Abox contains four red balls and eight black balls. two balls are randomly chosen from the box, and are not replaced. let event b be choosing a black ball first and event r be choosing a red ball second. what are the following probabilities? p(b) = p(r | b) = p(b ∩ r) = the probability that the first ball chosen is black and the second ball chosen is red is about percent.

black ball   and    red ball

8 out of 12   x     4 out of 11

                 x            

= = 0.24  = 24%

24%

arithmetic sequence: a sequnce in which the diffrence between any two consecutive terms is the same

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!