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3 years ago

H(x)=-4x+3;when h(x)=11

1 answer:

3 0

To solve this, since you know that h(x) is 11 and that it also equals -4x+3, you set them equal to one another which would look like this: 11=-4x+3.
Then, to solve for x, which is what I am assuming the question is asking, you would subtract 3 from both sides to isolate -4x, which would result in this:
-4x=8
Now, to solve for x, divide both sides by -4, and you get your answer which is x=-2

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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

An online shopping club has 10,000 members when it charges $7 per month for membership. For each $1 monthly

Sonja [21]

Answer:

The function that can be used in the online shopping club about its monthly revenue is:

$R = -400x^{2} +7200x+70000$

Step-by-step explanation:

First, we're gonna take into account the different values we have in the exercise:

10,000 members
$7 per month for membership
Loses of 400 members by each $1 monthly increase

How the variable $x$ represents the price increase, we can do the formula below:

$(10000 - 400x) * (7+x)$

In this formula, we represent in the first part that by each 1 in the variable $x$ , the total of members will be reduced in 400, in the second part, we mention that at the same time, the membership fee will be increased in the same value of $x$ . Now we must simplify this function: