LaShawn designs websites for local businesses. He charges $25 an hour to build a website, and charges $15 an hour to update websites once he builds them. He wants to earn at least $100 every week, but he does not want to work more than 6 hours each week. What is a possible weekly number of hours LaShawn can spend building websites x and updating websites y that will allow him to obtain his goals?

a) 33.33%)

b) 135 minutes

c) 8.66 min

d) 50%

Step-by-step explanation:

a) the probability for a uniform distribution is

P(b<X<a) = (a-b)/(c-d) , where c and d are the maximum and minimum values

therefore the probability that the flight is more than 140 minutes ( and less than 150 since it is the maximum value)

P(140<X<150) = (a-b)/(c-d) = (150-140)/(150-120) = 10/30 = 1/3 (33.33%)

b) the mean (expected value) for a uniform probability distribution is

E(X) = (c+d)/2 = (120+150)/2 = 135 minutes

c)  the standard deviation for a uniform probability distribution is

σ²(X)= (c-d)²/12 = (150-120)²/12 = 75 min²

σ = √75 min² = 8.66 min

b) following the same procedure as in a)

P(120<X<135) = (a-b)/(c-d) = (135-120)/(150-120) = 15/30 = 1/2 (50%)