Exercise 9 Page 341

Can we draw a trend line? If so, which direction does it point as we move from left to right on the graph?

Scatter Plot:

Equation: y=5.4x-10 493
Attendance: 334 millions

Practice makes perfect

To write the equation of a trend line, we should first plot the given points, then sketch a trend line that models the data. Finally, by substituting the given value into the obtained equation we can find the asked estimation.

Observing the Graph

By treating the table as a set of points, we can graph the given data as a scatter plot. Do you see any trends?

It looks like there is some kind of correlation. Let's draw a line of fit, or trend line. To do so, we will draw a line that appears to fit the data closely.

Equation for the Trend Line

To write an equation for the trend line we determined above, we first need to use two points on the line to find its slope. Let's use (1990,253) and (2004,328) in the Slope Formula.

m=y_2-y_1/x_2-x_1

SubstitutePoints

Substitute ( 1990,253) & ( 2004,328)

m=328- 253/2004- 1990

SubTerms

Subtract terms

m=75/14

CalcQuot

Calculate quotient

m=5.357142...

RoundDec

Round to 1 decimal place(s)

m≈ 5.4

Now, that we have the slope m ≈ 5.4, let's use the point ( 1990, 253) in the point-slope form to write and simplify an equation for our trend line.

y-y_1=m(x-x_1)

SubstituteValues

Substitute values

y- 253= 5.4(x- 1990)

▼

Simplify

Distr

Distribute 5.4

y-253=5.4x-10 746

AddEqn

LHS+253=RHS+253

y=5.4x-10 493

Calculating the Projected Value

To calculate the expected attendance at a given year, we can substitute the value into our equation of the trend line. In this case, we will substitute the year 2005 for x in our equation and solve for y.

y=5.4x-10 493

Substitute

x= 2005

y=5.4( 2005)-10 493

Multiply

y=10 827-10 493

SubTerm

Subtract term

y=334

The expected attendance in 2005 is about 334 millions of people.