Suppose two different methods are available for eye surgery. The probability that the eye has not recovered in a month is 0.002 if method A is used. When method B is used, the probability that the eye has not recovered in a month is 0.005. Assume that 40% of eye surgeries are done with method A and 60% are done with method B in a certain hospital. If an eye is recovered in a month after surgery is done in the hospital, what is the probability that method A was performed (write it up to second decimal place)? a) 0.5 b) 0.6 c) 0.8 d) 0.4 e) 0.7

d) 0.4

Step-by-step explanation:

P(Ra) = 0.998

P(Rb) = 0.995

P(A) = 0.4

P(B) = 0.6

The probability that method A was performed, given that the eye was recovered within a month of the surgery, is determined by the probability that the eye recovered within a month with method A (P(A) x P(Ra)),divided by the probability that the eye recovered within a month with either method (P(A) x P(Ra) + P(B) x P(Rb)):

The probability that method A was performed is 0.40.

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

try soting them out first

step-by-step explanation: