.2(2.50+4.95+6.00) how to solve with out distribution

Mathematics

1 answer:

nataly862011 [7]3 years ago

7 0

Add up inside the brackets and then multiply by 2

13.45x2 is 26.9

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A company uses three different assembly lines- A1, A2, and A3- to manufacture a particular component. Of thosemanufactured by li

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.

6 0

2 years ago

A family is driving 600 miles. Assuming they stop to get gas 3 times and eat 2 times, how many hours will it take them to reach

Umnica [9.8K]

Lets assume that they are driving at an average of 70mph, and getting gas takes 10 minutes, and stopping to eat takes 30.

t=d/s

t=600/70
t=8.57

Stopping to eat twice is 30 x 2 = 60 (another hour)

and Gas is 10 x 3 = 30 minutes (0.5 hours)

Total time is,

8.57+1+0.5 =10.07

So the trip will take approx 10.7 hours total.

(This is using my assumptions, from my experience!)

8 0

3 years ago

I need to know how to do these asap!!

ludmilkaskok [199]

Answer:

Step-by-step explanation:

y =4/3 x+4

slope intercept form

y = mx+b

the y intercept is 4 (0,4)

and the slope is 4/3 which is change in y over change in x

we go up 4 to the right 3