The average rate of change (AROC) of a function f(x) on an interval [a, b] is equal to the slope of the secant line to the graph of f(x) that passes through (a, f(a)) and (b, f(b)), a.k.a. the difference quotient given by
![f_{\mathrm{AROC}[a,b]} = \dfrac{f(b)-f(a)}{b-a}](https://tex.z-dn.net/?f=f_%7B%5Cmathrm%7BAROC%7D%5Ba%2Cb%5D%7D%20%3D%20%5Cdfrac%7Bf%28b%29-f%28a%29%7D%7Bb-a%7D)
So for f(x) = x² on [1, 5], the AROC of f is
![f_{\mathrm{AROC}[1,5]} = \dfrac{5^2-1^2}{5-1} = \dfrac{24}4 = \boxed{6}](https://tex.z-dn.net/?f=f_%7B%5Cmathrm%7BAROC%7D%5B1%2C5%5D%7D%20%3D%20%5Cdfrac%7B5%5E2-1%5E2%7D%7B5-1%7D%20%3D%20%5Cdfrac%7B24%7D4%20%3D%20%5Cboxed%7B6%7D)
This will be your formula:

cross multiply (32 times x) (120 times 800)
this is what you'll get:
Y = ax + b;
0.89 = a * 2 + b and 2.09 = a * 5 + b =>
=> 2.09 - 0.89 = a * 5 + a * 2 + b - b => 1.20 = 3 * a => a = 1.20 / 3 => a = 0.40;
b = 0.89 - 0.40 * 2 => b = 0.89 - 0.80 => b = 0.09;
y = 0.40x + 0.09 => 0.40x - y + 0.09 = 0 is the linear equation.