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natka813 [3]
3 years ago
7

Given the midpoint and one endpoint of a line segment, find the other endpoint.

Mathematics
1 answer:
neonofarm [45]3 years ago
3 0

Answer:

(-12 , 2)

Step-by-step explanation:

<u>GIVEN :-</u>

  • Co-ordinates of one endpoint = (-4 , -10)
  • Co-ordinates of the midpoint = (-8 , -4)

<u>TO FIND :-</u>

  • Co-ordinates of another endpoint.

<u>FACTS TO KNOW BEFORE SOLVING :-</u>

<em><u>Section Formula :-</u></em>

Let AB  be a line segment where co-ordinates of A = (x¹ , y¹) and co-ordinates of B = (x² , y²). Let P be the midpoint of AB . So , by using section formula , the co-ordinates of P =

(x , y) = (\frac{x^2 + x^1}{2} ,\frac{y^2 + y^1}{2} )

<u>PROCEDURE :-</u>

Let the co-ordinates of another endpoint be (x , y)

So ,

(-8 , -4) = (\frac{-4 + x}{2} , \frac{-10 + y}{2} )

First , lets solve for x.

=> \frac{x - 4}{2} = -8

=> x - 4 = -8 \times 2 = -16

=> x = -16 + 4 = -12

Now , lets solve for y.

=> \frac{y - 10}{2} = -4

=> y - 10 = -4 \times 2 = -8

=> y = -8 + 10 = 2

∴ The co-ordinates of another endpoint = (-12 , 2)

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

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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)}

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For this problem,

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

P(B2) = Probability of machine B2 = 0.2

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

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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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