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
6cm^3 =
6* 0.001dm^3=
6*10^(-3)dm^3=
6*10^(-3)*10^(-3)m^3=
6*10^(-6)m^3
or 0.000006 meters cubed
She has distributed the bracket incorrectly
(64 + 5) - 0.2(- 20 + 10)
= (69) - 0.2 (- 10)
=34.5 + 2
= 36.5
For the denominators (7, 13) the least common multiple (LCM<span>) is </span>91.
Therefore, the least common denominator (LCD<span>) is </span>91.
<span>Rewriting the original inputs as equivalent fractions with the </span>LCD:
<span>78/91, 49/91.</span>
Answer:
2(x+2)
Step-by-step explanation:
Factor out the 2.
