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aleksklad [387]
3 years ago
15

Find the value of y that makes the lines parallel. there is a photo ​

Mathematics
1 answer:
Eduardwww [97]3 years ago
5 0

Answer:

Y=20

Step-by-step explanation:

90= (4y+10)

90-10=4y

80=4y

Y=80/4= 20

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Using the graph , determine the coordinates of the vertex of the parabola
aivan3 [116]

Answer:

(-4,-4)

Step-by-step explanation:

Delta math

3 0
2 years ago
Signal mistakenly produced 1,000 defective cell phones. the phones cost $60 each to produce. a salvage company will buy the defe
zhannawk [14.2K]
We are asked to solve for the incremental net income from reworking the phones.
Number of items = 1000
Production Cost = 1000*$60 = $60,000
Salvage value = 1000*$30 = $30,000
Rework Cost = 1000 *80 = $80,000
Price when resold = 1000 * $120 = $120,000

Incremental Net = $120,000 + $30,000 - $60,000 -$80,000
Incremental Net = $10,000

The answer is $10,000. 

4 0
3 years ago
Read 2 more answers
-8(1+8n)-8(6-4n)=-24
Naily [24]

-8(1 + 8n) - 8(6 - 4n) = -24

-8 - 64n - 48 + 32n = -24

-32n - 56 = -24

-32n = 32

n = -1

3 0
3 years ago
What is the answer to the problem i need help with?
AveGali [126]

Answer:

C

Step-by-step explanation:

Recall the equation for a circle:

(x-h)^2+(y-k)^2=r^2, where the center is (h,k) and the radius is r.

The given equation is:

(x+5)^2+(y+7)^2=21^2

Another way to write this is:

(x-(-5))^2+(y-(-7))^2=21^2

Thus, we can see that h=-5 and k=-7.

The center is at (-5, -7).

6 0
3 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
3 years ago
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