Page 484 - Algebra 1

SOLUTION
a. Graph 3, as there is no correlation between the number of trucks on the road and the number of days in a month
b. Graph 1, as there is a positive correlation between time spent driving and distance traveled
c. Graph 2, as there is a negative correlation between the age of a car and its value
Estimations and predictions can be determined when there is a correlation, or a trend, between data values. Interpolation is a process of determining data points between given data points. Extrapolation is a process of determining data points that are beyond the given data points.
Application: Population Growth
The table shows the population of the United States from the year 1960 through the year 2000.
Example
5
U. S. Population
a. Draw a scatter plot and a trend line for the data.
Year
1960
1965
1970
1975
1980
1985
1990
1995
2000
Population (in millions)
180
194
205
215
228
238
250
267
282
Math Reasoning
Analyze Why might a prediction determined by an equation for the line of best fit, be better than a prediction determined by a trend line?
SOLUTION Plot the points with the year on the horizontal axis and the population on the vertical axis. Draw a straight line as near to as many of the points as possible.
b. Use the trend line to make a prediction for the population in the year 2005.
SOLUTION From the graph, the population in the year 2005 will be about 290 million people.
c. Find the equation for a line of best fit using a graphing calculator. Round the values for a and b to the nearest thousandth.
280 260 240 220 200 180
0
SOLUTION Use the LinReg (ax + b) feature on a graphing calculator to compute the values for the line of best fit.
The equation for the line of best fit is y = 2.467x - 4655.222.
d. Use the equation for the line of best fit to estimate the population in
the year 2005. Round the answer to the nearest million.
SOLUTION
y = 2.467x - 4655.222
y = 2.467(2010) - 4655.222 y ≈ 291 million
Year
Lesson 71 469
1960 1970 1980 1990 2000
Population (in millions)