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Komok [63]
3 years ago
7

An aquarium tank can hold 7200 liters of water. there are two pipes that can be used to fill the tank. the first pipe alone can

fill the tank in 48 minutes. the second pipe can fill the tank in 96 minutes by itself. when both pipes are working together, how long does it take them to fill the tank?
Mathematics
1 answer:
Arada [10]3 years ago
4 0

The first pipe pumps out

\dfrac{7200\text{ L}}{48\text{ min}}=\dfrac{150\text{ L}}{1\text{ min}}

while the second pipe pumps out

\dfrac{7200\text{ L}}{96\text{ min}}=\dfrac{75\text{ L}}{1\text{ min}}

So together, both pipes pump out

\dfrac{150+75\text{ L}}{1\text{ min}}=\dfrac{225\text{ L}}{1\text{ min}}=\dfrac{7200\text{ L}}{32\text{ min}}

That is, with both pipes filling the tank, it should take 32 minutes total.

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Step-by-step explanation:

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At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population pro
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Answer:

A sample of 1068 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

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In which

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The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population proportion?

We need a sample of n.

n is found when M = 0.03.

We have no prior estimate of \pi, so we use the worst case scenario, which is \pi = 0.5

Then

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.03 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.03\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.03}

(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.03})^{2}

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A sample of 1068 is needed.

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