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Vinvika [58]
2 years ago
12

A rental car company charges $77. 50 per day to rent a car and $0. 14 for every mile

Mathematics
1 answer:
muminat2 years ago
6 0

77.5d + .14m

m = 425

.14 * 425 = 59.5

77.5d + 59.5 ≤ 230

77.5d ≤ 170.5

d ≤ 170.5/77.5

d ≤ 2.2

if they only take payment for whole days, he has 2 whole days

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50=40%
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Equation of a line in point slope form passing through (8,2) and (4,5)
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3 0
3 years ago
10 points for this!!
MrMuchimi
This is the answer for your problem

7 0
2 years ago
The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in
Stolb23 [73]

Answer:

ME = 1.653 *\frac{7.9}{\sqrt{300}}= \pm 0.75

±0.75 ounce

Step-by-step explanation:

Assuming this complete question: The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in that state. To do this, they have selected a random sample of 300 people from the designated population. The following results were recorded: xbar=34.5 ounces, s=7.9 ounces Given this information, if the leaders wish to estimate the mean milk consumption with 90 percent confidence, what is the approximate margin of error in the estimate?

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=34.5 represent the sample mean

\mu population mean (variable of interest)

s=7.9 represent the sample standard deviation

n=300 represent the sample size  

Solution to the problem

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

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

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=300-1=299

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,199)".And we see that t_{\alpha/2}=1.653

And the margin of error is given by:

ME = 1.653 *\frac{7.9}{\sqrt{300}}= \pm 0.75

±0.75 ounce

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