Chauncey billups, a current shooting guard for the los angeles clippers, has a career free-throw percentage of 89.4%. suppose he shoots six free throws in tonight’s game. what is the probability that billups makes all six free throws?

0.5105

Step-by-step explanation:

We know that if events A and B are independent, then the probability of getting A and B is

P(A and B)= P(A) x P(B)  [Product of the probabilities of each event ]

Given :  Chauncey Billups, a current shooting guard for the Los Angeles Clippers, has a career free-throw percentage of 89.4%.

i.e. P( make free-throw)= 0.894

Since all the shots are independent.

Then , the probability that Billups makes all six free throws will be :

Hence, the probability that Billups makes all six free throws= 0.5105

b. would be the answer : )

step-by-step explanation:

step 1: add -16x to both sides.

16x−4y+−16x=64+−16x  

−4y=−16x+64

step 2: divide both sides by -4.

−4y /-4 = -16x + 64/-4

answer:   y=4x−16

hope i'm right