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Goryan [66]
3 years ago
5

HELP Please!!! I need this done quick.

Mathematics
2 answers:
Crazy boy [7]3 years ago
7 0
Hey ok so for A the answer is 1.14 and for b it’s 0.9 hope this helps
matrenka [14]3 years ago
6 0
What form do you need it in?
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Use the given data to find a regression line that best fits the price-demand data for price p in dollars as a function of the de
Rufina [12.5K]

Answer:

m=-\frac{7600}{8250}=-0.921

Nowe we can find the means for x and y like this:

\bar x= \frac{\sum x_i}{n}=\frac{550}{10}=55

\bar y= \frac{\sum y_i}{n}=\frac{1042}{10}=104.2

And we can find the intercept using this:

b=\bar y -m \bar x=104.2-(-0.921*55)=104.707

So the line would be given by:

y=-0.921 x +104.707

Step-by-step explanation:

For this case we have the following data given:

Demand (x): 10,20,30,40,50,60,70,80,90,100

Price (y): 141 , 133,126, 128,113,97, 90, 82,79,53

We want to construct a linear model like this:

y = mx +b

For this case we need to calculate the slope with the following formula:

m=\frac{S_{xy}}{S_{xx}}

Where:

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}

So we can find the sums like this:

\sum_{i=1}^n x_i =550

\sum_{i=1}^n y_i =1042

\sum_{i=1}^n x^2_i =38500

\sum_{i=1}^n y^2_i =115882

\sum_{i=1}^n x_i y_i =49710

With these we can find the sums:

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=38500-\frac{550^2}{10}=8250

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}=49710-\frac{550*1042}{10}=-7600

And the slope would be:

m=-\frac{7600}{8250}=-0.921

Nowe we can find the means for x and y like this:

\bar x= \frac{\sum x_i}{n}=\frac{550}{10}=55

\bar y= \frac{\sum y_i}{n}=\frac{1042}{10}=104.2

And we can find the intercept using this:

b=\bar y -m \bar x=104.2-(-0.921*55)=104.707

So the line would be given by:

p(x)=-0.921 x +104.707

4 0
3 years ago
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