The function graphed is . We know that it is strictly less than because the line is dotted (not solid). We also know that y is less than because the portion of the graph below the function is shaded.

To start, we want to write two equations that model this situation. Those equations are:

 (where c is the number of classes Miles teaches)

The graph of these two equations is below:

Finally, the intersection of these two lines is associated with the total number of classes Miles needs to teach to break even. We can see that the intersection is at (16, 400). Therefore, after teaching 16 classes, Miles will have made $400 from the gym and broke even on his investment. 

To start, we want to write two equations that model this situation. Those equations are:

 (where c is the number of classes Emeka teaches)

The graph of these two equations is below:

Finally, the intersection of these two lines is associated with the total number of classes Emeka needs to teach to break even. We can see that the intersection is at (20, 500). Therefore, after teaching 20 classes, Emeka will have made $500 from the gym and broke even on his investment. 

Recognize that the intersection of two lines occurs when the two lines meet at the same point - that is, where their x- and y-coordinates are the same. So you can find the point of intersection by simply solving the system. Here that is likely best done through the Elimination Method.  If you stack the equations:

3x - 2y = 0

4x+3y = -17

You should see that you have a negative y term in the first equation and a positive y term in the second. That means that if you can get the coefficients of y to be the same, you can sum the equations and eliminate the y-term altogether. To do that, multiply the top equation by 3 and the bottom equation by 2 so that your y-coefficients are each 6:

3(3x - 2y = 0) --> 9x - 6y = 0

2(4x + 3y = -17) --> 8x + 6y = -34

When you sum the equations, then, you arrive at:

17x = -34

Which means that x = -2.  

Then plug x = -2 into one of the equations.  If you use the first, you'll get:

3(-2) - 2y = 0

Meaning that:

-6 - 2y = 0

-6=2y

y = -3

Since the point of intersection is (-2, -3), with both coordinates negative, the point will lie in Quadrant III.

The expected value of each horse’s rank can be calculated as follows:

Expected value:

Horse 1:

Horse 2:

Horse 3:

The expected value predicts each horse’s rank in the upcoming race. Horse 1’s probability distribution yields the lowest expected value (i.e., the closest value to first place, represented by 1), so Horse 1 can be expected to rank most highly.

Expected value identifies the most likely outcome, but other outcomes may still result, including unlikely outcomes. Although the horse the spectator bet on had the highest likelihood among the other horses of ranking well in its next race, the race was still ultimately subject to chance. Additionally, the three horses’ expected ranks differed only slightly--2.1, 2.5 and 2.6--meaning that the horse with the highest expected rank had only a slightly greater likelihood of performing more highly than the other two horses to begin with. As a result, chance was a relatively significant determinant of outcomes.

The sample space for the die roll is {1, 2, 3, 4, 5, 6}. Student A gets a point if 2, 4 or 6 is rolled. Student B gets a point if 1, 2 or 3 is rolled. Student C gets a point if 2, 3 or 5 is rolled. Therefore, all three students have a 3/6 = ½ = 0/5 = 50% chance of getting a point on each die roll. The number of times the die is rolled does not affect each student’s probability of getting a point; the probability is the same in each roll so long as neither the die nor the point rules change.