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nekit [7.7K]
3 years ago
14

Ruben put an empty cup underneath a leaking faucet. After 1 1/2 hours, Ruben had collected 1/4 cup of water. What is the rate, i

n cups per hour, at which the water is leaking from the faucet? A 1/6 B 3/8 C 8/3 D 6/1
Mathematics
1 answer:
worty [1.4K]3 years ago
5 0

Answer:

The rate is \frac{1}{6} cups per hour

Step-by-step explanation:

It took the faucet 1 1/2 hours, i.e 1.5 or 3/2 hours to fill 1/4 cup by leaking

We need to find the rate in terms of cups that can be filled by water in 1 hour.

Using unitary method:

If it takes \frac{3}{2} h for \frac{1}{4} cup;

then it will take 1 h for how many cups?

= \frac{1 \times \frac{1}{4} }{\frac{3}{2} }

= \frac{2}{4 \times 3}

= \frac{2}{12}

= \frac{1}{6} cups

Therefore, the rate is \frac{1}{6} cups per hour

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There are 8 students on the school chess team. They need to raise at least $760 for a field trip. Their advisor says they still
frosja888 [35]

Answer:

Step-by-step explanation:

well first lets get 85% of 760 since thats how much tehy still need to raise

85% of 760 is 646

there are 8 kids so what we need to do is 646/8

646/8 is 80.75

I dont even know what diego did

5 0
2 years ago
Which fraction is equivalent to -3/7<br><br> 3/-7<br> -3/-7<br> -7/3<br> 7/3
Aneli [31]
3/-7 is the correct answer
8 0
3 years ago
If ( 2 x 1) (0 x -1) = ( 15 ), then the value of x is
adoni [48]

Answer: x = -15/2

Step-by-step explanation:

This answer IS simplified

7 0
3 years ago
A. A batch of 30 parts contains five defects. If two parts are drawn randomly one at a time without replacement, what is the pro
Zarrin [17]

Answer:

a) For this case on the first trial we have the following probability of selecting a defective part: 5/30. Because we have 5 in total defective at the begin and a total of 30 parts

For the second trial since the experiment is without replacement we have 29 parts left, and since we select 1 defective from the first trial we have in total 4 defective left, so then the probability of defective for the second trial is : 4/29

And then we can assume independence between the events and we have this probability:

(5/30)*(4/29) =0.16667*0.1379= 0.0230

b) For this case on the first trial we have the following probability of selecting a defective part: 5/30. Because we have 5 in total defective at the begin and a total of 30 parts

And since we replace the part selected is the same probability for the second trial and then the final probability assuming independence would be:

(5/30)*(5/30) =0.1667* 0.1667= 0.0278

Step-by-step explanation:

For this case we know that we have a batch of 30 parts with 5 defective.

Part a

If two parts are drawn randomly one at a time without replacement, what is the probability that both parts are defective?

For this case on the first trial we have the following probability of selecting a defective part: 5/30. Because we have 5 in total defective at the begin and a total of 30 parts

For the second trial since the experiment is without replacement we have 29 parts left, and since we select 1 defective from the first trial we have in total 4 defective left, so then the probability of defective for the second trial is : 4/29

And then we can assume independence between the events and we have this probability:

(5/30)*(4/29) =0.16667*0.1379= 0.0230

Part b

If this experiment is repeated, with replacement, what is the probability that both parts are defective?

For this case on the first trial we have the following probability of selecting a defective part: 5/30. Because we have 5 in total defective at the begin and a total of 30 parts

And since we replace the part selected is the same probability for the second trial and then the final probability assuming independence would be:

(5/30)*(5/30) =0.1667* 0.1667= 0.0278

3 0
3 years ago
Someone plz help this is due today
-Dominant- [34]

3x+5x+1=17

x=2

y=5•2+1

y=1

(x, y)=(2, 11)

3x+2(4x+5)=43

x=3

y=4•3+5

y=17

(x, y)=(3, 17)

I think I did it right and I hope I helped you

3 0
2 years ago
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