{Distance traveled by Tess in 2 days }=\frac{4}{3}\times 2

\text{Distance traveled by Tess in 2 days }=\frac{4\times 2}{3}

\text{Distance traveled by Tess in 2 days }=\frac{8}{3}

We are also told that Cam ran  \frac{2}{3} kilo-meters each day for 5 days.

Let us find total distance traveled by Cam in 5 days by multiplying \frac{2}{3} by 5 as:

\text{Distance traveled by Cam in 5 days }=\frac{2}{3}\times 5

\text{Distance traveled by Cam in 5 days }=\frac{2\times 5}{3}

\text{Distance traveled by Cam in 5 days }=\frac{10}{3}

Since \frac{10}{3} km is greater than \frac{8}{3} km, therefore, Cam ran more kilo-meters in total and option B is the correct choice.

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Answer:

Using a 90% confidence level

A. A sample size of 68 should be used.

B. A sample size of 98 should be used.

Step-by-step explanation:

I think there was a small typing mistake and the confidence level was left out. I will use a 90% confidence level.

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

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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.

A. If we want to estimate the population mean time for previews at movie theaters with a margin of error of 72 seconds, what sample size should be used?

We have the standard deviation in minutes, so the margin of error should be in minutes.

72 seconds is 72/60 = 1.2 minutes.

So we need a sample size of n, and n is found when M = 1.2. We have that \sigma = 6. So

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

1.2 = 1.645*\frac{6}{\sqrt{n}}

1.2\sqrt{n} = 6*1.645

\sqrt{n} = \frac{6*1.645}{1.2}

(\sqrt{n})^{2} = (\frac{6*1.645}{1.2})^{2}

n = 67.65

Rounding up.

A sample size of 68 should be used.

B. If we want to estimate the population mean time for previews at movie theaters with a margin of error of 1 minute, what sample size should be used?

Same logic as above, just use M = 1.

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

1 = 1.645*\frac{6}{\sqrt{n}}

\sqrt{n} = 6*1.645

(\sqrt{n})^{2} = (6*1.645)^2

n = 97.42

Rounding up

A sample size of 98 should be used.

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