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3 years ago

what is the 14 letter word with the definition of an example showing a statement is not true. it is c--n-e-t--m-l-

Mathematics

1 answer:

Zinaida [17]3 years ago

6 0

The word is counterexample.
Your second t is incorrect.
It should be an e.

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What is the volume of this rectangular prism?

Aleksandr-060686 [28]

Answer:

Step-by-step explanation:

4 times 5/2 times 1/2 is 5

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4 years ago

Which answer is the equation of the line represented in function notation?

blagie [28]

Answer: f(x)=3/2×-2

Step-by-step explanation:

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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

I really really need help with this please.

11111nata11111 [884]

The number of ancestors going back through the 5th generation, including Tle-nle and counting Tle-nle as the 1st generation is:

= 1 + 3 + 3^2 + 3^3 + 3^4
= (3^5 - 1) / (3 - 1)
= 242 / 2
= 121

Since we included Tle-nle as the 1st generation, we will only compute up to the 4th power. If it is until the 6th generation, add 3^5 to the equation.

8 0

3 years ago

Anyone know how to do this?!!

iren2701 [21]

Answer:

slope is -1/4

Step-by-step explanation:

$\frac{-4}{12}$ = $\frac{-1}{4}$