The water works commission needs to know the mean household usage of water by the residents of a small town in gallons per day. they would like the estimate to have a maximum error of 0.1 gallons. a previous study found that for an average family the variance is 5.76 gallons and the mean is 17.9 gallons per day. if they are using a 99% level of confidence, how large of a sample is required to estimate the mean usage of water?

19,652

Step-by-step explanation:

Error (E) = (t×sd)/√n

√n = (t×sd)/E

sd = √variance = √5.76 = 2.4, E = 0.1

Assuming the average family is a family of 4

n = 4, degree of freedom = n-1 = 4-1 = 3, t-value corresponding to 3 degrees of freedom and 99% confidence level is 5.841

√n = 5.841×2.4/0.1 = 140.184

√n = 140.184

n = 140.184^2 = 19651.55

n = 19,652

1. x> -5

2. x< 8

step-by-step explanation:

1. 2x> -10

divide by 2 from both sides of equation.

2x/2> -10/2

simplify, to find the answer.

-10/2=-5

x> -5 is the correct answer.

2. 3x< 24

divide by 3 from both sides of equation.

3x/3< 24/3

simplify, to find the answer.

24/3=8

x< 8 is the correct answer.

i hope this you, and have a wonderful day!