Afootball quarterback has two more chances to throw a touchdown before his team is forced to punt the ball. he misses the receiver on the first throw 25% of the time. when his first throw is incomplete, he misses the receiver on the second throw 15% of the time. what is the probability of not throwing the ball to a receiver on either throw?

The probability of not throwing the ball to a receiver on either throw is 0.0375 or 3.75%.

Step-by-step explanation:

Consider the provided information.

Let A is the event that he misses on the 1st throw. Thus the probability P(A) = 25% or 0.25

Let B is the event that he misses on the 2nd throw. Thus the probability P(B/A) = 15% or 0.15

The combined probability of the event is:

P(A) × P(B/A) = 0.25 × 0.15 = 0.0375

Hence, the probability of not throwing the ball to a receiver on either throw is 0.0375 or 3.75%.

Hello! The answer to your question would be 3%

A- missing the first throw     P (A) = 0.3

B- missing the second throw    P (B/A) = 0.1

Given the two possibilities, the probability of the quarterback not throwing the ball to a receiver is

P (A) x  P  (B/A) = 0.3 x 0.1 = 0.03 (3%)

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