[a]The best possible solution for allocating CPU time is to give terminal 2 twice as much time as terminal 1, and to give terminals 3 and 4 equal time. This can be represented by the equation C+Dx+Ex^2=0, where C=1, D=2, and E=-1.

(b)

The equation of a parabola that best fits the points 1,2,3,4,5,8 at -2,-1,0,1,2 is C+Dx+Ex^2=0, where C=1, D=2, and E=-1.

(c)

The best way to deliver the specified amount of current to ground is to use resistors with the following values: v1=1, v2=2, v3=3. This will minimize the total energy dissipated in the resistors.

 

Explanation:

 

(a)

 

The best possible solution for allocating CPU time is to give terminal 2 twice as much time as terminal 1, and to give terminals 3 and 4 equal time. This can be represented by the equation C+Dx+Ex^2=0, where C=1, D=2, and E=-1.

The reason this is the best solution is because it minimizes the amount of time spent by the programmer at terminal 1 while still ensuring that the programmer at terminal 2 finishes at the same time. This is because the programmer at terminal 2 types four times as fast as the programmer at terminal 1.

 

(b)

 

The equation of a parabola that best fits the points 1,2,3,4,5,8 at -2,-1,0,1,2 is C+Dx+Ex^2=0, where C=1, D=2, and E=-1.

This is the equation of a parabola that passes through the points (-2,-1), (0,1), (2,5), and (4,8). It is the equation of a parabola that best fits these points because it minimizes the sum of the squares of the distances between the points and the parabola.

The best way to deliver the specified amount of current to ground is to use resistors with the following values: v1=1, v2=2, v3=3. This will minimize the total energy dissipated in the resistors.

The reason this is the best solution is because it minimizes the amount of current flowing through each resistor. This in turn minimizes the amount of energy dissipated in each resistor, and thus minimizes the total energy dissipated in the resistors.