Math  /  Algebra

Question\begin{tabular}{|l|c|c|} \hline Time Line & Price(dollars) & Average rate of change \\ \hline 1970 & $0.20\$ 0.20 & XXXXXXXXXXXXXX X X X X X X X X X X X X \\ \hline 1980 & $0.50\$ 0.50 & 0.03 \\ \hline 2003 & $2.00\$ 2.00 & 0.0652 \\ \hline 2009 & $2.25\$ 2.25 & 0.0417 \\ \hline 2013 & $2.50\$ 2.50 & 0.0625 \\ \hline 2015 & $2.75\$ 2.75 & 0.125 \\ \hline \end{tabular}
Do these values suggest a linear trend? Explain. The average rates of change do not suggest a linear because their is no direct relation independent variable (time in years) and the dependent variable (cost in dollars).
Step 5: Linear Modeling (6 pts)
Assuming that the trend is linear, generate a linear model. To make the calculation easier, rescale the time values for 2009 through 2015 in the above table.
Let 2009 be the year 0 . \begin{tabular}{|c|c|} \hline t & P ( dollars) \\ \hline 0 & \\ \hline & \\ \hline & \\ \hline \end{tabular}

Studdy Solution
Generate a linear model using the rescaled time values:
- Use the formula for the slope m m of a line: m=ΔPΔt m = \frac{\Delta P}{\Delta t} . - Calculate the slope using the points (0, 2.25) and (6, 2.75):
m=2.752.2560=0.5060.0833 m = \frac{2.75 - 2.25}{6 - 0} = \frac{0.50}{6} \approx 0.0833
- Use the point-slope form of a line: P=mt+b P = mt + b . - Substitute m=0.0833 m = 0.0833 and the point (0, 2.25) to find b b :
2.25=0.0833×0+b 2.25 = 0.0833 \times 0 + b b=2.25 b = 2.25
- The linear model is:
P=0.0833t+2.25 P = 0.0833t + 2.25
The linear model for the rescaled data is P=0.0833t+2.25 P = 0.0833t + 2.25 .

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