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soldier1979 [14.2K]
3 years ago
12

PLEASE HELP NOW!! If K is the set of all letters in the word “equation”, and L is the set of all letters of the word “inequality

”, list all elements of the following sets
K∩L and K∪L
Mathematics
1 answer:
satela [25.4K]3 years ago
6 0

Answer:

e,q,u,a,t,i,n & EQUATIONYL

Step-by-step explanation:

K∩L:

First write both the words, equation and inequality down.

E Q U A T I O N

I N E Q U A L I T Y

You can see EQUATIN repeats in both of the words.

K∪L:

First write both the words, equation and inequality down.

E Q U A T I O N

I N E Q U A L I T Y

This time just write down all the letters EXCEPT the ones that repeat. EQUATIONYL

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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

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For this problem,

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

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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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