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expeople1 [14]
3 years ago
9

Find the value of x. Will give brainliest.

Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
3 0
Since a line is equal to 180 degree
you pick one that has the variable x in it.
For example:  5x=180
180/5=36
therefore x=36
2. 6x=180
180/6= 30
x=30

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Which of the following graphs represents a function?
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Step-by-step explanation:

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3 years ago
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Solve the following inequality.<br> -4(1 - 5a) &gt; -124
NemiM [27]

Answer:

a>−6

Step-by-step explanation:

−4(1−5a)>−124

Step 1: Simplify both sides of the inequality.

20a−4>−124

Step 2: Add 4 to both sides.

20a−4+4>−124+4

20a>−120

Step 3: Divide both sides by 20.

20a/20> −120/20

a>−6

8 0
3 years ago
Some adults and children went to a movie.
aalyn [17]
4 adults and 8 children
3 0
3 years ago
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Guys I need help on all three please​
kodGreya [7K]

Answer:

graph 1 proportional

graph 2&3 is non proportional

i think

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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