A bag contains five white balls and four black balls. Your goal is to draw two black balls. You draw two balls at random. Once you have drawn two balls, you put back any white balls, and redraw so that you again have two drawn balls. What is the probability that you now have two black balls? (Include the probability that you chose two black balls on the first draw.)

Probabilty of both Black

= 1/6

Step-by-step explanation:

A bag contains five white balls and four black balls.

Total number of balls= 5+4

Total number of balls= 9

Probabilty of selecting a black ball first

= 4/9

Black ball remaining= 3

Total ball remaining= 8

Probabilty of selecting another black ball without replacement

= 3/8

Probabilty of both Black

=3/8 *4/9

Probabilty of both Black

= 12/72

Probabilty of both Black

= 1/6

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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