Answer:
Step-by-step explanation:
From the information given,
Number of personnel sampled, n = 85
Mean or average = 6.5
Standard deviation of the sample = 1.7
We want to determine the confidence interval for the mean number of years that personnel spent in a particular job before being promoted.
For a 95% confidence interval, the confidence level is 1.96. This is the z value and it is determined from the normal distribution table. We will apply the following formula to determine the confidence interval.
z×standard deviation/√n
= 1.96 × 6.5/√85
= 1.38
The confidence interval for the mean number of years spent before promotion is
The lower end of the interval is 6.5 - 1.38 = 5.12 years
The upper end is 6.5 + 1.38 = 7.88 years
Therefore, with 95% confidence interval, the mean number of years spent before being promoted is between 5.12 years and 7.88 years
(4x+4y)+(8x+8y)
12x+12y
That should be your answers, you can't add them bc they're not like terms.
<span> x-4
--------- = 0
x-6
means x - 4 = 0
x = 4
answer
</span><span>B.)4</span>
Answer:
I would go with answer option A.
Step-by-step explanation:
Hope this helps. Also have an Amazing Day!
Answer:8.9
Step-by-step explanation: y=2.2 +3.5 =6.7 +0.18
y= 8.9
good luck