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AlladinOne [14]
3 years ago
13

Find the mileage for a car that travels 576 miles on 18 gallons of gas.

Mathematics
2 answers:
Reika [66]3 years ago
8 0

32 miles per gallon of gasoline.

Just divide the miles by gallons of gas.

lyudmila [28]3 years ago
5 0

Miles per Gallon means you have to divide, so in this case it would be

576/18=32

32 miles

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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
Nastasia [14]

Answer:

a) \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

b) ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.05 *\frac{17.45}{\sqrt{160}}=2.828

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:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.05 *\frac{17.45}{\sqrt{160}}=2.828

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}

5 0
2 years ago
ray uses a fourth of a gallon of gasoline to drive a distance of 5 1/2 miles at this rate how many miles can you drive on one ga
KIM [24]
On average you get like about 12 to 14 miles a gallon
4 0
3 years ago
Read 2 more answers
Plz help will mark brainliest
Kryger [21]

∆ABC=∆DEF

AB=DE –>String

BC=EF–> Rib

m<C=m<F=90° –>List

AC=DF

So m<A=m<D=35 (It is not clear whether the number is 35 or 36, but the same number)

8 0
2 years ago
If 60% of the registered voters cast 270,000 votes in an election how many registered voters are there. In another election 70%
quester [9]
60% of 270,000 is 162,000. 70% of 315,000 is 220,500. There is 58,500 more at the second election than the first.
3 0
3 years ago
Pls help ASAP ILL GIVE BRAINLYEST
docker41 [41]
Xavier

he charges $10 per hour

good luck :)
6 0
3 years ago
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