Question
Place the option that correctly completes the sentence.
Kaylen bought a new motorcycle for $5,800. After 4 years, the motorcycle is worth $4,926.21.
Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse.
If Kaylen assumes the value of the motorcycle depreciates according to an exponential decay function, with corresponding to the purchase date, the decay rate is Response area per year.
Kaylen bought a new motorcycle for $5,800. After 4 years, the motorcycle is worth $4,926.21.
Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse.
If Kaylen assumes the value of the motorcycle depreciates according to an exponential decay function, with corresponding to the purchase date, the decay rate is Response area per year.
Answers
GPT-4o mini
To find the decay rate for the motorcycle's value, we can use the formula for exponential decay, which is:
\[
V(t) = V_0 \cdot e^{-kt}
\]
Where:
- \( V(t) \) is the value after time \( t \),
- \( V_0 \) is the initial value (purchase price),
- \( k \) is the decay constant,
- \( t \) is the time in years,
- \( e \) is the base of the natural logarithm.
Given:
- \( V_0 = 5800 \)
- \( V(4) = 4926.21 \)
- \( t = 4 \)
We can plug in the values into the equation to solve for \( k \):
\[
4926.21 = 5800 \cdot e^{-4k}
\]
Now, divide both sides by 5800 to isolate the exponential term:
\[
\frac{4926.21}{5800} = e^{-4k}
\]
Calculating the left side:
\[
\frac{4926.21}{5800} \approx 0.8504
\]
Thus we have:
\[
0.8504 = e^{-4k}
\]
Now take the natural logarithm of both sides:
\[
\ln(0.8504) = -4k
\]
Then solve for \( k \):
\[
k = -\frac{\ln(0.8504)}{4}
\]
Calculating \( \ln(0.8504) \):
\[
\ln(0.8504) \approx -0.1625
\]
Substituting this back in, we get:
\[
k \approx -\frac{-0.1625}{4} \approx 0.040625
\]
Thus, the decay rate is approximately 0.040625, or 4.0625% per year.
Hence, the final response that completes the sentence is:
**4.0625%**
\[
V(t) = V_0 \cdot e^{-kt}
\]
Where:
- \( V(t) \) is the value after time \( t \),
- \( V_0 \) is the initial value (purchase price),
- \( k \) is the decay constant,
- \( t \) is the time in years,
- \( e \) is the base of the natural logarithm.
Given:
- \( V_0 = 5800 \)
- \( V(4) = 4926.21 \)
- \( t = 4 \)
We can plug in the values into the equation to solve for \( k \):
\[
4926.21 = 5800 \cdot e^{-4k}
\]
Now, divide both sides by 5800 to isolate the exponential term:
\[
\frac{4926.21}{5800} = e^{-4k}
\]
Calculating the left side:
\[
\frac{4926.21}{5800} \approx 0.8504
\]
Thus we have:
\[
0.8504 = e^{-4k}
\]
Now take the natural logarithm of both sides:
\[
\ln(0.8504) = -4k
\]
Then solve for \( k \):
\[
k = -\frac{\ln(0.8504)}{4}
\]
Calculating \( \ln(0.8504) \):
\[
\ln(0.8504) \approx -0.1625
\]
Substituting this back in, we get:
\[
k \approx -\frac{-0.1625}{4} \approx 0.040625
\]
Thus, the decay rate is approximately 0.040625, or 4.0625% per year.
Hence, the final response that completes the sentence is:
**4.0625%**