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EleoNora [17]
3 years ago
9

Find the y-intercept and x-intercept of the line. x-2y=-8

Mathematics
1 answer:
zimovet [89]3 years ago
6 0

Answer:

x-intercept = 4, y-intercept = -8

Step-by-step explanation:

To find the x-intercept, we have to set y equal to 0 and solve for x,

x - 2 (0) = -8

x = -8

So, the x-intercept is at (-8, 0).

The y of the x-intercept is always at 0 because it should be on the x-axis.

To find the y-intercept, we just have to set x equal to 0 and solve for y.

(0) - 2y = -8

-2y = -8

y = 4

So, the y-intercept is at (0, 4).

The x of the y-intercept is always at 0 because it should be on the y-axis.

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Y = -3(2) +22
Y = -6 +22
Y = 16
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If 4 times the sum of a number and 3 is six less than twice the number, what is the number?
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Answer:

-9

Step-by-step explanation:

4(n + 3) = 2n - 6

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2 years ago
In a certain assembly plant, three machines B1, B2, and B3, make 30%, 20%, and 50%, respectively. It is known from past experien
diamong [38]

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

P(B|a) is probability of event B not given event A  

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

P(B2) = Probability of machine B2 = 0.2

P(B3) = Probability of machine B3 = 0.5

Let P(D) = Probability of a defective product

P(N) = Probability of a Non-defective product

P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003

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

Likewise,

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

Hence the probability that a non-defective product is produced by machine B1 is 11.38%.

4 0
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Step-by-step explanation:

Edge, let me know if you got it right.

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2 years ago
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