A study shows that employees who begin their work day at 8:00 a. m. vary their times of arrival uniformly over the range 7:40 a. m. to 8:30 a. m. (in other words, their arrival time follows a uniform distribution over a 50-minute period). We wish to simulate how many minutes late or early the customer arrives. If we have the random number 0.571, what will be the simulated late or early time of the the employee? (An early time will be a negative number and a late time will be a positive number. So if the employee arrives 10 minutes early, input -10. If (s)he arrives 10 minutes late, input 10.) Give your answer to 2 decimal places.

The employees arrives 8 minutes 55 seconds late .

Step-by-step explanation:

A r.v is uniformly distributed if its density function is

f(x) = 1/b-a

In the given question the lower limit a= 7:40 a.m and the upper limit b= 8:30 am

Putting the values we get

f(x)= 1/ 8:40 - 7:30

f(x)= 1/ 50 = 0.02

Now the random number is 0.571

P(x)= base * height

0.571= k* 0.02

Dividing

k= 0.571/ 0.02 = 68.55 minutes

Adding 68.55 minutes to 7:30 gives

8: 38.55

Now subtracting from upper limit gives 8 minutes 55 seconds.

The employees arrives 8 minutes 55 seconds late .

it is contrapositive

i could not see the question too well so check yourself again too . this is because i want you to get it correct.

step-by-step explanation:

the shape,the size and the distance