Asurvey conducted by the u. s. department of labor found the 48 out of 500 heads of households were unemployed. compute a 99% confidence interval for the proportion of unemployed heads of households in the population. round to three decimal places.

(0.062, 0.130)

Step-by-step explanation:

Sample size = n = 500

Number of heads that were unemployed = x = 48

Proportion of heads that were unemployed = p =

Proportion of heads that were not unemployed = q = 1 - p = 1 - 0.096 = 0.904

Confidence Level = 99%

z-value for 99% confidence level = z = 2.58

The confidence interval about a population proportion is calculated as:

Using the values, we get:

Thus, 99% confidence interval for the proportion of unemployed heads of households in the population is (0.062, 0.130)

step-by-step explanation:

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18