During a recent drought, a water utility in a certain town sampled 100 residential water bills and found that 75 of the residences had reduced their water consumption over that of the previous year. Find a 95% confidence interval for the proportion of residences that reduced their water consumption. Round the answers to three decimal places.

The 95% confidence interval for the proportion of residences that reduced their water consumption is (0.665, 0.835).

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 .

Sampled 100 residential water bills and found that 75 of the residences had reduced their water consumption over that of the previous year.

This means that

95% confidence level

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

The 95% confidence interval for the proportion of residences that reduced their water consumption is (0.665, 0.835).

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option

the answer is c hope it

step-by-step explanation: