A bag contains 8 red crayons, 14 purple crayons, 6 yellow crayons, and 4 Green crayons. A crayon is selected, NOT replaced, then another is selected. Find the probability of two reds.
P(two reds)

A bag contains 8 red crayons, 14 purple crayons, 6 yellow crayons, and 4 Green crayons.

=> Number of red crayons: 8

=> Total number of crayons: N = 8 + 14 + 6 + 4 = 32

=> The probability of selecting 2 red ones sequentially (without replacement):

P = (8/32) x (7/31) = 0.06

Hope this helps!

:)

expression #3 is not equivalent. the others are.

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!