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Katen [24]
3 years ago
8

Suppose the linear regression line y = 2.1x + 130 predicts sales based on the money spent on advertising. If x represents the do

llars spent in advertising, and y represents the company sales in dollars, about how much can the company expect in sales if it spends $50 in advertising?
Mathematics
1 answer:
hoa [83]3 years ago
4 0
We have here a simple linear regression since it involves 1 dependent variable and 1 independent variable. Linear regression is used to present the relationship between variables. In the problem, given the equation y = 2.1x + 130, we are looking for the amount  the company expect in sales if it spends $50 in advertising? We will have a direct substitution on the given equation using x = $50.The answer will be $235  - it is the expected sales at $50 spent on advertising .
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Elenna [48]

Since he started with Nothing

He had zero initially

After x Months...

He'll have 10x naxvips

in slope intercept form

y=10x

To get the y values ... substitute the corresponding x value

For x=0

y=10(0)

y=0. Which is true cos he had no naxvip initially

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y=10(1)

=10. Which is true cos he buys 10 more monthly... So 10 + 0 = 10.

x=2

y=10(2) = 20. This is true cos he added an additional 10 after each month

x=7

y=70.

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-2,-1,0,1 whichave pair of numbers has a sum of zero
Verizon [17]
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How do you write 53% as a fraction in lowest terms
Vaselesa [24]
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-4.7 divided by 1\4 ????????????????
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In an examination of purchasing patterns of shoppers, a sample of 16 shoppers revealed that they spent, on average, $54 per hour
Lesechka [4]

Answer:

54-1.64\frac{21}{\sqrt{16}}=45.39    

54 +1.64\frac{21}{\sqrt{16}}=62.61    

So on this case the 90% confidence interval would be given by (45.39;62.61)    

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=54 represent the sample mean  

\mu population mean (variable of interest)

\sigma=21 represent the sample standard deviation

n=16 represent the sample size  

Solution to the problem

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

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

Since the Confidence is 0.90 or 90%, the value of \alpha=1-0.9=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: "=-NORM.INV(0.05,0,1)".And we see that z_{\alpha/2}=1.64

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

54-1.64\frac{21}{\sqrt{16}}=45.39    

54 +1.64\frac{21}{\sqrt{16}}=62.61    

So on this case the 90% confidence interval would be given by (45.39;62.61)    

4 0
3 years ago
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