Question

a) graph the data on the scatter plot and draw a line of best fit for the data.
b) write an equation for the line of best fit for this data. let x represent the years since 2007 and y represent the sales, in thousands of dollars.
c) according to your equation, in what year will the sales reach about $500 (in thousands)? use mathematics to explain how you determined your answer.

Explanation:

Step1: Find the slope

The formula for slope \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Let's take two points \((x_1,y_1)=(0,305)\) (corresponding to 2007) and \((x_2,y_2)=(5,420)\) (corresponding to 2012). Then \(m=\frac{420 - 305}{5-0}=\frac{115}{5}=23\).

Step2: Use the slope - intercept form \(y = mx + b\)

The slope - intercept form of a line is \(y=mx + b\), where \(m\) is the slope and \(b\) is the \(y\) - intercept. When \(x = 0\) (year 2007), \(y=305\), so \(b = 305\). The equation of the line of best fit is \(y=23x + 305\).

Step3: Solve for \(x\) when \(y = 500\)

Substitute \(y = 500\) into the equation \(y=23x+305\). Then \(500=23x + 305\). Subtract 305 from both sides: \(500-305=23x\), so \(195 = 23x\). Then \(x=\frac{195}{23}\approx8.48\). Since \(x\) represents the number of years since 2007, the year is \(2007 + 8.48\approx2016\).

Answer:

B) The equation of the line of best fit is \(y = 23x+305\). C) According to the equation \(y = 23x + 305\), when \(y = 500\), \(x=\frac{500 - 305}{23}=\frac{195}{23}\approx8.48\). So the sales will reach about \(500\) (in thousands) around the year \(2007+8.48\approx2016\).