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Thepotemich [5.8K]
3 years ago
8

A standard fire hose is 50 feet long. What is the length of the hose in yards and feet?

Mathematics
2 answers:
Tpy6a [65]3 years ago
3 0
50 feet = 16.67 yards
Elodia [21]3 years ago
3 0
16.7 rounded if you round the number to a tens place I think if I’m not mistaken
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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
I really need help with this guys please help it’s due tomorrow and I’m so frikin dumb
il63 [147K]

Possible 24 impossible 3

5 0
2 years ago
Solve 5c + 4 = -26
mr_godi [17]
<em>1) 5 c + 4 = - 26
5 c = -26 -4
5 c = -30
c = -30 / 5
c = -6    so correct option is B..

2) 3 x - x +2 = 12
2 x +2 = 12
2x = 12-2
2x = 10
x = 10/2
x= 5    so correct option is D

3 ) 3 ( x + 1 )+ 6 = 33
3x + 3 + 6 = 33
3x + 9 = 33
3x = 33-9
3x = 24
x = 24/3
x = 8   so correct option is B

4) y/-6=9
y=9 x -6
y= - 54   there is no such option i guess question is missing

5)(x +  4) /2 = 7
x +4 = 7 x 2
x + 4 = 14
x = 14-4
x = 10  so correct option is D

6)1/3 ( 2x - 8) = 4
2x/ 3 - 8 /3 = 4
2x - 8 / 3 = 4
2x - 8 = 4 x 3
2x - 8 = 12
2x = 12 + 8
2x = 20
x = 20/2
x = 10   so correct option is C


</em>
6 0
3 years ago
A gene carries the______ for a trait.
MrRissso [65]

Answer:

A gene carries the <u>information</u> for a trait.

Step-by-step explanation:

Just got it right on edge

8 0
3 years ago
The kelso building costs 84.5 million to build, the jones building costs 33.8 million
Dominik [7]
Kelso / jones = 84.5/33.8 = 5/2 = 2 1/2
==> c
7 0
3 years ago
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