Subtract  from both sides:

Subtract 7 from both sides:

Divide both sides by :

Remember to switch the inequality when dividing by a negative number:

Since is not an answer, we must find an answer that, at the very least, does not contradict the fact that  is less than (approximately) 4.67.  Since any number that is less than 4.67 is also less than any number that is bigger than 4.67, we can be sure that  is less than 5.

To estimate the mean number of errors made, first calculate the mean of each bar in the histogram.

Recall to calculate a mean of a histogram bar use the formula:

The red bar's mean:

The yellow bar's mean:

The green bar's mean:

The teal bar's mean:

The blue bar's mean:

The purple bar's mean:

Now to estimate the mean of the number of errors, add the means of the bars together and divide by the total number of frequencies.

Mrs. Smith assigns homework to her class once a week and records each student's results. For last week's assignment, the histogram depicts these results. If the homework set was worth twenty points, what is the range of grades for last week's assignment?

Note: For each error, one point is subtracted from the total points possible. 

To solve this problem, first identify what the highest and lowest grade on the assignment was. The highest grade will be the one with the least amount of errors made and the lowest grade will be the one with the most amount of errors made.

The red bar in the histogram says that students made between zero and two errors. In other words, either: 0, 1, 2 errors.

The yellow bar in the histogram says that the students made between two and four errors. Since the red bar includes two errors the yellow bar only includes 3 or 4 errors.

The green bar in the histogram says that students made 5 or 6 errors.

The teal bar in the histogram says that students made 7 or 8 errors.

The blue bar in the histogram says that students made 9 or 10 errors.

The purple bar in the histogram says that students made 11 or 12 errors.

This means that at least one student made zero errors on the assignment and therefore did not loose any points. Therefore, the highest grade on the assignment was twenty.

The histogram shows that at least one student made twelve errors on the test. Therefore, the lowest grade on the assignment would be:

From here, calculate the range.

First, subtract 3 from both sides in order to obtain an equation that equals 0:

The left side can be factored. We need factors of  that add up to .  and  work:

Set both factors equal to 0 and solve:

To solve the left equation, add 1 to both sides. To solve the right equation, subtract 3 from both sides. This yields two solutions:

Only one of these solutions is negative, so the answer is 1.

To begin, we analyze the equation given: the base equation,  is shifted left one unit and vertically stretched by a factor of 2. The graph of the equation  is:

To solve the inequality, we need to take a test point and plug it in to see if it matches the inequality. The only points that cannot be used are those directly on our parabola, so let's use the origin . If plugging this point in makes the inequality true, then we shade the area containing that point (in this case, outside the parabola); if it makes the inequality untrue, then the opposite side is shaded (in this case, the inside of the parabola). Plugging the numbers in shows:

Simplified as:

Which is not true, so the area inside of the parabola should be shaded, resulting in the following graph:

This question tests one's ability to recognize algebraic characteristics of a graph. This particular question examines a linear function.

Looking at the graph, it is seen that the line passes through the points (-8,-5) and (8,5).

The slope of a line through the points  and  can be found by setting 

:

in the slope formula:

This question is testing one's ability to compare the properties of functions when they are illustrated in different forms. This question specifically is asking for the examination and interpretation of two quadratic functions for which one is illustrated in a table format and the other is illustrated graphically.

Step 1: Identify the minimum of the table.

Using the table find the time value where the lowest distance exists. 

Recall that the time represents the  values while the distance represents the  values. Therefore the ordered pair for the minimum can be written as .

Step 2: Identify the minimum of the graph

Recall that the minimum of a cubic function is known as a local minimum. This occurs at the valley where the vertex lies.

For this particular graph the vertex is at .

Step 3: Compare the minimums from step 1 and step 2.

Compare the  value coordinate from both minimums.

Therefore, the graph has the lowest minimum.

To solve this problem, add the two equations together:

The only answer choice that satisfies this equation is 0, because 0 is less than 4.

We need 2 equations, because we have 2 unkown variables. Let  = the entrance fee, and  = the fee per ride. One ride costs  dollars. We know that Lisa spent 120 dollars in total. Since Lisa went on 6 rides, she spent  dollars on rides. Her only other expense was the entrance fee, : 

Apply similar logic to Tom:

Subtracting the second equation from the first equation results in:

Divide both sides by 2:

So every ride costs 12.5 dollars. Plugging 12.5 back into one of the original equations allows us to solve for the entrance fee:

Subtract 50 from both sides: