Answer:
At least 547 records need to be studied.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence interval
, we have the following confidence interval of proportions.

In which
Z is the zscore that has a pvalue of
.
And the margin of error is:

95% confidence interval
So
, z is the value of Z that has a pvalue of
, so
.
In this problem, we have that:






At least 547 records need to be studied.
The answer is in decimal formal in the file below
Answer:
B ( 432 units^2)
Step-by-step explanation:
Total surface area= 2*(6*6)+ 4*(15*6)
There are 2 squares at each end of the same size so total area will be 2*(6*6)
If we look at the rectangles( bottom, top and the sides), we can see that all are of same dimensions and there are 4 of them so it's 4*(15*6)
In short, total surface area= 2*(6*6)+ 4*(15*6) = 432 units^2
The answer is A. I plugged it into a TI-83 calculator. If you need to know how I did that I will be happy to help
Answer:
C Cos32
Step-by-step explanation: