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KengaRu [80]
4 years ago
13

A dripping faucet loses about 2.5×10−42.5×10^−4 liters of water each minute. There are approximately 3×1053×10^5 minutes in a mo

nth. About how many liters of water are lost because of the dripping faucet in a month? Drag and drop the values to the boxes to express the answer in scientific notation.
Mathematics
2 answers:
Inessa [10]4 years ago
5 0
For this case we can start from the definition of the flow.
 Flow rate is defined as the quotient between volume and time.
 So, we have
 Q = V / t
 By clearing the volume, we have:
 V = Q * t
 Substituting the values.
 V = ((2.5 * 10 ^ -42.5) * (10 ^ -4)) * ((3 * 10 ^ 53) * (10 ^ 5))
 V = 2.37E + 12
 answer
 Approximately 2.37E + 12 liters of water are lost due to the dripping faucet every month.
777dan777 [17]4 years ago
3 0

Answer:

the answer is 7.5 times 10 to the power of 1

Step-by-step explanation:

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The amount of coffee that a filling machine puts into an 8 dash ounce 8-ounce jar is normally distributed with a mean of 8.2 oun
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Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theore.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

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Z = \frac{X - \mu}{\sigma}

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Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 8.2, \sigma = 0.18, n = 100, s = \frac{0.18}{\sqrt{100}} = 0.018

What is the probability that the sampling error made in estimating the mean amount of coffee for all 8 dash ounce 8-ounce jars by the mean of a random sample of 100​ jars, will be at most 0.02​ ounce

That is, probability of the sample mean between 8.2-0.02 = 8.18 and 8.2 + 0.02 = 8.22, which is the pvalue of Z when X = 8.22 subtracted by the pvalue of Z when X = 8.18.

X = 8.22

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{8.22 - 8.2}{0.018}

Z = 1.11

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X = 8.18

Z = \frac{X - \mu}{s}

Z = \frac{8.18 - 8.2}{0.018}

Z = -1.11

Z = -1.11 has a pvalue of 0.1335.

0.8665 - 0.1335 = 0.7330

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