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:
Dilation
Hope that helps!
Step-by-step explanation:
The answer is 5r^2 -1
hope this helps !
Hey buddy
The d part is correct
Hope it helps :)
The answer would be -5 because then you can do the opposite, 0+5 which would equal 5
