Let's begin by identifying key information given to us:
We have the best line of fit that is a straight line.
The equation of a straight line is given by the equation: y = mx + b
We will pick two coordinates that lie on the straight line to form the equation:

We will proceed to calculate the slope of the equation using the coordinates above. We have:

Since slope (m) = 8/5, the equation of the line becomes:

The y-intercept (where the straight line touches the y-axis) of the graph equals 22: b = 22
Therefore, the equation of the straight line thus becomes:
