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OlgaM077 [116]
3 years ago
5

For his long distance phone service, Keith pays a $3 monthly fee plus 11 cents per minute. Last month, Keith's long distance bil

l was $16.42. For how many minutes was Keith billed?​
Mathematics
1 answer:
LenaWriter [7]3 years ago
3 0

Answer:

122 Minutes

Step-by-step explanation:

Start with $16.42

subtract $3 for the monthly fee, 16.42-3= $13.42

he pays $0.11 per minute so that's 13.42 x (1min/0.11) = 122minutes

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Answer:

C and E

Step-by-step explanation:

The variable "y" has to be by itself so here is the breakdown to that:

-3x + 6y + 5 = - 7

    6y + 6 = 3x - 7

       6y = 3x - 7 - 6

    6y = 3x - 12

         6y/6 = 3x/6 - 12/6

y = 1/2 x - 2

Then you would graph -2 --> (0, -2) and use 1/2 to navigate upwards to get your answers.

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What's the average rate of change?<br> f(x) over 4&lt;=x&lt;=7
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Step-by-step explanation:

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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

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Step-by-step explanation:

7 0
4 years ago
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