Arandomly selected sample of 100 horse owners found that 72 of them feed two flakes of grass hay in the morning and one flake of alfalfa plus one flake of grass hay in the evening to their horses while the rest feed two flakes of grass hay in the morning and one flake of alfalfa plus oat hay in the evening. the estimated probability that horse owners feed grass hay in the a. m. and alfalfa plus oat hay in the p. m. is:

a. 0.72.
b. 0.50.
c. 0.28.
d. 0.75.

The correct answer is B. 0.5

Step-by-step explanation:

Total number of horse owners =100

72 of the horse owners feed their horses in the morning with

•2 flakes of grass hay

•1 flake of alfalfa

72 of the horse owners feed their horses in the evening with

28 of the horse owners feed their horses in the morning with

•2 flakes of grass hay  

•1 flake of alfalfa

28 of the horse owners feed their horses in the evening with

There are two categories of farmer

P(72) =72/100 =0.72

P(28) = 28/100 = 0.28

P(grass hay in A.M for 72 farmers) = 1/2*72/100  =0.36

P (grass hay in A.M for 28 farmers) =1/2*28/100 = 0.14

P (grass hay in A.M for both farmers) = 0.36 + 0.14 = 0.5

P(alfalfa plus oat hay in P.M for 72 farmers) =0

P(alfalfa plus oat hay in P.M for 28 farmers) =0

P(alfalfa plus oat hay in P.M for both farmers) = 0+ 0 = 0

The estimated probability that horse owners feed grass hay in the AM and alfalfa plus oat hay in the PM is

P (E) = P (grass hay in AM for both farmers) + P(alfalfa plus oat hay in PM for both farmers)

  P(E)    =0.5 + 0 = 0.5

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )