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aliina [53]
4 years ago
15

A random sample of 30 lunch orders at noodles and company showed a mean bill of $10.36 with a standard deviation of $5.31. find

the 95 percent confidence interval for the mean bill of all lunch orders.
Business
1 answer:
Paladinen [302]4 years ago
4 0

The formula for calculating the Confidence Interval is as follows:

Confidence Interval = x +- (z*s)/√N

Where:

x = mean = 10.36

z = taken from standard normal distribution table based on 95% confidence level = 1.96

s = standard deviation = 5.31

N = sample size = 30

Substituting know values on the equation:

Confidence Interval = 10.36 +- ( 1.96 * 5.31) / √30

Confidence Interval = 8.46 and 12.26

Hence the bill of lunch orders ranges from 8.46 to 12.26.

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Holding all else constant, an increase in preferences by Mexicans for U.S. goods will ______ the demand for dollars in the forei
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D. Increase; increase

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3 years ago
George Jefferson established a trust fund that provides $170,500 in scholarships each year for worthy students. The trust fund e
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Answers

let <em>C</em> be the capital, then :

C\times4 \%  = 170500\\C\times\frac{4}{100}= 170500\\C=170500\times\frac{100}{4}\\C=4262500

The capital contributed by Mr. Jefferson was <em>$4,262,500</em>

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4 years ago
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He should use a limit order.

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3 years ago
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Explanation:

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