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Juliette [100K]
3 years ago
11

A large container holds 8 gallons of water. It begins leaking at a constant rate. After 48 minutes, the container has 5 gallons

of water left.
Part 1 out of 2
At what rate is the water leaking? Express your answer as a positive value.

The water is leaking at the rate of gal/min.
Mathematics
1 answer:
Juliette [100K]3 years ago
5 0

Answer:

<em>The water is leaking at the rate of 0.0625 gal/min.</em>

Step-by-step explanation:

<u>Rate of change</u>

The rate of change (ROC) is a measure that compares two quantities, usually to know how fast one variable changes in time.

The container holds 8 gallons of water and is leaking at a constant rate to be calculated later.

When 48 minutes have passed, the container has 5 gallons of water left. This means it has leaked 8 gallons - 5 gallons = 3 gallons.

Calculating the rate at which it's leaking:

\displaystyle r=\frac{3\ gallons}{48\ min}

r = 0.0625 gal/min

The water is leaking at the rate of 0.0625 gal/min.

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The mean consumption of bottled water by a person in the United States is 28.5 gallons per year. You believe that a person consu
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Answer:

t=\frac{27.8-28.5}{\frac{4.1}{\sqrt{100}}}=-1.707    

p_v =P(t_{99}

If we compare the p value with a significance level for example \alpha=0.1 we see that p_v so we can conclude that we reject the null hypothesis, so there is not enough evidence to conclude that the mean for the consumption is less than 28.5 gallons at 0.1 of significance, so we can reject the claim that person consume more than 28.5 gallons.

Step-by-step explanation:

Data given and notation    

\bar X=27.8 represent the mean for the account balances of a credit company

s=4.1 represent the population standard deviation for the sample    

n=1000 sample size    

\mu_o =28.5 represent the value that we want to test  

\alpha=0.1 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)    

p_v represent the p value for the test (variable of interest)

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to determine if the mean for the person consume is more than 28.5 gallons, the system of hypothesis would be:    

Null hypothesis:\mu \geq 28.5    

Alternative hypothesis:\mu < 28.5    

We don't know the population deviation, so for this case we can use the t test to compare the actual mean to the reference value, and the statistic is given by:    

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic    

We can replace in formula (1) the info given like this:    

t=\frac{27.8-28.5}{\frac{4.1}{\sqrt{100}}}=-1.707    

Calculate the P-value    

First we need to calculate the degrees of freedom given by:

df=n-1=100-1=99

Since is a one-side lower test the p value would be:    

p_v =P(t_{99}

In Excel we can use the following formula to find the p value "=T.DIST(-1.707,99)"  

Conclusion    

If we compare the p value with a significance level for example \alpha=0.1 we see that p_v so we can conclude that we reject the null hypothesis, so there is not enough evidence to conclude that the mean for the consumption is less than 28.5 gallons at 0.1 of significance, so we can reject the claim that person consume more than 28.5 gallons.

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<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E%7B%5Cfrac%7B5%7D%7B6%7D%20%7D%20%7D%7Bx%5E%7B%5Cfrac%7B1%7D%7B6%7D%20%7D%20%7D"
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simplify 4/6, you get 2/3.

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