Suppose you have a litter of mice which consists of 8 males and 4 females. if you randomly grab two of them, what is the probability that they are both male? you may think of this questions as without replacement. keep your answer in decimal form and round to three decimal places.

The probability that they are both male is 0.424 (3 d.p.)

Step-by-step explanation:

The first step is to find the probability of the first selection being male. This is calculated as number of male mice divided by total number of mice in the litter

Prob (1st male) = 8 ÷ 12 = 0.667

Next is to find the probability of the second selection also being male. Note that the question states that the first mice was selected without replacement. This means the first mouse taken results in a reduction in both the number of male mice and total number of mice in the litter.

Prob (2nd male) = (8 - 1) ÷ (12 - 1) = 7/11 = 0.636

Therefore,

Prob (1st male & 2nd male) = 0.667 × 0.636 = 0.424

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