The points given are (1,1), (2,6), (3,4), (4,0), (5,1), and (6,5). We can use linear regression to find the line of best fit, or we can eyeball it. The points (2,6), (3,4), (4,0), and (6,5) seem to form a rough line. Using the points (4,0) and (6,5), we can find the slope: $m = \frac{5-0}{6-4} = \frac{5}{2}$. Using the point-slope form, we have $y - 0 = \frac{5}{2}(x-4)$, which simplifies to $y = \frac{5}{2}x - 10$. This line doesn't quite fit the data, so let's try eyeballing a line that follows the general trend of the points. It appears that a slightly less steep slope would fit the data better, perhaps passing through (1,1) and (6,5). The slope of this line is $m = \frac{5-1}{6-1} = \frac{4}{5}$. Using the point-slope form, we have $y - 1 = \frac{4}{5}(x-1)$, which simplifies to $y = \frac{4}{5}x + \frac{1}{5}$. This line seems to fit the data better visually.
The correlation coefficient, denoted by $r$, measures the strength and direction of a linear relationship between two variables. We can use a calculator or software to calculate the correlation coefficient for the given set of points. Using the points (1,1), (2,6), (3,4), (4,0), (5,1), and (6,5), we get $r \approx 0.3273$.
- $y = \frac{4}{5}x + \frac{1}{5}$
- $r \approx 0.3273$
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