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omeli [17]
3 years ago
8

A fuel tank contains 1 3/4 gallons of gasoline. Casey adds 33 1/3 gallons of

Mathematics
1 answer:
romanna [79]3 years ago
5 0

Answer:

The gallons of gas the tank contain now is <u>35 1/12</u>.

Step-by-step explanation:

Given:

Fuel tank contains 1 3/4 gallons of gasoline.

Casey adds 33 1/3 gallons of  gasoline to the tank.

Now, to find the total gallons of gas the tank contain.

Converting the mixed fractions into improper.

Gallons of gasoline in fuel tank =1\frac{3}{4}=\frac{7}{4}.

Gallons of gasoline Casey adds =33\frac{1}{3} =\frac{100}{3}.

<u>According to question:</u>

By adding we get the total gallons of gasoline now tank contain:

\frac{7}{4} +\frac{100}{3}

=\frac{7\times 3+100\times 4}{12}

=\frac{21+400}{12}

=\frac{421}{12}

=35\frac{1}{12}

Therefore, the gallons of gas the tank contain now is 35 1/12.

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Step-by-step explanation:

Incomplete question:

There is no point to complete the equation.

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Which equation is true when m = 4?
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In a random sample of 80 teenagers, the average number of texts handled in a day is 50. The 96% confidence interval for the mean
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Answer:

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And the margin of error is given by:

ME= \frac{54-46}{2}= 4

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And as we can see that the margin of error would be lower than the original value of 4, the margin of error would be reduced by a factor \sqrt{2}

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)  

\sigma represent the population standard deviation  

n=80 represent the sample size  

Solution to the problem

Part a

The confidence interval for the mean is given by the following formula:  

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)  

For this case we can calculate the mean like this:

\bar X =\frac{46+54}{2}=50

And the margin of error is given by:

ME= \frac{54-46}{2}= 4

The confidence level is 0.96 and the significance level is \alpha=1-0.96=0.04 and the value of \alpha/2 =0.02 and the margin of error is given by:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

We can calculate the critical value and we got:

z_{\alpha/2} = 2.05

And if we solve for the deviation like this:

\sigma = ME * \frac{\sqrt{n}}{z_{\alpha/2}}

And replacing we got:

\sigma =4 *\frac{\sqrt{80}}{2.05} =17.45

Part b

For this case is the sample size is doubled the margin of error would be:

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And as we can see that the margin of error would be lower than the original value of 4, the margin of error would be reduced by a factor \sqrt{2}

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Answer:

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