Suppose the line of best fit is being found for some data points that have an r-value of 0.793. If the standard deviation of the x-coordinates is 5.591, and the standard deviation of the y-coordinates is 2.772, what is the slope of the line to three decimal places?

Answer: 0.393

Explanation: If the given quantities are r-value (or Pearson's coefficient) and the standard deviations of x and y coordinates, we'll use the following formula for slope:

[tex]m = \frac{rs_y}{s_x} [/tex]

where:
m = slope of the best fit line
r = r-value = 0.793
[tex]s_x[/tex] = standard deviation of x-coordinates = 5.591
[tex]s_y[/tex] = standard deviation of y-coordinates = 2.772

So, the slope is calculated as follows:

[tex]m = \frac{rs_y}{s_x} \\ = \frac{(0.793)(2.772)}{5.591} \\ \boxed{m \approx 0.393}[/tex]

Hence, the slope in 3 decimal places is 0.393.