Principles of Mathematics - MPM1DC
Did you know?High school students participating in the 30-Hour Famine in 2006 raised $11.6 million. Check out the web site at www.30hourfamine.org.
A number of schools across the province of Ontario participate in the 30 Hour Famine each year. In one school, each participating student must contribute a minimum of $25 to the campaign. In addition, student collect pledges from the community. The average amount collected from the community last year was $8 per person.
You can model this fundraising activity using a table of values, a graph and an algebraic equation.
Table of Values showing First Differences
| Number of Pledges (n) |
Amount Contributed (C) |
(n, C) | First Difference |
|---|---|---|---|
| 0 | 25 | (0, 25) | |
| 10 | 10(8) = 25 + 105 | (10, 105) | 105 - 25 = 80 |
| 20 | 20(8) + 25 = 185 | (20, 185) | 185 - 105 = 80 |
The table of values gives us the data needed to graph this relation. First differences can be calculated if the independent variable increases by the same amount from row to row. Since the first difference values are all equal, we can conclude that this is a linear relation. We can also calculate the rate of change from a table.
By plotting the points found in the table of values and finding the line of best fit, we can create a graphical model of the situation.
A quick look at the line of best fit indicates a very strong linear relation. All data points are on the line. Therefore, we have a linear relation. Information can be read from the graph. Using interpolation, find the amount of money raised if a student has 15 pledges. The amount raised would be approximately $140. Ues extrapolation to find how much is raised with 30 pledges. The amount is approximately $260.
An algebraic model is an equation that represents the relation.
C = 8n + 25
where C represents the amount of money raised and n represents the number of pledges. The number 8 is the average amount pledged by each person. 25 is the minimum amount the student must contribute.
Since all the variables have an exponent value of 1, this is a linear relation. We can use this equation or model for find specific values. For example: how much money would be raised by a student with 40 pledges.
Solution
C = 8n + 25
C = 8(40) + 25
C = 320 + 25
C = 345
Therefore, the student would raise $345.
In SummaryRelations can be modelled using tables of values, graphs, and algebraic equations. We can determine the type of relation by analysing these models. Each of these models can be used to gain more information about the relationship.
Linear relations in a table of values are represented by first differences that are equal. In a graphical representation of a linear relation, all data points lie on a straight line. Linear relations represented by a algebraic equation contain variables with exponents of 1.
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