The owner of a meat market has an assistant who has determined that the weights of roasts are normally distributed, with a mean of 3.2 pounds and a standard deviation of 0.8 pounds. if a sample of 25 roasts yields a mean of 3.6 pounds, what is the z-score for this ample mean?

2.5

Step-by-step explanation:

We are given that

Mean= pounds

Standard deviation= pounds

n=25

We have to find the Z-score if the mean of  a sample of 25 roasts is 3.6 pounds.

We know that Z-score formula

We have

Substitute the values then we get

Z- score=

Z- score=

Hence, the Z-score value for the given sample mean=2.5

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).