A professional basketball player is a poor free-throw shooter. Consider situations in which he shoots a pair of free throws. The probability that he makes the first free throw is 0.51. Given that he makes the first, suppose the probability that he makes the second is 0.59. Given that he misses the first, suppose the probability that he makes the second one is 0.36. What is the probability that he makes one of the two free throws?

38.55% probability that he makes one of the two free throws

Step-by-step explanation:

We have these following probabilities:

51% probability of making the first free throw.

100-51 = 49% probability of missing the first free throw.

If he makes the first free throw, a 59% probability of making the second free throw and a 100-59 = 41% probability of missing the second free throw.

If he misses the first free throw, a 36% probability of making the second free throw and a 100-36 = 64% probability of missing the second free throw.

What is the probability that he makes one of the two free throws?

Makes first(0.51) and misses second(0.41)

Misses first(0.49) and makes second(0.36)

So

38.55% probability that he makes one of the two free throws

eveidence i thing that

step-by-step explanation:

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!