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choli [55]
2 years ago
11

The hours a week people spend answering and sending emails is normally distributed with a standard deviation of Ï = 150 minutes.

How large of a sample size would you need to estimate the mean number of minutes people spend answering and sending emails with an error of 1 minutes with 99% confidence? What would be the practical problem with attempting to find this confidence interval?
Mathematics
1 answer:
storchak [24]2 years ago
4 0

Answer:

sample size n would be 149305 large

Value of n (149305) is too high, this will be the practical problem with attempting to find this confidence interval

Step-by-step explanation:

Given that;

standard deviation α = 150 min

confidence interval = 99%

since; p( -2.576 < z < 2.576) = 0.99

so z-value for 99% CI is 2.576

E = 1 minutes

Therefore

n = [(z × α) / E ]²

so we substitute

n = [(2.576 × 150) / 1 ]²

n = [ 386.4 ]²

n = 149304.96 ≈ 149305

Therefore sample size would be 149305 large

Value of n is too high, that would be the practical problem with attempting to find this confidence interval

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An isosceles triangle has an angle that measures 92° which other angles could be in that isosceles triangle choose all that appl
nignag [31]

Answer:

44

Step-by-step explanation:

The two base angles cannot both be 92 degrees because they would add up to 184 which is more degrees than any triangle has. The apex angle (the top angle) therefore has 92 degrees.

The base angles are equal and found as follows.

2*b + 92 = 180                 Subtract 92 from both sides

2b +92-92 = 180-92

2b = 88                           divide by 2

2b/2=88/2

b = 44

The only angle measurement that fits is 44

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Answer:

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Step-by-step explanation:

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2 years ago
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