In a class of 29 students, 20 have a brother and 8 have a sister. There are 3 students who do not have any siblings. What is the probability that a student chosen randomly from the class is an only child?

1/4 = 0.25 = 25%

Step-by-step explanation:

Let x be the number of students having both brother and sister.

Brother only is 9-x

Sister only is 16-x

Brother only + sister only + both + neither = 28

(9-x) + (16-x) + x + 7 = 28

32 - x = 28

x = 4

The probability per the diagram is 4/16 = 1/4

Per definition of conditional probability:

prob( brother | sister) = prob ( brother AND sister) / prob ( sister)

            = 4/16 = 1/4

1/4 = 0.25 = 25%

-22

times both sides by 2, times () by 8, add alile terms, move terms, add alike and caculate divide both sides by 3

b. would be the answer : )

step-by-step explanation:

step 1: add -16x to both sides.

16x−4y+−16x=64+−16x  

−4y=−16x+64

step 2: divide both sides by -4.

−4y /-4 = -16x + 64/-4

answer:   y=4x−16

hope i'm right