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The graph shows the number of cakes sold at Karen's Cake Shoppe for each of their 7 weeks in business: A scatter plot is shown with the title Karens Cake Shoppe. The x axis is labeled Weeks in Business, and the y axis is labeled cakes sold. The data points are located at 1 and 2, 2 and 4, 3 and 5, 4 and 4, 5 and 6, 6 and 5, and 7 and 8. A line of best fit passes through the y axis at 1 and through the point 10 and 10. If her current pattern continues, how many cakes will Karen most likely sell in her 10th week of business?

10, because approximately y = 9 over 10.x + 1
11, because approximately y = 9 over 10.x + 1
8, because approximately y = 1x − 1
12, because approximately y = 1x + 2

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prosborough198p2ub41
Hello! 

So given your problem here we need to find the slope value and that will tell us everything we need. 

slope = y2-y1/x2-x1  so using both end points given to get most accurate value we have;

slope= (10-1)/10-0)= 9/10

so our rise over run value or slope is 9/10. so this gives us the equation;

y=(9/10)x+1

therefore we can see the most likely number of pies sold in her 10th week is;

y=9/10(10)+1 = 9+1 = 10

so the answer is 10, because approximately y=(9/10)x + 1

Hope this helps! Any questions please ask! Thank you so much!!


lhmarianateixeira

In this exercise we have to deal with the graphs and their corresponding function, so we find that:

[tex]y=(9/10)x + 1[/tex]

First, we have that the slope value is given by:

[tex]y_2-y_1/x_2-x_1=\\(10-1)/10-0)= 9/10[/tex]

so our rise over run value or slope is 9/10. so this gives us the equation;

y=(9/10)x+1

Knowing that the number of pieces sold was 10, then:

[tex]x=10\\y=9/10(10)+1 = 9+1 = 10[/tex]

See more about slope at brainly.com/question/3605446

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