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Mathematics 13 Online
OpenStudy (anonymous):

Part 1: Create a scatter plot with the predicted line of best fit drawn on it. Determine the type of correlation (if any), and predict the model that will be used. Part 2: Find the line of best fit for the data either by hand or using technology. Explain your method. Find the predicted score for each time listed in the table. Part 3: Find the residuals, and decide if your model is a good fit. Explain your method. (If your model is not a good fit, complete Part 2 again with a different set of points or choose a different model.)

OpenStudy (kirbykirby):

Using technology will be easiest especially if you have a lot of data points..

OpenStudy (anonymous):

i did number 1 i need a lot of help with 2

OpenStudy (kirbykirby):

Did you do it using technology or by hand?

OpenStudy (anonymous):

tech

OpenStudy (kirbykirby):

There should be a function integrated into your software that can find the regression line... um what software are you using?

OpenStudy (anonymous):

windows 7

OpenStudy (kirbykirby):

Um that's the operating system :) I mean more specifically, which program did you use to do the scatter plot ?

OpenStudy (kirbykirby):

I don't see your plot

OpenStudy (kirbykirby):

Well regardless... you could actually use Wolfram Alpha to find the regression line http://www.wolframalpha.com/input/?i=regression+line%7B%7B4%2C+10%7D%7B5%2C+13%7D%7D If you change the input command to all the (x,y) pairs you have in this format: regression line{{x1, y1}, {x2, y2}, ..., {xn, yn}} it should give it to you like in that small example in the link

OpenStudy (anonymous):

how would you find it out with multiple points

OpenStudy (kirbykirby):

just add more pairs {x, y} Like say you have this data: age (x): 16, 20, 45 eye vision (y): 1.0 2.25 4.2 Then you can write: regression line{{16, 1.0}, {20, 2.25}, {45, 4.2}}

OpenStudy (kirbykirby):

if you need more points, keep adding \(, \{ ~ , ~\}\)

OpenStudy (anonymous):

thanks for the help

OpenStudy (kirbykirby):

:) For the predicted score... you will just need to plug in the "x" value into the fitted line equation, and the y-value that comes out will be the predicted value

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