The annual revenue of Amazon.com rose from approximately $10.7 billion in 2006 to $34.2 billion in 2010.†
(a)
Use this information to find both a linear model and an exponential model for Amazon.com's annual revenue I (in billions of dollars) as a function of time t in years since 2000. (Round all coefficients to three significant digits.)

Let's begin by finding the linear and exponential models for Amazon.com's annual revenue $ I $ as a function of time $ t $ in years since 2000. We are given:

- In 2006 ($ t = 6 $), the revenue was $ 10.7 $ billion dollars.
- In 2010 ($ t = 10 $), the revenue was $ 34.2 $ billion dollars.

### Part (a) - Linear Model

The general form of a linear model is:

\[
I(t) = mt + b
\]

where $ m $ is the slope, and $ b $ is the intercept.

#### Step 1: Calculate the slope $ m $

We can calculate the slope $ m $ as:

\[
m = \frac{I(10) - I(6)}{10 - 6} = \frac{34.2 - 10.7}{10 - 6} = \frac{23.5}{4} = 5.875
\]

So, $ m = 5.875 $.

#### Step 2: Find the intercept $ b $

To find $ b $, we can use one of the points, say $ (6, 10.7) $:

\[
10.7 = 5.875(6) + b
\]
\[
10.7 = 35.25 + b
\]
\[
b = 10.7 - 35.25 = -24.55
\]

Thus, the linear model is:

\[
I(t) = 5.88t - 24.6
\]

(Rounding to three significant digits)

### Part (b) - Exponential Model

The general form of an exponential model is:

\[
I(t) = I_0 e^{kt}
\]

where $ I_0 $ is the initial value (at $ t = 0 $), and $ k $ is the growth rate.

#### Step 1: Write two equations using the given data

We know:

- $ I(6) = 10.7 $
- $ I(10) = 34.2 $

This gives us two equations:

\[
10.7 = I_0 e^{6k}
\]
\[
34.2 = I_0 e^{10k}
\]

#### Step 2: Solve for $ k $

Dividing the second equation by the first equation:

\[
\frac{34.2}{10.7} = \frac{I_0 e^{10k}}{I_0 e^{6k}} \implies \frac{34.2}{10.7} = e^{4k}
\]
\[
e^{4k} = \frac{34.2}{10.7} = 3.196
\]
\[
4k = \ln(3.196) \implies k = \frac{\ln(3.196)}{4} = \frac{1.162}{4} = 0.2905
\]

Thus, $ k = 0.2905 $.

#### Step 3: Solve for $ I_0 $

Substitute $ k = 0.2905 $ into one of the original equations, say $ 10.7 = I_0 e^{6k} $:

\[
10.7 = I_0 e^{6 \times 0.2905} = I_0 e^{1.743}
\]
\[
10.7 = I_0 \times 5.717
\]
\[
I_0 = \frac{10.7}{5.717} = 1.872
\]

Thus, the exponential model is:

\[
I(t) = 1.87 e^{0.291t}
\]

(Rounding to three significant digits)

### Summary:
- **Linear model**: $ I(t) = 5.88t - 24.6 $
- **Exponential model**: $ I(t) = 1.87 e^{0.291t} $

Would you like more details on any part of the solution?

### 5 Related Questions:
1. How do linear and exponential growth models differ in predicting long-term behavior?
2. How can you graph both models to compare their fit to the data?
3. How would changes in $ k $ affect the exponential growth rate?
4. What are some limitations of using a linear model for predicting revenue?
5. How can you determine which model better fits the data over a longer period?

### Tip:
When choosing between linear and exponential models, always consider how the data behaves over time—exponential models often provide a better fit for rapidly growing trends.

The annual revenue of Amazon.com rose from approximately $10.7 billion in 2006 to $34.2 billion in 2010.
(a) Use this information to find both a linear model and an exponential model for Amazon.com's annual revenue I (in billions of dollars) as a function of time t in years since 2000. (Round all coefficients to three significant digits.)

Find the linear and exponential models for Amazon.com's revenue growth from $10.7 billion in 2006 to $34.2 billion in 2010. Learn how to calculate the slope, growth rate, and model functions, with all coefficients rounded to three significant digits. Ideal for students in advanced algebra courses.

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