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alexdok [17]
3 years ago
13

I need to know this I don’t understand and it is last one

Mathematics
1 answer:
Len [333]3 years ago
4 0

Answer:

The answer is 4x

Step-by-step explanation:

Hope this helped

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What is the equation of the midline for the function f(x)? f(x)=1/2 cos(x)+5
ivolga24 [154]

Answer:

y = 5

Step-by-step explanation:

That "+5" elevates the graph of this cosine function 5 units above the x axis, and thus the "midline" you wanted is  y = 5.

6 0
2 years ago
If I read a whole dictionary would you donate 1$ to see it?
Monica [59]

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I<em>t D</em>e<em>p</em>en<em>d</em>s

Step-by-step explanation:

7 0
2 years ago
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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
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
Solve for x<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%7D%7B2%7D%20%20%2B%20%20%5Cfrac%7B3%7D%7B4%7D%20%20%3D%2010" id=
Degger [83]

Answer:

x = 18.5

Step-by-step explanation:

Given

\frac{x}{2} + \frac{3}{4} = 10

Multiply through by 4 to clear the fractions

2x + 3 = 40 ( subtract 3 from both sides )

2x = 37 ( divide both sides by 2 )

x = \frac{37}{2} = 18.5

6 0
3 years ago
Read 2 more answers
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