What is the greatest common denominator

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

Simplify the expression

ollegr [7]

Answer:

Step-by-step explanation:

$\dfrac{x^2-8x+12}{x^2-7x+10}\\\\x^2-7x+10\ne0\ \iff\ x=\frac{7\pm\sqrt{49-40}}{2}\ne0\ \iff\ x\ne5\ \wedge\ x\ne2\\\\\\\dfrac{x^2-8x+12}{x^2-7x+10}=\dfrac{x^2-2x-6x+12}{x^2-2x-5x+10}=\dfrac{x(x-2)-6(x-2)}{x(x-2)-5(x-2)}=\\\\\\ =\dfrac{(x-2)(x-6)}{(x-2)(x-5)}=\dfrac{x-6}{x-5}\\\\\\f(x)=x-6\\\\g(x)=x-5$

4 0

3 years ago

Brad had a small gathering at a local steakhouse. The steakhouse offers three dinner platters which vary by size and price. They

marissa [1.9K]

Answer: 4($9.95)+2($12.95)+($15.95)=115

Step-by-step explanation:

7 0

3 years ago

What is the area of the real bedroom?

stiks02 [169]

Answer:

area of real bedroom = 32 cm²

hope it helps..

6 0

3 years ago

Read 2 more answers

Quality control finds on average that 0.026% of the items from a certain factory are detective. One month, 4000 items are checke

telo118 [61]

Given that 4000 items are checked in one month, let the number of detective items be represented as y.

Quality control finds on average that 0.026% of the items in the factory are detective. This implies that

$\text{Number of }\det ective\text{ items = }\frac{\text{0.026}}{100}\times\text{Total number of items checked}$

When 4000 items are checked, we have the number of detective items to be evaluated as

$\begin{gathered} \text{Number of }\det ective\text{ items = }\frac{\text{0.026}}{100}\times\text{Total number of items checked} \\ y=\frac{0.026}{100}\times4000 \\ \Rightarrow y=1.04 \end{gathered}$

Hence, the number of detective items is 1.04.

7 0

1 year ago