Answer:
Following are the responses to the given question:
Explanation:
Its machine slows for each further level of evaluation by an n/m factor. Therefore are the times for implementation for levels 2, 3, and 4 are [tex]\frac{kn}{m} \ and\ \frac{kn^2}{m^2} \ and \ \frac{kn^3}{m^3}[/tex].
So, the level values are:
[tex]level \ 2 = \frac{kn}{m}\\\\level \ 3 = \frac{kn^2}{m^2}\\\\level \ 4 = \frac{kn^2}{m^2}\\\\[/tex]