Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 387 drivers and find that 298 claim to always buckle up. Construct a 84% confidence interval for the population proportion that claim to always buckle up.

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

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 .

They randomly survey 387 drivers and find that 298 claim to always buckle up.

This means that

84% 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 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15