Jalen Brunson is shooting free throws. Making or missing free throws doesn't change the probability that he will make his next one, and he makes his free throws 90% percent of the time. What is the probability of Jalen Brunson making his next 2 free throw attempts?

The probability is 0.81

Step-by-step explanation:

In this question, we are tasked with calculating the probability that Jalen makes his next two free throws.

P(of making free throw) = 90% = 90/100 or 0.9

P(of missing free throw) = 1 - P( of making free throw) = 1-0.9 = 0.1

Now we want to calculate that probability of making the next two free throws.

That would be = p( of making next throw) * p(of making next two throws) = 0.9 * 0.9 = 0.81

the mean and mad

step-by-step explanation:

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18