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2 years ago

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PLEASE HELP A barrel in Jim's yard contains 60 gallons of water. Water leaks out of the barrel at a rate of 1 gallon every 10 mi

nutes.
Create and graph the solution set of the equation for the gallons of water, y, remaining in the barrel in terms of the number of minutes elapsed, x.

1 answer:

Murrr4er [49]2 years ago

3 0

Answer:

$y=-0.10x+60$

Step-by-step explanation:

Let

y -----> gallons of water remaining in the barrel

x-----> number of minutes elapsed

we know that

Water leaks out of the barrel at a rate of 1 gallon every 10 minutes

so

$1/10=0.10\frac{gal}{min}$

The linear equation that represent this situation is

$y=-0.10x+60$

The graph in the attached figure

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Answer: The answer is $\textup{x}=\dfrac{2}{3}\textup{h}-5.$

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So, for that, let 'x' be the number of hours she babysat this week. Then, according to the question, we can write

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Also, it is given that

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2 years ago

The results of a mathematics placement exam at two different campuses of Mercy College follow: Campus Sample Size Sample Mean Po

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Answer:

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$p_v =P(Z>3.37)=1-P(Z$

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Step-by-step explanation:

Data given

Campus Sample size Mean Population deviation

1 330 33 8

2 310 31 7

$\bar X_{1}=33$ represent the mean for sample 1

$\bar X_{2}=31$ represent the mean for sample 2

$\sigma_{1}=8$ represent the population standard deviation for 1

$\sigma_{2}=7$ represent the population standard deviation for 2

$n_{1}=330$ sample size for the group 1

$n_{2}=310$ sample size for the group 2

$\alpha$ Significance level provided

z would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for Campus 1 is higher than the mean for Campus 2, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}\leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

We have the population standard deviation's, and the sample sizes are large enough we can apply a z test to compare means, and the statistic is given by:

$z=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

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

$z=\frac{(33-31)-0}{\sqrt{\frac{8^2}{330}+\frac{7^2}{310}}}}=3.37$

P value

Since is a one right tailed test the p value would be:

$p_v =P(Z>3.37)=1-P(Z$

Comparing the p value with a significance level for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the mean for the Campus 1 is significantly higher than the mean for the group 2.

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Answer:

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And for the margin of error we have the following estimation:

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Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

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$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

$\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75$

And for the margin of error we have the following estimation:

$ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62$

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