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Reil [10]
1 year ago
7

Part 1: Kevin works at an Internet Café. The café charges customers based on the amount of time they use on the Internet. There’

s a flat fee of three dollars for all users. In addition, users are charged .20 cents for each 10 minute. They use the Internet.
Part 2: A customer at the café is charged $4.60 for her time on the Internet. How many hours was she on the Internet? Show your work.
Mathematics
1 answer:
tia_tia [17]1 year ago
8 0

Answer:

46mins

EXPLANATION

20 cents=10mins

customers working hrs=$4.60*10

$4.60*10=46 minutes

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3 years ago
As part of their business promotional package, the Milwaukee Chamber of Commerce would like an estimate of the mean cost per mon
Morgarella [4.7K]

Answer:

a) 884-2.426\frac{50}{\sqrt{40}}=864.82    

884+2.426\frac{50}{\sqrt{40}}=903.179    

So on this case the 98% confidence interval would be given by (864.82;903.179)

b) For this case the upper bound of the confidence interval is lower than 950 so we don't have a significant evidence to conclude that the true mean is 950 per month

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=884 represent the sample mean

\mu population mean (variable of interest)

s=50 represent the sample standard deviation

n=40 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 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=40-1=39

Since the Confidence is 0.98 or 98%, the value of \alpha=0.02 and \alpha/2 =0.01, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,39)".And we see that t_{\alpha/2}=2.426

Now we have everything in order to replace into formula (1):

884-2.426\frac{50}{\sqrt{40}}=864.82    

884+2.426\frac{50}{\sqrt{40}}=903.179    

So on this case the 98% confidence interval would be given by (864.82;903.179)

Part b

For this case the upper bound of the confidence interval is lower than 950 so we don't have a significant evidence to conclude that the true mean is 950 per month

3 0
3 years ago
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