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 0.130.13 gallons. A previous study found that for an average family the variance is 2.562.56 gallons and the mean is 16.316.3 gallons per day. If they are using a 80%80% level of confidence, how large of a sample is required to estimate the mean usage of water? Round your answer up to the next integer.

Therefore, a sample that is required to estimate the mean usage of water must be at least 249.

Step-by-step explanation:

We are given that they would like the estimate to have a maximum error of 0.13 gallons.

A previous study found that for an average family the variance is 2.56 gallons and the mean is 16.3 gallons per day.  

As we know that; Margin of error formula is given by;

   Margin of error  =  

where, = level of significance = 1 - 0.80 = 20% and   = 10%

Standard of error =  

where, = standard deviation = = 1.6

            n = sample size

Also, the critical value of z at 10% significance level is given in the z table as 1.282.

SO,  Margin of error  =  

                 0.13           =  

                   =  

                    n  =  

                    n  =  248.95  ≈ 249.

Therefore, a sample that is required to estimate the mean usage of water must be at least 249.

stew reads at the fastest rate of 265 words per minute.

step-by-step explanation:

it's a given that felicia reads at 260 word per minute and on connie's graph we can see that she reads 200 words per minute. since stew's table tells us that he read 795 in 3 minutes, we can divide 795 by 3 to find out how many words he reads per minute.

795/3 = 265

stew reads 265 words per minute - he reads at the fastest rate since felecia reads at 260 words per minute and connie reads at 200 words per minute.

stew (265) > felecia (260) > connie (200)

16 is the absolute value

step-by-step explanation: