5. At the local burger joint, 45% of the customers purchase cheeseburgers and 80% purchase fries. If a customer is selected at randomlet the event that they bought a burger be A and the event that they purchased fries be B. If A and B are independent what is P(A or B)?

P (A or B) =  0.53

Step-by-step explanation:

P(A or B) is the probability where one of the events happened but not both of them.

Probability of event A = 0.45

Probability of not event A = 1 - 0.45 = 0.55

Probabiity of event B = 0.80

Probability of not event B = 1 - 0.80 = 0.20

Then, the probability that one customer buys a burger but does not buy fries is:

P1 = 0.45*0.20 = 0.09

Probability of purchasing fries but not a burger:

P2 = 0.80*0.55 = 0.44

P (A or B) = P1 + P2 = 0.09 + 0.44 = 0.53

i believe the answer is 68cm

step-by-step explanation:

i just took 34 x 2 because the space between v and y is the same as y to k.