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larisa [96]
3 years ago
12

1/3y+1=1/2t-3 what is y?

Mathematics
2 answers:
gizmo_the_mogwai [7]3 years ago
7 0
1/3y+1=1/2t-3
1/3y=1/2t-4
y=3/2t-12
Oduvanchick [21]3 years ago
7 0
I see it as 1/3y = 1/2t - 3 - 1

1/3y = 1/2t - 4
y = 1/2t - 4 / 1/3 
    
I got stuck there

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hammer [34]

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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