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prohojiy [21]
2 years ago
14

(HELP) A Boeing 747 airplane carries approximately 467 passengers. How many trips would it take the airplane to transport 100,00

0 people?
Mathematics
2 answers:
Finger [1]2 years ago
6 0
215. If you divide 100,000 by 467 you get 214.132. Since you can’t have .132 of a person you need to round up to 215
netineya [11]2 years ago
5 0

Answer:

You'll need to divide 100,000 by 467 which is 214.132, then round up so no one is left behind, so you would have to make <u>215 trips</u>

Step-by-step explanation:

I hope I helped ^-^

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37.362 is 7 4/5 % of what number?
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Step-by-step explanation:

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1) B

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You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confide
FrozenT [24]

Question:

You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If​ convenient, use technology to construct the confidence intervals. A random sample of 45 home theater systems has a mean price of ​$114.00. Assume the population standard deviation is ​$15.30. Construct a​ 90% confidence interval for the population mean.

Answer:

At the 90% confidence level, confidence interval = 110.2484 < μ < 117.7516

At the 95% confidence level, confidence interval = 109.53 < μ < 118.48

The 95% confidence interval is wider

Step-by-step explanation:

Here, we have

Sample size, n = 45

Sample mean, \bar x = $114.00

Population standard deviation, σ = $15.30

The formula for Confidence Interval, CI is given by the following relation;

CI=\bar{x}\pm z\frac{\sigma}{\sqrt{n}}

Where, z is found for the 90% confidence level as ±1.645

Plugging in the values, we have;

CI=114\pm 1.645 \times \frac{15.3}{\sqrt{45}}

or CI: 110.2484 < μ < 117.7516

At 95% confidence level, we have our z value given as z = ±1.96

From which we have CI=114\pm 1.96 \times \frac{15.3}{\sqrt{45}}

Hence CI: 109.53 < μ < 118.48

To find the wider interval, we subtract their minimum from the maximum as follows;

90% Confidence level: 117.7516 - 110.2484 = 7.5

95% Confidence level: 118.47503 - 109.5297 = 8.94

Therefore, the 95% confidence interval is wider.

8 0
3 years ago
Read 2 more answers
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