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
Answer:
5^5-7^2= 3076
Step-by-step explanation:
5^5=3125
5*5*5*5*5
7^2=49
3125-49=3076
Hope this helps! Remember you are smart and you got this! Have a great day -Alycia
Answer:
Step-by-step explanation:
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Happy to help :)
Answer:
32-28=4
Step-by-step explanation:
See the attachment :)
The answer to your question is 54325
