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antiseptic1488 [7]
3 years ago
10

Two hoses of different sizes are used to fill a pool. The smaller hose can fill the pool in 1.5 times as long as the larger hose

. If both hoses
are used it takes 9 hours to fill the pool. If only the larger hose is used, how many hours will it take to fill the pool?
Mathematics
1 answer:
Rzqust [24]3 years ago
7 0

Answer:

  15

Step-by-step explanation:

Let x represent the number of hours it takes the larger hose to fill the pool. Then each hour, it contributes 1/x of the pool volume. Similarly, the other hose contributes a volume of 1/(1.5x) each hour.

The two hoses together contribute a volume of 1/9 of the pool volume each hour:

  1/x +1/(1.5x) = 1/9

  2.5/(1.5x) = 1/9 . . . . combine the fractions

  22.5/1.5 = x . . . . . . multiply by 9x

  15 = x

It takes the larger hose 15 hours to fill the pool alone.

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Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean \mu and standard deviation \sigma, we have these following probabilities

Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827

Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545

Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So \mu = 53, \sigma = 11

So:

Pr(53-11 \leq X \leq 53+11) = 0.6827

Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545

Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973

-----------

Pr(42 \leq X \leq 64) = 0.6827

Pr(31 \leq X \leq 75) = 0.9545

Pr(20 \leq X \leq 86) = 0.9973

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have Pr(31 \leq X \leq 75) = 0.9545 = 0.9545 subtracted by Pr(42 \leq X \leq 64) = 0.6827 is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.

To find just the percentage above the mean, we divide this value by 2

So:

P = {0.9545 - 0.6827}{2} = 0.1359

The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.

4 0
4 years ago
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