Using the equation: \(\displaystyle y=mx+b\) you know that the slope is m.

When you look at the equation:

\(\displaystyle \small y=90x+1\) you see that the value of \(\displaystyle \small m\) is \(\displaystyle \small 90\). 

\(\displaystyle 6 + 18 + 30 + 40= 94\) students achieved scores between 200 and 600, and Clark outscored all of them.

\(\displaystyle 6 + 18 + 30 + 40 + 16= 110\) students achieved scores between 200 and 700; Clark did not outscore all of them. 

Therefore, his score fell in the range from 601 to 700, making 635 the only plausible choice of the five.

   \(\displaystyle 18\) students scored in the 301-400 range.

   \(\displaystyle 30\) students scored in the 401-500 range.

   \(\displaystyle 40\) students scored in the 501-600 range.

\(\displaystyle \underline{+16}\) students scored in the 601-700 range. Add these:

\(\displaystyle 104\) students in all scored anywhere from 301-700.

Recall the slope formula:

\(\displaystyle \small \small m=\frac{rise}{run}=\frac{y2-y1}{x2-x1}\)

Plug in the two given points:

\(\displaystyle \small m = \frac{y2-y1}{x2-x1}= \frac{7-4}{4-2}\)

\(\displaystyle \small m = \frac{7-4}{4-2} = \frac{3}{2}\)

The slope of the line is \(\displaystyle \small \frac{3}{2}\).

30 students achieved a score of 401-500; 40 students achieved a score of 501-600; 16 achieved a score of 601-700; 10 achieved a score of 701-800. Add these:

\(\displaystyle 30 + 40 + 16 + 10 = 96\)

The number of students who took the test is the sum of the students who finished in the six ranges:

\(\displaystyle 6 + 18 + 30 + 40 + 16 + 10 = 120\)

The question is now to find out what percent 96 is of 120, which can be calculated as follows:

\(\displaystyle \frac{96}{120} \times 100 \% = 80 \%\)

Using the equation:

 \(\displaystyle \frac{y_{1}-y_{2}}{x_{1}-x_{2}}\) 

you plug in your points (2,4) and (3,4)

\(\displaystyle \frac{4-4}{3-2}\)  \(\displaystyle or\)   \(\displaystyle \frac{0}{1}\) 

\(\displaystyle 105+83+64+45+39 = 336\) students voted; of those students, \(\displaystyle 64+45+39 = 148\) voted for a candidate other than Phillip or Young. To convert this to a percent, use this proportion and solve for \(\displaystyle p\) :

\(\displaystyle \frac{p }{100}= \frac{148}{336}\)

\(\displaystyle \frac{p }{100} \cdot 100 = \frac{148}{336} \cdot 100\)

\(\displaystyle p = \frac{14,800}{336} \approx 44.0 \%\)

By comparing the sizes of the sectors, it can be seen that Mills won the largest share of the vote - but not at least one half of it - and Jones won the second-largest share. Therefore, they will face each other in a runoff.

Since we are asking for an estimate, one way to work this problem is to round each candidate's vote tally to the nearest thousand, then adding the rounded numbers:

\(\displaystyle 4,178 + 2,177 + 5,813 + 1,578\)

\(\displaystyle \approx 4,000 + 2,000 + 6,000 + 2,000 = 14,000\)

Johnson's share of the vote is about 6,000 of those approximately 14,000 votes, or \(\displaystyle \frac{6,000}{14,000} = \frac{6}{14} = \frac{3}{7}\)

As a percent, \(\displaystyle \frac{3}{7}\) is equal to:

\(\displaystyle \frac{3}{7} \cdot 100 = \frac{300}{7}\)

\(\displaystyle 300 \div 7 \approx 43\)

Therefore, 40% is the best estimate of the choices we are given.

The sector representing Lyle is roughly one-tenth to one-eighth of the circle - in other words, 10-12 % of it. This is the approximate percent of the vote won by Lyle.