in 2009, 2012, 2015 the number (in millions) of smartphones sold in the world was 167.4, 675.1, and 1418.9 respectively. Let t represent time in years since 2009, and let S represent the number of smartphones sold in millions. determine M, A, and k for a logistic model, S(t) = M/1+Ae^-kt that fits the given data points. Approximate A and k to three decmila places and M to one place.

We are tasked with fitting a logistic model to the given data, where the model is of the form:

\[
S(t) = \frac{M}{1 + A e^{-kt}}
\]

where:
- \( S(t) \) is the number of smartphones sold (in millions) at time \( t \) years since 2009.
- \( t \) is the time in years since 2009.
- \( M \), \( A \), and \( k \) are parameters to be determined.

### Step 1: Data Points
The given data points are:
- \( t = 0 \) (in 2009), \( S(0) = 167.4 \)
- \( t = 3 \) (in 2012), \( S(3) = 675.1 \)
- \( t = 6 \) (in 2015), \( S(6) = 1418.9 \)

### Step 2: Asymptotic Behavior
The logistic function approaches \( M \) as \( t \to \infty \), meaning \( M \) represents the upper limit (the maximum possible number of smartphones sold). From the data, we know that smartphone sales have increased over time, and we assume that sales will eventually level off, reaching a maximum \( M \). Let's approximate \( M \) based on the data trend.

Looking at the values, we see significant growth between 2009 and 2015. A reasonable estimate for the upper bound of smartphone sales is approximately \( M \approx 1600 \) million.

### Step 3: Determine A and k
We now use the data points to solve for \( A \) and \( k \).

At \( t = 0 \):
\[
S(0) = \frac{M}{1 + A e^{-k(0)}} = \frac{M}{1 + A} = 167.4
\]
Substituting \( M = 1600 \):
\[
\frac{1600}{1 + A} = 167.4
\]
\[
1 + A = \frac{1600}{167.4} \approx 9.557
\]
\[
A \approx 8.557
\]

At \( t = 3 \) (in 2012):
\[
S(3) = \frac{1600}{1 + 8.557 e^{-3k}} = 675.1
\]
Solving for \( k \):
\[
\frac{1600}{1 + 8.557 e^{-3k}} = 675.1
\]
\[
1 + 8.557 e^{-3k} = \frac{1600}{675.1} \approx 2.37
\]
\[
8.557 e^{-3k} = 1.37
\]
\[
e^{-3k} = \frac{1.37}{8.557} \approx 0.16
\]
Taking the natural logarithm of both sides:
\[
-3k = \ln(0.16) \approx -1.833
\]
\[
k \approx \frac{1.833}{3} \approx 0.611
\]

### Step 4: Verify with \( t = 6 \) (in 2015)
At \( t = 6 \):
\[
S(6) = \frac{1600}{1 + 8.557 e^{-6(0.611)}} = 1418.9
\]
We can check this calculation to ensure that the model fits the data point at \( t = 6 \).

### Summary of Parameters
- \( M \approx 1600 \) (maximum number of smartphones in millions)
- \( A \approx 8.557 \)
- \( k \approx 0.611 \)

These are the values for \( M \), \( A \), and \( k \) that fit the logistic model to the given data.

Would you like more details on the verification or the next steps?

---

Here are 5 related questions:
1. How would the model behave if \( M \) were much larger or smaller than 1600?
2. What would happen to the model if \( k \) were smaller, indicating slower growth?
3. Can the logistic model predict when smartphone sales will peak?
4. How accurate is this model for data beyond 2015?
5. Could a different mathematical model, like exponential growth, fit the data better?

**Tip:** Logistic models are often used to describe growth that slows over time, such as population growth or product adoption in markets.

In 2009, 2012, 2015 the number (in millions) of smartphones sold in the world was 167.4, 675.1, and 1418.9 respectively. Let t represent time in years since 2009, and let S represent the number of smartphones sold in millions. Determine M, A, and k for a logistic model, S(t) = M/(1+Ae^-kt), that fits the given data points. Approximate A and k to three decimal places and M to one decimal place.

Learn how to fit a logistic model to smartphone sales data from 2009 to 2015. This problem involves determining the parameters M, A, and k for the logistic growth formula S(t) = M/(1 + A e^-kt), using data points of global smartphone sales over the years. Approximate values of M, A, and k are calculated with detailed steps.

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