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erik [133]
4 years ago
7

I don't get this at all​

Mathematics
1 answer:
dezoksy [38]4 years ago
6 0

Answer:

-5=n

Step-by-step explanation:

75=3(-6n-5)

Distribute the 3

75=-18n-15

Add 15 to both sides

90=-18n

Divide both sides by -18 to get n alone

-5=n

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Answer:

The probability that a randomly selected component needs rework when it came from line A₁ is 0.3623.

Step-by-step explanation:

The three different assembly lines are: A₁, A₂ and A₃.

Denote <em>R</em> as the event that a component needs rework.

It is given that:

P (R|A_{1})=0.05\\P (R|A_{2})=0.08\\P (R|A_{3})=0.10\\P (A_{1})=0.50\\P (A_{2})=0.30\\P (A_{3})=0.20

Compute the probability that a randomly selected component needs rework as follows:

P(R)=P(R|A_{1})P(A_{1})+P(R|A_{2})P(A_{2})+P(R|A_{3})P(A_{3})\\=(0.05\times0.50)+(0.08\times0.30)+(0.10\times0.20)\\=0.069

Compute the probability that a randomly selected component needs rework when it came from line A₁ as follows:

P (A_{1}|R)=\frac{P(R|A_{1})P(A_{1})}{P(R)}=\frac{0.05\times0.50}{0.069}  =0.3623

Thus, the probability that a randomly selected component needs rework when it came from line A₁ is 0.3623.

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