Answer:
We need a sample of at least 1797 if we wish to be 95% confident that the sample percentage of those equating success with personal satisfaction is within 2.3% of the population percentage.
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 zscore that has a pvalue of
.
The margin of error is:

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

How large a sample is needed if we wish to be 95% confident that the sample percentage of those equating success with personal satisfaction is within 2.3% of the population percentage?
We have to find n for which
. So







We need a sample of at least 1797 if we wish to be 95% confident that the sample percentage of those equating success with personal satisfaction is within 2.3% of the population percentage.
Answer:488
<em>Multiply</em><em> </em><em>12.20</em><em> </em><em>and</em><em> </em><em>4</em><em>0</em>
Answer:
2x*2x*2x/5y
Step-by-step explanation:
Expand the exponets number out by that many times
Answer:
11/20
Step-by-step explanation:
3/4 - 1/5 = 15/20 - 4/20 = 11/20
Yup D Is Right Because if you divide by 3.8% * 5000 you get d