The equation of the line given is of the from y \approx ax + b where a is the slope of the line and b is the y-intercept.

Take any two points on the graph to find the slope of the line.

We use the slope formula to calculate the slope of the line. Let (29.25) and (95,92) be the two points on the line.

Let x_{1} = 29, x_{2} = 25, x_{2} = 95, y_{2} = 92.

We plug this in the formula \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{92-25}{95-29} = \frac{67}{66} \approx 1.01

Now we plug in this slope, a = 1.01, x = 29, y = 25 in y = ax + b.

25 = 1.01 \times 29 + b

Solve for b to get b = -4.09.

So, the equation of the line will be y \approx 1.01 x - 4.09.

This linear equation best describes the given model.

Suppose we want to find the exam score 2 of the student based on the fact that the exam score 1 was 88, just plug in x = 88 in the same equation to get,

y \approx 88.88 - 4.09 \approx 84.79.

So, the exam score 2 for the student would be approximately 85 points.