Answer:
The answer is "".
Step-by-step explanation:
Please find the complete question in the attached file.
We select a sample size n from the confidence interval with the mean
and default
, then the mean take seriously given as the straight line with a z score given by the confidence interval

Using formula:
The probability that perhaps the mean shells length of the sample is over 4.03 pounds is

Now, we utilize z to get the likelihood, and we use the Excel function for a more exact distribution
the required probability:

For this case, the first thing you should do is the following sum:
income = Annual Salary + benefits
Substituting values we have:
income = (48500) + (2500 + 5000)
income = 56000
Answer:
She can expect a total amount of income with benefits from:income = 56000
(option 4)
Answer:
The percent of change is 44.7%
Step-by-step explanation:
Given that:
38 staples to 55 staples
As the number of staples are increased, therefore, the percent will also increase.
Difference = 55 - 38 = 17 staples
Percent of change = 
Percent of change = 
Percent of change = 0.447*100
Percent of change = 44.7%
Hence,
The percent of change is 44.7%