The probability of getting two good coils when two coils are randomly selected if the first selection is replaced before the second is made is 0.7744

Solution:

Total number of coils = number of good coils + defective coils = 88 + 12 = 100

p(getting two good coils for two selection) = p( getting 2 good coils for first selection ) $\times$ p(getting 2 good coils for second selection)

p(first selection) = p(second selection) = $\frac{\text { number of good coils }}{\text { total number of coils }}$

Hence, p(getting 2 good coil for two selection) = $\frac{88}{100} \times \frac{88}{100} =\bold{0.7744}$

5 0

3 years ago

Plzz help me Solve -24 = 6x. Show your work.

ZanzabumX [31]

Answer:

x = -4

Step-by-step explanation:

-24 = 6x

________

Switch sides:

6x = -24

________

Divide both sides by 6:

6x/6 = -24/6

________

Simplify:

X = -4

3 0

3 years ago