Sarah Anderson is running for mayor in a small rural town and would like to estimate the age of women voters. The town has 636 registered women voters on the records. Sarah found that a random sample of 70 women voters had an average age of 42.7 years and a standard deviation of 7.5 years. Determine the 90% confidence interval for the average of women voters in this town.

The 90% confidence interval for the average of women voters in this town is between 30.193 years and 55.207 years

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 70 - 1 = 69

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 69 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.6676

The margin of error is:

M = T*s = 1.6676*7.5 = 12.507

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 42.7 - 12.507 = 30.193 years

The upper end of the interval is the sample mean added to M. So it is 42.7 + 12.507 = 55.207 years

The 90% confidence interval for the average of women voters in this town is between 30.193 years and 55.207 years

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

cretdit to source

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