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scoundrel [369]
3 years ago
15

A rental car cost $25 plus a fixed charge per mile driven. The total charge for 210 miles of use was $67. Write an equation for

the cost, C (in dollars), in terms of the miles driven, x.
Mathematics
1 answer:
blagie [28]3 years ago
4 0
To start, we know the basic equation from what us given:

c = 25 + mx
And we want to find m. Fortunately, we have some numbers to plug in to find m. We can make an equation like this
67 = (210)m + 25
by taking the amount charged and the amount of miles driven to get that charge given in the original problem.

Now we solve.
67 - 25 = 210m + 25 - 25 \\ 42 = 210m \\  \frac{42}{210}  =  \frac{210}{210} m \\  \frac{1}{5}  = m
Now we can put that in our equation to finish it.
c =  \frac{1}{5} x + 25
And that is the final equation
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Step-by-step explanation:

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Suppose you know the length of a confidence interval of a population mean is 8.4 and the sample mean (x bar) is 10. Find the ​ma
SOVA2 [1]

Answer:

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

The lenght of the interval correspond to:

8.4 = 2ME

ME= \frac{8.4}{2}= 4.2

And since we know the margin of error we can find the limits for the confidence interval:

Lower = 10 -4.2=5.8

Upper = 10 +4.2=14.2

Step-by-step explanation:

Previous concepts

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".  

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".

\bar X=10 represent the sample mean  

\mu population mean (variable of interest)  

\sigma represent the population standard deviation  

n represent the sample size  

Solution to the problem

Assuming the X follows a normal distribution  

X \sim N(\mu, \sigma)

The sample mean \bar X is distributed on this way:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})  

The confidence interval on this case is given by:

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

The margin of error is given by:

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

The lenght of the interval correspond to:

8.4 = 2ME

ME= \frac{8.4}{2}= 4.2

And since we know the margin of error we can find the limits for the confidence interval:

Lower = 10 -4.2=5.8

Upper = 10 +4.2=14.2

4 0
3 years ago
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