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den301095 [7]
3 years ago
5

PLEASEEEE HELPPP MEEE!!!! (15 PTS!)

Mathematics
1 answer:
vazorg [7]3 years ago
8 0

Answer:

332324

Step-by-step explanation:

b4hbf3hkrbf;h3bfh3bfc;khjbr3fkhbn;vb 4;khvcj3nc

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Drone INC. owns four 3D printers (D1, D2, D3, D4) that print all their Drone parts. Sometimes errors in printing occur. We know
USPshnik [31]

Answer:

Step-by-step explanation:

Hello!

There are 4 3D printers available to print drone parts, then be "Di" the event that the printer i printed the drone part (∀ i= 1,2,3,4), and the probability of a randomly selected par being print by one of them is:

D1 ⇒ P(D1)= 0.15

D2 ⇒ P(D2)= 0.25

D3 ⇒ P(D3)= 0.40

D4 ⇒ P(D4)= 0.20

Additionally, there is a chance that these printers will print defective parts. Be "Ei" represent the error rate of each print (∀ i= 1,2,3,4) then:

P(E1)= 0.01

P(E2)= 0.02

P(E3)= 0.01

P(E4)= 0.02

Ei is then the event that "the piece was printed by Di" and "the piece is defective".

You need to determine the probability of randomly selecting a defective part printed by each one of these printers, i.e. you need to find the probability of the part being printed by the i printer given that is defective, symbolically: P(DiIE)

Where "E" represents the event "the piece is defective" and its probability represents the total error rate of the production line:

P(E)= P(E1)+P(E2)+P(E3)+P(E4)= 0.01+0.02+0.01+0.02= 0.06

This is a conditional probability and you can calculate it as:

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

To reach the asked probability, first, you need to calculate the probability of the intersection between the two events, that is, the probability of the piece being printed by the Di printer and being defective Ei.

P(D1∩E)= P(E1)= 0.01

P(D2∩E)= P(E2)= 0.02

P(D3∩E)= P(E3)= 0.01

P(D4∩E)= P(E4)= 0.02

Now you can calculate the probability of the piece bein printed by each printer given that it is defective:

P(D1/E)= \frac{P(E1)}{P(E)} = \frac{0.01}{0.06}= 0.17

P(D2/E)= \frac{P(E2)}{P(E)} = \frac{0.02}{0.06}= 0.33

P(D3/E)= \frac{P(E3)}{P(E)} = \frac{0.01}{0.06}= 0.17

P(D4/E)= \frac{P(E4)}{P(E)} = \frac{0.02}{0.06}= 0.33

P(D2)= 0.25 and P(D2/E)= 0.33 ⇒ The prior probability of D2 is smaller than the posterior probability.

The fact that P(D2) ≠ P(D2/E) means that both events are nor independent and the occurrence of the piece bein defective modifies the probability of it being printed by the second printer (D2)

I hope this helps!

8 0
3 years ago
Three planes that intersect in one line
guapka [62]
They are collinear points and are called planes. 
5 0
3 years ago
What angle corresponds to angle AFG?
viva [34]

Look at the image below :)

In case you didn't know, corresponding angles are equal to each other and are on the same side of the traversal

Therefore the answer is angle CGH

Hope this helped!  

7 0
3 years ago
Read 2 more answers
24<br> 23<br> 22<br> 21十<br> 20-<br> 4<br> 8<br> 12<br> 16
DochEvi [55]

Answer:

counted down

Step-by-step explanation:

it was counting up and went down by couting down in 2

7 0
2 years ago
Which term best describes the graph of yy = x - 1
dimaraw [331]

Answer:

C

Step-by-step explanation:

Its C because its x (a variable) and is costanly changing

4 0
3 years ago
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