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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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Answer:

Step-by-step explanation:

Suppose the cost C(x), to build a football stadium of x thousand square feet is approximated by C(x) = 7,250,000/x + 60. Given the function, we can substitute values for x to determine the cost of a particular size of stadium or we can substitute values for C(x) to determine the number of square feet.

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Answer:

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Step-by-step explanation:

* Lets talk about some transformation

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- If the function f(x) translated horizontally to the left  

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- If the function f(x) translated vertically up  

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Answer:

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<em></em>

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