Let's go step by step to solve the situation:

Given information:

• In July 2007, there were 3,580 purchases. In terms of months since January 2007, July 2007 is month 6 (since January 2007 = month 0).
• In March 2008, there were 3,420 purchases. March 2008 is month 14 (since January 2007 = month 0).

The purchases follow a linear model of the form:
$p = S(t) = r \cdot t + i$ where:

• $p$ is the number of purchases in thousands.
• $t$ is the number of months since January 2007.
• $r$ is the rate of change of purchases per month.
• $i$ is the initial number of purchases when $t = 0$ (January 2007).

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Part A: Find the rate of change of sales

We need to find the rate of change of the number of purchases per month. This is essentially the slope $r$ of the linear model.

We are given two points:

• In month 6 (July 2007), $p = 3.580$.
• In month 14 (March 2008), $p = 3.420$.

The slope formula is: $r = \frac{{\Delta p}}{{\Delta t}} = \frac{{p_2 - p_1}}{{t_2 - t_1}}$ Substituting the values: $r = \frac{{3.420 - 3.580}}{{14 - 6}} = \frac{{-0.160}}{{8}} = -0.02 \text{ (thousand purchases per month)}$

So, the rate of change of purchases is -0.02 thousand purchases per month.

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Part B: Write out the linear model

The slope $r = -0.02$, and now we need to find the intercept $i$.

Using the equation $p = r \cdot t + i$ and one of the points (for example, July 2007, where $t = 6$ and $p = 3.580$):

$3.580 = -0.02 \cdot 6 + i$ $3.580 = -0.12 + i$ $i = 3.580 + 0.12 = 3.700$

Thus, the linear model is: $p = S(t) = -0.02t + 3.700$ This is the equation that represents the number of purchases (in thousands) as a function of the number of months since January 2007.

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Part C: Predict total purchases in October 2010

To predict the purchases in October 2010, we first find how many months after January 2007 this corresponds to. October 2010 is month 45 (since January 2007).

Substitute $t = 45$ into the linear model: $p = S(45) = -0.02 \cdot 45 + 3.700$ $p = -0.90 + 3.700 = 2.800$

So, the model predicts there will be 2,800 purchases in October 2010.

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Part D: Find the month and year when purchases will be 2,580

We want to find when $p = 2.580$. Using the linear model: $2.580 = -0.02t + 3.700$ Solve for $t$: $-0.02t = 2.580 - 3.700 = -1.120$ $t = \frac{{-1.120}}{{-0.02}} = 56$

So, t = 56 months after January 2007. This corresponds to May 2011.

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Final answers:

• A. Sales are decreasing at a constant rate of -0.02 thousand purchases per month.
• B. The linear model is:
  $p = S(t) = -0.02t + 3.700$
• C. The model predicts there will be 2,800 purchases in October 2010.
• D. The model predicts that sales will be 2,580 purchases in May 2011.

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Do you want more details or have any questions?

Here are 5 follow-up questions:

1. How do you interpret a negative rate of change in this context?
2. How can this model help the company in forecasting future sales?
3. Could a non-linear model fit better in certain business situations? Why or why not?
4. What assumptions might be limiting in this linear sales model?
5. How would you adjust the model if the rate of sales decline increased over time?

Tip: Linear models work well when the relationship between variables is constant, but always check if this assumption holds in real-world data.