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evablogger [386]
2 years ago
14

A sample of 75 information systems managers had an average hourly income of $40.75 with a standard deviation of $7.00. The 95% c

onfidence interval for the average hourly wage (in $) of all information system managers is
a. 37.54 to 43.96.
b. 38.61 to 42.89.
c. 39.14 to 42.36.
d. 39.40 to 42.10.
Mathematics
1 answer:
IrinaVladis [17]2 years ago
6 0

Answer: c. 39.14 to 42.36

Step-by-step explanation:

We want to determine a 95% confidence interval for the average hourly wage (in $) of all information system managers

Number of sample, n = 75

Mean, u = $40.75

Standard deviation, s = $7.00

For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean +/- z ×standard deviation/√n

It becomes

40.75 +/- 1.96 × 7/√75

= 40.75 ± 1.96 × 0.82

= 40.75 + 1.6072

The lower end of the confidence interval is 40.75 - 1.6072 =39.14

The upper end of the confidence interval is 40.75 + 1.6072 =42.36

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question 3

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Step-by-step explanation:

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question 4

Jimthompson5910Expert

Part A

Yes y is a function of x. Specifically the equation is y = x^5

For any given input, there is exactly one output

Notice how if x = 1 then y = x^5 = (1)^5 = 1

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=====================================================

Part B

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f(4) = 2(4)+12

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What does it mean? It is the total cost of renting the bowling lane for 4 hours. If you rent it for four hours, then you'll pay a total of $20. 

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In total, variableCost+fixedCost = 8+12 = 20 = totalCost

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