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mrs_skeptik [129]
3 years ago
7

Pump 1 can empty a pool in 8 hours while Pump 2 can empty the same pool in 9 hours. If the two pumps worked together, how long w

ould it take them to empty the pool?
Mathematics
1 answer:
yan [13]3 years ago
7 0
<span>Rate of pump A: 1/8 of a pool per hour 
Rate of pump B: 1/9 of a pool per hour 
Combined rate: 1/8+1/9 = 17/72 +1/9 = 25/72 

So if they work together, the two pumps have a combined rate of 25/72 of a pool per hour (i.e in one hour, the two pumps will empty 25/72 of the pool) 

</span><span>But we want to empty ONE pool (not 25/72 of one). So we need to multiply 25/72 by some number x to get 1.
 </span>
<span>Now solve for x 
x=2.88
</span><span>It will take the two pumps 2.88 hours to empty the pool.
2 hours 52 minutes 50 seconds</span>
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A tobacco company claims that the amount of nicotene in its cigarettes is a random variable with mean 2.2 and standard deviation
Aleksandr-060686 [28]

Answer:

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

Step-by-step explanation:

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

Normal probability distribution

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

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem

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

In this problem, we have that:

\mu = 2.2, \sigma = 0.3, n = 100, s = \frac{0.3}{\sqrt{100}} = 0.03

What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?

This is 1 subtracted by the pvalue of Z when X = 3.1. So

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

By the Central Limit Theorem

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

Z = \frac{3.1 - 2.2}{0.03}

Z = 30

Z = 30 has a pvalue of 1.

1 - 1 = 0

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

4 0
4 years ago
In ΔABC, if the lengths of sides a and c are 8 centimeters and 16 centimeters, respectively, and the measure of is 35°, what is
Wittaler [7]
Sin(A)/a= sin(C)/c, therefore sin(A) would = asin(C)/c= sin(35deg)/2. So, A then = arc sin(sin(35deg)/2)= 16.67 
7 0
3 years ago
Read 2 more answers
Can someone help me with this question, please. The problem is attached.
fomenos

It is a reflection across the y axis, then a translation to the right 2 units.  


I hope this helps!  If you have more you want help with I can help.  

3 0
3 years ago
A survey of 25 randomly selected customers found the ages shown (in years). The mean is 32.64 years and the standard deviation i
____ [38]

Answer:

See below for answers and explanations

Step-by-step explanation:

<u>Part A</u>

Assuming that conditions have been met for the interval, we use the formula \displaystyle CI=\bar{x}\pm t\frac{s}{\sqrt{n}} where \bar{x} represents the sample mean, t represents the critical value, s represents the sample standard deviation, and n is the sample size.

The critical value of t for an 80% confidence level with degrees of freedom df=n-1=25-1=24 is equivalent to t=1.317836

Thus, we can compute the confidence interval:

\displaystyle CI=\bar{x}\pm t\frac{s}{\sqrt{n}}\\\\CI=32.64\pm1.317836\biggr(\frac{9.39}{\sqrt{25}}\biggr)\\\\CI\approx\{30.17,35.11\}

Therefore, we are 80% confident that the true mean age of all customers is between 30.17 and 35.11 years.

<u>Part B</u>

The margin of error is \displaystyle t\frac{s}{\sqrt{n}}=1.317836\biggr(\frac{9.39}{\sqrt{25}}\biggr)\approx2.47

6 0
2 years ago
A local hamburger shop sold a combined total of 813 hamburgers and cheeseburgers on Saturday. There were 63 more cheeseburger so
Naya [18.7K]

9514 1404 393

Answer:

  375 hamburgers

Step-by-step explanation:

Let h represent the number of hamburgers. Then h+63 is the number of cheeseburgers, and the combined total is ...

  h +(h +63) = 813

  2h = 813 -63 = 750

  h = 750/2 = 375

375 hamburgers were sold on Saturday.

_____

<em>Additional comment</em>

In a "sum and difference" problem like this, the smaller number is half the difference between the sum and the difference. h = (813 -63)/2 = 375. This generic solution applies to all "sum and difference" problems.

8 0
3 years ago
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