Suppose there is a box containing 9 white balls, 3 black balls, and 4 red balls. is it possible to add the same number of balls of each color to the box so that when a ball is drawn at random the probability of the event that is a black ball is the following? a) 1/3
b) 0.24
explain what number needs to be added to get this answer!

it is contrapositive

i could not see the question too well so check yourself again too . this is because i want you to get it correct.

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

you already gave the answer

step-by-step explanation:

5+2x=a

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