Suppose the manager agrees to pay each employee a $50 bonus if they meet a certain goal. On a typical Saturday, the oil-change facility will perform 4545 oil changes between 10 A. M. and 12 P. M. Treating this as a random sample, at what mean oil-change time would there be a 10% chance of being at or below? This will be the goal established by the manager. There would be a 10% chance of being at or below nothing minutes. (Round to one decimal place as needed.)

The answer is given below

Step-by-step explanation:

The mean (μ) = 21.2 minutes and the standard deviation σ = 3.5 minutes.

the oil-change facility will perform 45 oil changes between 10 A.M. and 12 P.M, therefore the sample size n = 45

there be a 10% chance of being at or below. From the normal distribution table, The z score corresponding to a probability of 10% (= 0.1)  is -1.28.

z = -1.28

To calculate the mean oil-change time, we use the formula:

y = - 6

step-by-step explanation:

given the point (- 2, y) lies on the graph then it is a solution of it's equation.

substitute x = - 2 into the equation for value of y

y = 3(- 2) = 3 × - 2 = - 6

the point on the graph is (- 2, 6)

d? maybe

step-by-step explanation: