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This week your student will consider what makes a graph useful and then use graphs, equations, tables, and descriptions to compare two different situations. There are many successful ways to set up and add a scale to a pair of axes when creating a graph of a situation. Sometimes we choose specific ranges for the axes in order to see specific information.
For example, if two large, cylindrical water tanks are being filled at a constant rate, we could show the amount of water in them using a graph like this:
While the first graph is accurate, it only shows up to 10 liters, which isn’t that much water. Let’s say we wanted to know how long it would take each tank to have 110 liters. With 110 as a guide, we could set up our axes like this:
Notice how the vertical scale goes beyond the value we are interested in. Also notice how each axis has values that increase by 10, which, along with numbers like 1, 2, 5, and 25, is a friendly number to count by.
Here is a task to try with your student:
This table shows some lengths measured in inches and the equivalent length in centimeters. Complete the table.
| length (inches) | length (centimeters) |
|---|---|
| 1 | 2.54 |
| 2 | |
| 10 | |
| 50.8 |
Sketch a graph of the relationships between inches and centimeters. Scale the axes so that all the values in the table can be seen on the graph.
Solution:
| length (inches) | length (centimeters) |
|---|---|
| 1 | 2.54 |
| 2 | 5.08 |
| 10 | 25.4 |
| 20 | 50.8 |
This week your student will learn how to write equations representing linear relationships. A linear relationship exists between two quantities when one quantity has a constant rate of change with respect to the other. The relationship is called linear because its graph is a line.
For example, say we are 5 miles into a hike heading toward a lake at the end of the trail. If we walk at a speed of 2.5 miles per hour, then for each hour that passes we are 2.5 miles further along the trail. After 1 hour, we would be 7.5 miles from the start. After 2 hours, we would be 10 miles from the start (assuming no stops). This means there is a linear relationship between miles traveled and hours walked. A graph representing this situation is a line with a slope of 2.5 and a vertical intercept of 5.
Here is a task to try with your student:
The graph shows the height in inches, \(h\), of a bamboo plant \(t\) months after it has been planted.
Solution:
This week your student will work with scatter plots. Scatter plots show us how two different variables are related. In this example, each plotted point corresponds to a dog, and its coordinates tell us the height and weight of that dog. The point on the lower left of the graph, for example, might represent a dog that is 8 inches tall and weighs about 5 pounds. The plot shows that, generally speaking, taller dogs weigh more than shorter dogs.
Since a larger value for one characteristic (height) generally means a larger value for the other characteristic (weight), we say that there is a positive association between dog height and dog weight.
In the next example, each point corresponds to a car, and its coordinates tell us the weight (kilograms) and fuel efficiency (miles per gallon) of the car.
This time, we see that larger values for one characteristic (car weight) generally go along with lower values for the other characteristic (fuel efficiency), and so we say that there is a negative association between car weight and fuel efficiency.
Here is a task to try with your student:
This scatter plot shows the relationship between average temperature and gas usage in a building.
Solution:
This week your student will use two-way tables. Two-way tables are a way of comparing two variables. For example, this table shows the results of a study of the relation between meditation and state of mind of athletes before a track meet.
| meditated | did not meditate | total | |
|---|---|---|---|
| calm | 45 | 8 | 53 |
| agitated | 23 | 21 | 44 |
| total | 68 | 29 | 97 |
23 of the people who meditated were agitated, while 21 of the people who did not meditate were agitated. Does this mean that meditation has no impact or even a slight negative association with mood? Probably not. When we look for associations between variables, it can be more informative to know the approximate percentages in each category, like this:
| meditated | did not meditate | |
|---|---|---|
| calm | 66% | 28% |
| agitated | 34% | 72% |
| total | 100% | 100% |
Of the people who meditated, about 66% were calm (\(45 \div 68 ≅ 0.66\)), and about 34% were agitated (\(23 \div 68 ≅ 0.34\)). When we compare that to the percentages for people who did not meditate, we can now see more easily that the group of people who meditated has a lower percentage of athletes who are agitated. The percentages in this table are called relative frequencies.
Here is a task to try with your student:
This table contains data about whether people in various age groups use their cell phone as their main alarm clock.
| use cell phone as alarm | do not use cell phone as alarm | total | |
|---|---|---|---|
| 18 to 29 years old | 47 | 16 | 63 |
| 30 to 49 years old | 66 | 21 | 87 |
| 50+ years old | 31 | 39 | 70 |
| total | 144 | 76 | 220 |
| use cell phone as alarm | do not use cell phone as alarm | total | |
|---|---|---|---|
| 18 to 29 years old | \(75\%\), since \(\frac{47}{63}≅0.75\) | 100% | |
| 30 to 49 years old | |||
| 50+ years old |
Solution:
| use cell phone as alarm | do not use cell phone as alarm | total | |
|---|---|---|---|
| 18 to 29 years old | \(75\%\), since \(\frac{47}{63}≅0.75\) | \(25\%\), since \(\frac{16}{63}≅0.25\) | 100% |
| 30 to 49 years old | \(76\%\), since \(\frac{66}{87}≅0.76\) | \(24\%\), since \(\frac{21}{87}≅0.24\) | 100% |
| 50+ years old | \(44\%\), since \(\frac{31}{70}≅0.44\) | \(56\%\), since \(\frac{39}{70}≅0.56\) | 100% |
This week your student will work with systems of equations. A system of equations is a set of 2 (or more) equations where the letters represent the same values. For example, say Car A is traveling 75 miles per hour and passes a rest area. The distance in miles it has traveled from the rest area after \(t\) hours is \(d=75t\). Car B is traveling toward the rest area, and its distance from the rest area at any time is \(d=14-65t\). We can ask if there is ever a time when the distance of Car A from the rest area is the same as the distance of Car B from the rest area. If the answer is “yes,” then the solution will correspond to one point that is on both lines, such as the point \((0.1, 7.5)\) shown here. This means that 0.1 hours after Car A passes the rest area, both cars will be 7.5 miles from the rest area.
We could also answer the question without using a graph. Because we are asking when the \(d\) values for each car will be the same, we are asking for what \(t\) value, if any, makes \(75t=14-65t\) true. Solving this equation for \(t\), we find that \(t=0.1\) is a solution, and at that time the cars are 7.5 miles away because \(75t=75\boldcdot 0.1=7.5\). This finding matches the graph.
Here is a task to try with your student:
Lin and Diego are biking the same direction on the same path, but start at different times. Diego is riding at a constant speed of 18 miles per hour, so his distance traveled in miles can be represented by \(d\) and the time he has traveled in hours by \(t\), where \(d=18t\). Lin started riding a quarter hour before Diego at a constant speed of 12 miles per hour, so her total distance traveled in miles can be represented by \(d\), where \(d=12\left(t+\frac14 \right)\). When will Lin and Diego meet?
Solution:
To find when Lin and Diego meet, that is, when they have traveled the same total distance, we can set the two equations equal to one another: \(18t=12\left(t+\frac14 \right)\). Solving this equation for \(t\), \(18t=12t+3\), \(6t=3\), \(t=\frac12\).
They meet after Diego rides for one half hour and Lin rides for three quarters of an hour. The distance that they each travel before meeting is 9 miles, because we can use \(\frac{1}{2}\) for \(t\) in Diego’s equation and see that \(9=18 \boldcdot \frac12\). Another way to find a solution would be to graph both \(d=18t\) and \(d=12\left(t+\frac14 \right)\) on the same coordinate plane and interpret the point where these lines intersect.
This week your student will investigate linear relationships with slopes that are not positive. Here is an example of a line with negative slope that represents the amount of money on a public transit fare card based on the number of rides taken:
The slope of the line graphed here is -2.5, since \(\text{slope}=\frac{\text{vertical change}}{\text{horizontal change}}=\frac{\text-40}{16}=\text-2.5\), and corresponds to the cost of 1 ride. The vertical intercept is 40, which means the card started out with \$40 on it.
One possible equation for this line is \(y=\text-2.5x+40.\) It is important to understand that every pair of numbers \((x,y)\) that is a solution to the equation representing the situation is also a point on the graph representing the situation. Points not on the graph representing the situation will not be a solution to the equation representing the situation.
Here is a task to try with your student:
A length of ribbon is cut into two pieces. The graph shows the length of the first piece with respect to the length of the second piece\(\).
Solution: