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 of the motorcycle's value over time using an exponential decay function, we can use the formula for exponential decay:
\[
V(t) = V_0 e^{-kt}
\]
where:
- \( V(t) \) is the value at time \( t \),
- \( V_0 \) is the initial value (purchase price),
- \( k \) is the decay rate,
- \( t \) is the time in years,
- \( e \) is the base of the natural logarithm (approximately equal to 2.71828).
Given:
- \( V_0 = 5800 \)
- \( V(4) = 4926.21 \)
- \( t = 4 \)
We can substitute these values into the formula and solve for \( k \):
\[
4926.21 = 5800 e^{-4k}
\]
Now, divide both sides by 5800:
\[
\frac{4926.21}{5800} = e^{-4k}
\]
Calculate the left side:
\[
0.84935 \approx e^{-4k}
\]
Next, take the natural logarithm of both sides:
\[
\ln(0.84935) = -4k
\]
Calculate \( \ln(0.84935) \):
\[
\ln(0.84935) \approx -0.1625
\]
Now, solve for \( k \):
\[
-0.1625 = -4k
\]
\[
k = \frac{0.1625}{4} \approx 0.040625
\]
Thus, the decay rate is approximately \( 0.040625 \) per year.
In response to your task, the decay rate is approximately **0.040625**.
\[
V(t) = V_0 e^{-kt}
\]
where:
- \( V(t) \) is the value at time \( t \),
- \( V_0 \) is the initial value (purchase price),
- \( k \) is the decay rate,
- \( t \) is the time in years,
- \( e \) is the base of the natural logarithm (approximately equal to 2.71828).
Given:
- \( V_0 = 5800 \)
- \( V(4) = 4926.21 \)
- \( t = 4 \)
We can substitute these values into the formula and solve for \( k \):
\[
4926.21 = 5800 e^{-4k}
\]
Now, divide both sides by 5800:
\[
\frac{4926.21}{5800} = e^{-4k}
\]
Calculate the left side:
\[
0.84935 \approx e^{-4k}
\]
Next, take the natural logarithm of both sides:
\[
\ln(0.84935) = -4k
\]
Calculate \( \ln(0.84935) \):
\[
\ln(0.84935) \approx -0.1625
\]
Now, solve for \( k \):
\[
-0.1625 = -4k
\]
\[
k = \frac{0.1625}{4} \approx 0.040625
\]
Thus, the decay rate is approximately \( 0.040625 \) per year.
In response to your task, the decay rate is approximately **0.040625**.