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Alina [70]
3 years ago
7

The height (in inches) of adult men in the United States is believed to be Normally distributed with mean μ . The average height

of a random sample of 25 American adult men is found to be ¯ x = 69.72 inches, and the standard deviation of the 25 heights is found to be s = 4.15 . A 90% confidence interval for μ is: a. 69.72 ± 1.37 b. 69.72 ± 1.42 c. 69.72 ± 1.09
Mathematics
1 answer:
Jlenok [28]3 years ago
5 0

Answer: a. 69.72 ± 1.37

Step-by-step explanation:

We want to determine a 90% confidence interval for the mean height (in inches) of adult men in the United States.

Number of sample, n = 25

Mean, u = 69.72 inches

Standard deviation, s = 4.15

For a confidence level of 90%, the corresponding z value is 1.645. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean ± z × standard deviation/√n

It becomes

69.72 ± 1.645 × 4.15/√25

= 69.72 ± 1.645 × 0.83

= 69.72 ± 1.37

The lower end of the confidence interval is 69.72 - 1.37 = 68.35

The upper end of the confidence interval is 69.72 + 1.37 = 71.09

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Answer:

  •   <u> 7,692 moose.</u>

Explanation:

The table that shows the pattern for this question is:

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  0                   40

  1                     62

  2                    96

  3                   149

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   P(0) = 40 = P₀ (r)⁰ = P₀ (1) = P₀

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<u>  2) r</u>

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You can verify that, for any other two consecutive terms you get the same result: 96/62 ≈ 149/96 ≈ 231/149 ≈ 1.55

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<span><u>Answer </u>
31.12 in

<u>Explanation </u>
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a/sin⁡A =b/sin⁡B Where a and b are the length opposite to angle A and B respectively.
DE/sin75=22/sin⁡43
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</span>
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