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
Here you go. Hope this helps! :)
The answer is a. 5 x 10³.
3.75 x 10^5 = 375,00.( population of Greenville)
375,000 ÷ 75 = 5,000 ( population of Fairview)
5 x 10³(1,000)= 5,000.
Use the proportion formula:
5mm/32= Xmm/160
=22.22 mm
Answer:
∠ QPR = 125°
Step-by-step explanation:
the angles subtended by 2 congruent arcs are congruent , that is
5x = 4x + 25 ( subtract 4x from both sides )
x = 25
Then
∠ QPR = 4x + 25 = 4(25) + 25 = 100 + 25 = 125°
