- The margin of error of the proportion given is of 2.9%.
- Applying the margin of error, the confidence interval is (29.1%, 34.9%).
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The margin of error of a confidence interval of a proportion
in a sample of size n, with a confidence level of
, is:

In which z is the z-score that has a p-value of
.
The confidence interval is:

In this problem:
- Poll of 1012 people, thus
. - 32% keep a dog, thus

95% confidence level
Thus
, z is the z-score that has a p-value of
, so
.
The margin of error is:

As a percent, 2.9%, as 0.029 x 100% = 2.9%.
Now for the confidence interval, which is <u>percentage plus/minus margin of error</u>, thus:


The confidence interval is (29.1%, 34.9%).
A similar problem is given at brainly.com/question/16807970
Answer:
The expected value of the money she selects is 6.65.
Step-by-step explanation:
Expected value:
Sum of each outcome multiplied by its probability.
We have:
8 + 5 + 4 + 3 = 20 total bills.
8 are ones, so P(X = 1) = 8/20 = 0.4.
5 are fives, so P(X = 5) = 5/20 = 0.25.
4 are tens, so P(X = 10) = 4/20 = 0.2
3 are twenties, so P(X = 20) = 3/20 = 0.15.
What is the expected value of the money she selects?

The expected value of the money she selects is 6.65.
No, it is equal to the decimal -.3333... 1/3^-1 however, would be equal to three since the negative exponent could be brought to the top and made positive, maybe that's what you were thinking of?
7.1 and 7.5 because any integer is a rational number