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.
Common factors are 5 and 15 and 3
Answer:
60
Step-by-step explanation:
Sum means add. Then we get twice. This means to multiply by 2. Twice a number means we have 2x. And is what we need to add to that 2x, so it is 2x + 13. Is means equals, so answer is 2x + 13 = 75.
Area = length x width
60 = (x+5) (x-2)
60 = x^2 + 3x -10
Subtract 60 from both sides:
x^2 +3x -70 = 0
Solve for x by finding 2 numbers when added together = 3 and when multiplied by each other = -70
x = -10 and 7
Since the sides have to be a positive number we need to use 7
The length = x+5 = 7+5 = 12 cm
The width = x-2 = 7-2 = 5 cm