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saw5 [17]
3 years ago
9

NOTE: I FINISHED THE FIRST PART, I JUS NEED HELP W THE SECOND PART BELOW!!!

Mathematics
1 answer:
bagirrra123 [75]3 years ago
4 0

Answer:

  see below

Step-by-step explanation:

<em>Which of the equations from part A represent adding two rational numbers?</em>

  Equations A, C, E

<em>What hypothesis can you make about the sum of two rational numbers?</em>

  The sum of two rationals will always be rational

<em>Will the addition result in a rational or an irrational number?</em>

  Our hypothesis is that the result is always rational. This can be justified by the fact that the sum of two rationals a/b + c/d, where a, b, c, d are integers and bd≠0, is (ad+bc)/(bd), a rational, based on closure of integers for multiplication and addition.

<em>Which equations represent the sum of a rational and an irrational number?</em>

  Equations B, F

<em>What hypothesis can you make about the sum of an irrational and a rational number?</em>

  The sum of a rational and irrational number is always irrational.

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Answer:

The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.

Step-by-step explanation:

Using Bayes' Theorem

P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|a)P(a)}

where

P(B|A) is probability of event B given event A

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and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.

For this problem,

Let P(B1) = Probability of machine B1 = 0.3

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P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006

P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010

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P(N|B1) be probability of a non-defective product produced by machine 1 = 1 - P(D|B1) = 1 - 0.003 = 0.997

P(N|B2) be probability of a non-defective product produced by machine 2  = 1 - P(D|B2) = 1 - 0.006 = 0.994

P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990

For the probability of a finished product produced by machine B1 given it's non-defective; represented by P(B1|N)

P(B1|N) =\frac{P(N|B1)P(B1)}{P(N|B1)P(B1) + P(N|B2)P(B2) + (P(N|B3)P(B3)} = \frac{(0.297)(0.3)}{(0.297)(0.3) + (0.994)(0.2) + (0.990)(0.5)} = 0.1138

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Step-by-step explanation:

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