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hoa [83]
3 years ago
13

A sample of 64 was taken from a population, with the mean of the sample being 43 and the standard deviation of the sample being

6. What is the 95% confidence interval for the mean of the population?
A. (42.25, 43.75)
B. (37.3, 48.7)
C. (41.5, 44.5)
D. (37, 49)
Mathematics
1 answer:
zzz [600]3 years ago
5 0

Answer:

C. (41.5, 44.5)

Step-by-step explanation:

Point estimate of population variance:

64(6²)/63

256/7

xbar +/- [z × sqrt(var/n)]

43 +/- [1.96 × sqrt(256/7 ÷ 64l]

43 +/- 1.481620735

[41.51837927 , 44.48162073]

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nekit [7.7K]

Answer:68.3 degrees

Step-by-step explanation:

The diagram of the triangle ABC is shown in the attached photo. We would determine the length of side AB. It is equal to a. We would apply the cosine rule which is expressed as follows

c^2 = a^2 + b^2 - 2abCos C

Looking at the triangle,

b = 75 miles

a = 80 miles.

Angle ACB = 180 - 42 = 138 degrees. Therefore

c^2 = 80^2 + 75^2 - 2 × 80 × 75Cos 138

c^2 = 6400 + 5625 - 12000Cos 138

c^2 = 6400 + 5625 - 12000 × -0.7431

c^2 = 12025 + 8917.2

c = √20942.2 = 144.7

To determine A, we will apply sine rule

a/SinA = b/SinB = c/SinC. Therefore,

80/SinA = 144.7/Sin 138

80Sin 138 = 144.7 SinA

SinA = 53.528/144.7 = 0.3699

A = 21.7 degrees

Therefore, theta = 90 - 21.7

= 68.3 degees

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Equation

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Answer

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