A manufacturing firm employs three analytical plans for the design and development of a particular product. For cost reasons, all three are used at varying times. In fact, plans 1, 2, and 3 are used for 30%, 20%, and 50% of the products, respectively. The defect rate is different for the three procedures as follows: P(D|P1) = 0.01, P(D|P2) = 0.03, P(D|P3) = 0.02,
where P(D|Pj) is the probability of a defective product, given plan j. If a random product was observed and found to be defective, which plan was most likely used and thus responsible?

it's d.

step-by-step explanation:

because: -1(4x + 1y) +2 (1x-3y) distribute each number into the parenthesis.

then you would get: -4x - 1y + 2x -6y combine like terms

and get: 2x -7y

hope my answer has you if not then i'm sorry.

i can not answer the question due to the fact that you spelled "carton" as "cartoon", a cartoon is something you watch on tv, a carton is a box that holds things. and also you've been waiting for this question for a while so why don't you get off your lazy a** and try to figure it out for yourself for once!

step-by-step explanation:

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look above, and leave your comments down below, and don't for get a five stars and a and brainliest xd

sorry for the stars in the middle of words its because it wouldn't let me submit. xd