A machine fill bags with cement. Assume the actual masses of the bags have a normal distribution. If the standard deviation of the amount dispensed is 1.1 lb and the mean is set to 50 lb. What proportion of bags have a mass more than 50.4 lb? (2 p

0.35808

Step-by-step explanation:

Given a normal distribution :

Mean (m) = 50 lb

Standard deviation (σ) = 1.1 lb

Proportion of bags with mass more than 50.4 lbs

X : X > 50.4 lbs

Using the Zscore relation :

Zscore = (x - m) / σ

Zscore = (50.4 - 50) / 1.1

Zscore = (0.4) / 1.1

Zscore = 0.3636363

P(Z > 0.3636) = 1 - P(Z < 0.3636)

P(Z < 0.3636) = 0.64192

1 - P(Z < 0.3636) = 1 - 0.64192 = 0.35808

  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).