Question 1190382
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Data table:
<table border = "1" cellpadding = "5"><tr><td>x</td><td>0</td><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>8</td><td>9</td><td>10</td></tr><tr><td>y</td><td>1.7</td><td>1.7</td><td>2.3</td><td>2.9</td><td>3</td><td>3.5</td><td>4</td><td>11</td><td>9.5</td><td>9.7</td><td>9.9</td></tr></table>which is the same as writing<table border = "1" cellpadding = "5"><tr><td>x</td><td>y</td></tr><tr><td>0</td><td>1.7</td></tr><tr><td>1</td><td>1.7</td></tr><tr><td>2</td><td>2.3</td></tr><tr><td>3</td><td>2.9</td></tr><tr><td>4</td><td>3</td></tr><tr><td>5</td><td>3.5</td></tr><tr><td>6</td><td>4</td></tr><tr><td>7</td><td>11</td></tr><tr><td>8</td><td>9.5</td></tr><tr><td>9</td><td>9.7</td></tr><tr><td>10</td><td>9.9</td></tr></table>


The linear regression equation is roughly 
y = 1.0164x + 0.3


The correlation coefficient is roughly
r = 0.8962


I used the GeoGebra calculator to find the regression equation and correlation coefficient value. I strongly recommend using a calculator rather than do it by hand. 
If your teacher requires you to use a formula, then please let me know so I can change my response. 


Since r is fairly close to 1, I'd consider this to be strong positive correlation. 
A linear model seems to be a good fit.
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