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2 years ago

Help #3??? I really need help!

Mathematics

2 answers:

Over [174]2 years ago

6 0

Answer:

It’s A

Step-by-step explanation:

QveST [7]2 years ago

4 0

Answer:

The answer is C

Step-by-step explanation:

When you expand the 2 binomials, you get

(-5x + 3)(6x - 2) you get the following 4 terms.

-30x^2 + 18x + 10x + 6 Combine the middle 2 terms

-30x^2 + 28x + 6

The answer is either A or C. I can't tell you why B and D are there. I don't know why anyone would choose - 15x^2 but that is not correct.

So you now have to find out which is true for the domain. There are no values that this quadratic can't have for its domain, so A isn't true. The answer is C

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PLEASE HELP THIS IS DUE TODAY!!!!!!

Mnenie [13.5K]

Answer:

Step-by-step explanation:

A does not work, because you stop on the way to the meeting (which you don't)

B does not work, because you speed up on the way to the meeting, when in reality you slow down

C does not work, because it does not show that you went back home or changed speed at all

D is correct!

5 0

2 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

Any help on this please ?

dlinn [17]

Answer:

(0,-5), I believe.

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

What is the vertex of this parabola? y= 2x² + 4x +5