Answer :
To determine which model (linear or exponential) most accurately predicts the value of a smartphone that is 2 years old and worth [tex]$375, we need to first understand the nature of each model and then evaluate their fit.
### Linear Model:
A linear model assumes that the value of the smartphone decreases at a constant rate over time. Mathematically, a linear depreciation model can be expressed as:
\[ V(t) = V_0 - kt \]
where:
- \( V(t) \) is the value of the smartphone at time \( t \),
- \( V_0 \) is the initial value of the smartphone (value at \( t = 0 \)),
- \( k \) is the constant rate of depreciation,
- \( t \) is the time in years.
### Exponential Model:
An exponential model assumes that the value of the smartphone decreases by a constant percentage over time. Mathematically, an exponential depreciation model can be expressed as:
\[ V(t) = V_0 \cdot e^{-kt} \]
where:
- \( V(t) \) is the value of the smartphone at time \( t \),
- \( V_0 \) is the initial value of the smartphone,
- \( k \) is the constant percentage rate of depreciation,
- \( t \) is the time in years,
- \( e \) is the base of the natural logarithm.
### Step-by-Step Solution:
1. Gather Information and Make Assumptions:
- Assume the initial value \( V_0 \) of the smartphone when it was new (at \( t = 0 \)) is known (for example, let’s say it was $[/tex]1000 when new).
2. Linear Model Analysis:
- Let's find the rate of depreciation [tex]\( k \)[/tex].
- Given [tex]\( V(2) = 375 \)[/tex].
[tex]\[ 375 = 1000 - 2k \][/tex]
[tex]\[ 2k = 1000 - 375 \][/tex]
[tex]\[ 2k = 625 \][/tex]
[tex]\[ k = 312.5 \][/tex]
- So, the linear depreciation model is:
[tex]\[ V(t) = 1000 - 312.5t \][/tex]
3. Exponential Model Analysis:
- Let's find the rate of depreciation [tex]\( k \)[/tex].
- Given [tex]\( V(2) = 375 \)[/tex].
[tex]\[ 375 = 1000 \cdot e^{-2k} \][/tex]
[tex]\[ e^{-2k} = \frac{375}{1000} \][/tex]
[tex]\[ e^{-2k} = 0.375 \][/tex]
[tex]\[ -2k = \ln(0.375) \][/tex]
[tex]\[ k = -\frac{\ln(0.375)}{2} \][/tex]
[tex]\[ k \approx 0.4906 \][/tex] (using natural logarithm values)
- So, the exponential depreciation model is:
[tex]\[ V(t) = 1000 \cdot e^{-0.4906t} \][/tex]
4. Evaluate Models at Different Points:
- Let's compare the value of the smartphone at different time points (such as at [tex]\( t = 1 \)[/tex] year).
Linear Model:
[tex]\[ V(1) = 1000 - 312.5(1) \][/tex]
[tex]\[ V(1) = 687.5 \][/tex]
Exponential Model:
[tex]\[ V(1) = 1000 \cdot e^{-0.4906(1)} \][/tex]
[tex]\[ V(1) \approx 1000 \cdot 0.612 \][/tex]
[tex]\[ V(1) \approx 612 \][/tex]
5. Determine the Most Accurate Model:
- If we know the actual value at [tex]\( t = 1 \)[/tex] year (let’s assume it was [tex]$650), we can see which model predicts more accurately. In this case: - Linear Model predicts $[/tex]687.5 which is closer to [tex]$650. - Exponential Model predicts $[/tex]612 which is further from $650.
Given this comparison, the linear model might more accurately predict the value of the smartphone at 2 years old if the depreciation is indeed more consistent with linearly decreasing value. However, the exponential model could be more suitable in cases where items depreciate by a percentage consistently over time, which is often the case with electronics.
### Conclusion:
Justification: Generally, smartphones and other electronics depreciate by a certain percentage over time, making the exponential model a more accurate representation. Therefore, even though the linear model might be closer at a specific point, the natural depreciation pattern of smartphones supports the use of an exponential model for long-term predictions.
In conclusion, while the linear model might more closely align with the given data point, the exponential model is likely a better long-term predictor because it reflects the typical behavior of electronic devices' depreciation.
2. Linear Model Analysis:
- Let's find the rate of depreciation [tex]\( k \)[/tex].
- Given [tex]\( V(2) = 375 \)[/tex].
[tex]\[ 375 = 1000 - 2k \][/tex]
[tex]\[ 2k = 1000 - 375 \][/tex]
[tex]\[ 2k = 625 \][/tex]
[tex]\[ k = 312.5 \][/tex]
- So, the linear depreciation model is:
[tex]\[ V(t) = 1000 - 312.5t \][/tex]
3. Exponential Model Analysis:
- Let's find the rate of depreciation [tex]\( k \)[/tex].
- Given [tex]\( V(2) = 375 \)[/tex].
[tex]\[ 375 = 1000 \cdot e^{-2k} \][/tex]
[tex]\[ e^{-2k} = \frac{375}{1000} \][/tex]
[tex]\[ e^{-2k} = 0.375 \][/tex]
[tex]\[ -2k = \ln(0.375) \][/tex]
[tex]\[ k = -\frac{\ln(0.375)}{2} \][/tex]
[tex]\[ k \approx 0.4906 \][/tex] (using natural logarithm values)
- So, the exponential depreciation model is:
[tex]\[ V(t) = 1000 \cdot e^{-0.4906t} \][/tex]
4. Evaluate Models at Different Points:
- Let's compare the value of the smartphone at different time points (such as at [tex]\( t = 1 \)[/tex] year).
Linear Model:
[tex]\[ V(1) = 1000 - 312.5(1) \][/tex]
[tex]\[ V(1) = 687.5 \][/tex]
Exponential Model:
[tex]\[ V(1) = 1000 \cdot e^{-0.4906(1)} \][/tex]
[tex]\[ V(1) \approx 1000 \cdot 0.612 \][/tex]
[tex]\[ V(1) \approx 612 \][/tex]
5. Determine the Most Accurate Model:
- If we know the actual value at [tex]\( t = 1 \)[/tex] year (let’s assume it was [tex]$650), we can see which model predicts more accurately. In this case: - Linear Model predicts $[/tex]687.5 which is closer to [tex]$650. - Exponential Model predicts $[/tex]612 which is further from $650.
Given this comparison, the linear model might more accurately predict the value of the smartphone at 2 years old if the depreciation is indeed more consistent with linearly decreasing value. However, the exponential model could be more suitable in cases where items depreciate by a percentage consistently over time, which is often the case with electronics.
### Conclusion:
Justification: Generally, smartphones and other electronics depreciate by a certain percentage over time, making the exponential model a more accurate representation. Therefore, even though the linear model might be closer at a specific point, the natural depreciation pattern of smartphones supports the use of an exponential model for long-term predictions.
In conclusion, while the linear model might more closely align with the given data point, the exponential model is likely a better long-term predictor because it reflects the typical behavior of electronic devices' depreciation.
