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tensa zangetsu [6.8K]
2 years ago
9

Which set of ordered pairs represents y as a function of x?

Mathematics
2 answers:
ella [17]2 years ago
6 0

Answer:

b

Step-by-step explanation:

Aloiza [94]2 years ago
4 0
B












(0,0)(1,1)(1'0)(2,1)
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Answer: the slope is 2/3
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The Ross family had 100 relatives at their big Thanksgiving dinner of all the guest nine he had French heritage 80 had English h
il63 [147K]
Hi, Deedee. I am quite confused with the numbers that you've given because they don't add up to 100. Nevertheless, if by 'this', you mean the Thanksgiving culture, then you would just count the English and the Native American heritage. This is because the first Thanksgiving dinner was shared between the English colonists and the Native American tribes.

I hope I was able to help you in a way. Have a good day.
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2 years ago
Find the x-intercept and the y-intercept of the line above
Liono4ka [1.6K]

Answer:

(a): x-intercept: -4

(b): y-intercept: -1

Step-by-step explanation:

5 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
Need help quick!! <br> WILL GIVE BRAINLIEST
Murljashka [212]

Answer:

3 and 4 x=25, y=29

Step-by-step explanation:

If you need a step by step explanation, I will edit this later.

5 0
3 years ago
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