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When dealing with linear equations, the slope is a crucial element. It indicates how steep the line is or how quickly the y-values are changing as you move from one x-value to another. This change is calculated as "rise over run," or the change in y divided by the change in x. To find the slope, use the formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] In our scenario with Timberland Co.'s cash flow data, we use the years they provided: 1995, which we consider year 5, and 2003, which is year 13. Now, the cash flows (or y-values) for those years are \\(0.18 and \\)4.04, respectively. By substituting these numbers into the formula, you can determine how cash flow per share changes per year. This slope tells us the rate of increase in cash flow per year between 1995 and 2003. It's like finding the speed of growth for the cash flow.

The y-intercept is the value where the line crosses the y-axis. In simpler terms, it is the starting point of your equation when all other values are zero. In our context, it shows the starting cash flow per share when considering the year variable.To find the y-intercept using the point-slope form of a line, which is:\[ y - y_1 = m(x - x_1) \]You substitute the known slope \( m \), and one point's coordinates \((x_1, y_1)\) into the formula. Here, we've already figured the slope and can use the point from the year 1995 with a cash flow of \$0.18 (translating to coordinates \(x_1 = 5, y_1 = 0.18\)). By isolating \( y \), you can determine the equation of the line that describes cash flow growth over the years. The y-intercept essentially sets the baseline cash flow when the year count is zero, establishing a starting point for subsequent predictions.

Once the equation of a line is determined, it can be used for prediction. This prediction can tell us what the cash flow might be in future years. By simply substituting the desired year into the equation where \( t \) represents years since 1990, you can calculate the expected cash flows for those years.To predict the cash flows for the years 2008 and 2010:- Identify the corresponding \( t \) values as 18 (for 2008) and 20 (for 2010).- Substitute these \( t \) values into your linear equation.The resulting y-values will give you the projected cash flow per share for the years 2008 and 2010.

In this exercise, the year is cleverly turned into a variable, \( t \), to fit within the linear equation framework. This transformation allows us to conveniently use years as a numeric variable, simplifying calculations and predictions.Representing years in this way replaces the more cumbersome calendar years with easy-to-use integers. In our example, the year 1995 corresponds to \( t = 5 \). By setting 1990 as year zero, we effectively create a timeline where each subsequent year is increasingly larger by one unit, making future predictions scalable and straightforward.When solving real-world problems like this one, turning the year into such a variable helps streamline the computations and interpretations involved in analyzing patterns over time.