
Benjamin G. answered 02/11/16
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Notice that the number of knots is increasing by the same constant amount, one, from line to line, but that the rope length decreases by an amount that varies: 10, 10, 13, 12, 11, 9. So this is not an exact line.
The formula for slope is m = Δy / Δx = (y2 - y1) / (x2 - x1), and since in our problem, successive lines have Δx=1, the Δy values are the slopes. Since these are not constant, we don't have a single line through the data. (This would be what's called a piecewise linear function, a connected graph of line segments of varying slope; but that doesn't have a single equation for the whole domain.)
Are you supposed to use least squares to fit a line? Is this a statistics class? Plotting, you might be able to eyeball a line. If the same pattern of rope length decreasing by 10 for each unit increase in number of knots, you could use the equation y = 100 - 10x, for example, where x=# of knots and y=rope length in cm. But the actual "best fit" line will differ since the rope lengths are not 100, 90, 80, 70, 60, 50, 40 (all decreasing by 10).
If you were taking a statistics class (a big if!!), and were trying to calculate a least squares regression line, I would point you to a resource such as [1] to fit the line (find the formula with least sum-of-squares predictive error) using the free statistics software R (see [2]):
y = 100.71 - 11.14 x
This line will always return an approximate answer, and not go through the data points, but rather, will have the least amount of "total" error if we sum up this total in a specific way (by taking the squares of the difference between the formula for y and the observed values for y at each corresponding value of x in the table). Along with this model of the data, there is a (linear) correlation coefficient of -0.9990153, which, because it is so close to 1 in absolute value, shows that the data is very nearly, but not quite perfectly, linear.
[1] http://www.cyclismo.org/tutorial/R/linearLeastSquares.html
[2] http://www.r-project.org