Class records at rockwood college indicate that a student selected at random has probability 0.75 of passing french 101. for the student who passes french 101, the probability is 0.91 that he or she will pass french 102. what is the probability that a student selected at random will pass both french 101 and french 102? (round your answer to three decimal places.)

clock

step-by-step explanation:

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!