Write an expression to match the statement 1/5 of the difference between 18 and 11

A box contains 10 cards numbered 1 through 10. Which events are mutually exclusive? A. A card is drawn from the box and the numb

coldgirl [10]

The answer is A. 5 is not an even number, so both events can't take place together.

6 0

3 years ago

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

25. The perimeter of a rectangle is equal to twice the sum of its length and its width. One

AlexFokin [52]

Answer:

The equation that can be used to find the width is 2(13 + w) = 42

The width of the rectangle is 8 inches

Step-by-step explanation:

The formula of the perimeter of a rectangle is P = 2(l + w), where l is its length and w is its width

∵ One rectangle has a length of 13 inches

∴ l = 13 inches

∵ Its perimeter is 42 inches

∴ P = 42 inches

- Substitute the values of l and P in the formula of the perimeter

∵ 42 = 2(13 + w)

The equation that can be used to find the width is 2(13 + w) = 42

To find W divide both sides by 2

∴ 21 = 13 + w

- Subtract 13 from both sides

∴ 8 = w

The width of the rectangle is 8 inches

7 0

2 years ago

Find the values of the variables.

Leokris [45]

Answer:

x=3 & y=162

Step-by-step explanation:

it may help you

6 0

3 years ago

Please solve this <br>with each step

grandymaker [24]

Step-by-step explanation:

please mark me as brainlest

6 0

2 years ago