This would vary on the perspective of the solver. In this case, the solution would be
A A A B B B B C C C
A A A B B B B C C C
A A A B B B B C C C
A A A B B B B C C C
A A A B B B B C C C
A A A B B B B C C C
A A A B B B B C C C
A A A B B B B C C C
A A A B B B B C C C
A A A B B B B C C C
such that Employee A would constitute 30%, Employee B would constitute 40% and Employee C would constitute 30% of the work.
I hope I was able to answer your question. Have a good day.
Answer:
A sample of 801 is needed.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.
In which
z is the z-score that has a p-value of .
The margin of error is of:
95% confidence level
So , z is the value of Z that has a p-value of , so .
25% of U.S. homes have a direct satellite television receiver.
This means that
How large a sample is necessary to estimate the true population of homes which do with 95% confidence and within 3 percentage points?
This is n for which M = 0.03. So
Rounding up:
A sample of 801 is needed.
<span>60 divide 9840 = </span>0.00609756
<span>542.396828571 that is your answer </span>
2,600/4 =650
650•5 =3,250
I found the unit rate of the 2,600 by dividing the per hour and then times it by the 5 hours
Hopefully that was what you were looking for