Copy or type the data above into Excel. Select the data and then under the Insert tab, in the graph and chart section, insert a scatterplot. By right-clicking on any of the points you select "Add trendline". In that menu that pops up on the right, select "Display equation on graph" and "R-squared value". Now you will see both on your graph. Use that equation to put your answer in below. \[ y=\square x+\square \] \begin{tabular}{|c|r|} \hline \( \mathbf{x} \) & \multicolumn{1}{|c|}{\( \mathbf{y} \)} \\ \hline 2 & 13.01 \\ \hline 3 & 12.08 \\ \hline 4 & 11.55 \\ \hline 5 & 9.02 \\ \hline 6 & 9.49 \\ \hline 7 & 6.86 \\ \hline 8 & 4.83 \\ \hline 9 & 5.9 \\ \hline \end{tabular}
Added by Joel W.
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Step 1
Label the first column as "x" and the second column as "y". Enter the values as follows: - x: 2, 3, 4, 5, 6, 7, 8, 9 - y: 13.01, 12.08, 11.55, 9.02, 9.49, 6.86, 4.83, 5.9 Show more…
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(a) Draw a scatter plot. (b) Select two points from the scatter plot, and find an equation of the line containing the points selected. (c) Graph the line found in part (b) on the scatter plot. (d) Use a graphing utility to find the line of best fit. (e) What is the correlation coefficient $r$ ? (f) Use a graphing utility to draw the scatter plot and graph the line of best fit on it. $$ \begin{array}{|c|ccccccc|} \hline x & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ y & 4 & 6 & 7 & 10 & 12 & 14 & 16 \\ \hline \end{array} $$
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Copy and paste the following data into Excel. Create a scatterplot of the data and use the scatterplot to determine if the data would best be represented by a linear, quadratic, or exponential trendline. Then use Excel to find the equation of the appropriate trendline. x,y 15,-909.24 14,-894.48 18,-175.56 20,293.25 17,-385.56 12,-1,097.67 8,-1,062.48 12,-1,109.4 7,-992.48 Equation of trendline:
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