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Natasha_Volkova [10]
2 years ago
6

Is the graphs (both) misleading?

Mathematics
1 answer:
Alecsey [184]2 years ago
6 0

Answer:

B is misleading

Step-by-step explanation:

by making the Y numbers smaller it looks like there was a "big" increase" when theere really wasnt

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konstantin123 [22]
The answer is -31 2/5<span />
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3 years ago
A certain firm has plants A, B, and C producing respectively 35%, 15%, and 50% of the total output. The probabilities of a non-d
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Answer:

There is a 44.12% probability that the defective product came from C.

Step-by-step explanation:

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

P = \frac{P(B).P(A/B)}{P(A)}

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

-In your problem, we have:

P(A) is the probability of the customer receiving a defective product. For this probability, we have:

P(A) = P_{1} + P_{2} + P_{3}

In which P_{1} is the probability that the defective product was chosen from plant A(we have to consider the probability of plant A being chosen). So:

P_{1} = 0.35*0.25 = 0.0875

P_{2} is the probability that the defective product was chosen from plant B(we have to consider the probability of plant B being chosen). So:

P_{2} = 0.15*0.05 = 0.0075

P_{3} is the probability that the defective product was chosen from plant B(we have to consider the probability of plant B being chosen). So:

P_{3} = 0.50*0.15 = 0.075

So

P(A) = 0.0875 + 0.0075 + 0.075 = 0.17

P(B) is the probability the product chosen being C, that is 50% = 0.5.

P(A/B) is the probability of the product being defective, knowing that the plant chosen was C. So P(A/B) = 0.15.

So, the probability that the defective piece came from C is:

P = \frac{0.5*0.15}{0.17} = 0.4412

There is a 44.12% probability that the defective product came from C.

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