Answer: The answer is 0.18
Step-by-step explanation:
I took the test
R= 5
You divide -7 by both sides to 1. Cancel out the -7 2. To simplify the -35. So simplified -35/-7=5
Answer:
The 99% confidence interval for the true population proportion of people with kids is (0.293, 0.547).
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 
.
Out of 100 people sampled, 42 had kids.
This means that 
99% confidence level
So 
, z is the value of Z that has a p-value of 
, so 
.
The lower limit of this interval is:
The upper limit of this interval is:
The 99% confidence interval for the true population proportion of people with kids is (0.293, 0.547).
Solve the equation of the function for n.
a(n) = 3n - 20
a = 3n - 20
a + 20 = 3n
(a + 20)/3 = n
n = (a + 20)/3
n(a) = (a + 20)/3
The 90% confidence interval for the population mean is 78.1905<x<81.8095
<h3>Confidence interval</h3>
The formula for calculating the confidence interval is expressed as:
CI = x ± z*s/√n
Given the following parameters
mean "x" = 80
z = 1.645
s = 5.5
n = 25
Substitute
CI = 80 ± 1.645*5.5/√25
CI = 80 ± 1.8095
CI = {78.1905, 81.8095}
Hence the 90% confidence interval for the population mean is 78.1905<x<81.8095
Learn more on confidence interval here: brainly.com/question/2141785
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