8t=28 how to solve it
2 answers:
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Answer:
The average is exactly 26,375 gallons of gas per hour.
How I found this is buy subtracting each numbers like 422,000 - 369,250 = 52,750. 369,250 - 316,500 = 52, 750 and so on. They all ended up being 52,750, but then I realized that they are all 2 hours apart and the question asks of gallons of gas per 1 hour so then I divided 52,750 in half and got my final answer, 26,375.
Step-by-step explanation:
This is how much gas the jet uses per 1 hour
my question doesn't ask me to create a linear model but it asks for gasoline per hour, and it's just like this question with the same graph so if you have the same as me this is the answer.
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:
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(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!
mark must leave it for 5.5 months or 5 and half moths to gain 5600 in interest .
<u>Step-by-step explanation:</u>
Here we have , mark invests 8000 in an account that pays 12% interest and 2000 in one that pays 8%. if he leaves the money in the accounts for the same length of time, We need to find how long must he leave it to gain 5600 in interest . Let's find out:
Let mark invests 8000 in an account that pays 12% interest and 2000 in one that pays 8% for time x months , So total interest gain is 5600 i.e.
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Therefore , mark must leave it for 5.5 months or 5 and half moths to gain 5600 in interest .