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Nat2105 [25]
3 years ago
9

Solve the system by the substitution method 8x-3y=-22 and y=5x+19

Mathematics
1 answer:
Travka [436]3 years ago
6 0

The answer is: x=-5\\y=-6

The explanation is shown below:

1. In substitution method you can solve one of the equation for one of the variables, then you must substitute it into the other equation and solve it to calculate the other variable. Once you do this, you must substitute the value obtained into one of the original equations to calculate the second variable.

2. Keeping this on mind, you can see that the second equation is already solved for y. Therefore, you can substitute this into the first one and solve for x:

8x-3y=-22\\8x-3(5x+19)=-22\\8x-15x-57=-22\\-7x=35\\x=-5

3. Let's substitute the value obtained into the second equation to calculate y:

y=5x+19\\y=5(-5)+19\\y=-6

4. The result is:

x=-5\\y=-6

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In a certain assembly plant, three machines B1, B2, and B3, make 30%, 20%, and 50%, respectively. It is known from past experien
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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

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%.

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Billy is hiking in Colorado. He walks eastward four miles, then turns $60$ degrees northward and walks six miles. How far is he
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He is  2√19 miles far way from starting point.

Suppose Billy starts at point A, turns at point B, and ends at point D, as shown below.

If Billy turns 60◦ northward and walks six miles, then we can draw a 30 − 60 − 90 triangle whose hypotenuse is 6 miles

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