Question
The radius of a large balloon after it is punctured is represented by the following table:
Time (seconds) Radius (cm)
\[0\]
\[401\]
\[5\]
\[352\]
\[10\]
\[302\]
\[15\]
\[250\]
\[20\]
\[199\]
\[25\]
\[149\]
Which model for
\[R(t)\], the radius of the balloon
\[t\] seconds after it's punctured, best fits the data?
Choose 1 answer:
Choose 1 answer:
(Choice A)
\[R(t)=401\cdot(0.75)^t\]
A
\[R(t)=401\cdot(0.75)^t\]
(Choice B)
\[R(t)=401-50t\]
B
\[R(t)=401-50t\]
(Choice C)
\[R(t)=401\cdot(0.88)^t\]
C
\[R(t)=401\cdot(0.88)^t\]
(Choice D)
\[R(t)=401-10t\]
D
\[R(t)=401-10t\]
Time (seconds) Radius (cm)
\[0\]
\[401\]
\[5\]
\[352\]
\[10\]
\[302\]
\[15\]
\[250\]
\[20\]
\[199\]
\[25\]
\[149\]
Which model for
\[R(t)\], the radius of the balloon
\[t\] seconds after it's punctured, best fits the data?
Choose 1 answer:
Choose 1 answer:
(Choice A)
\[R(t)=401\cdot(0.75)^t\]
A
\[R(t)=401\cdot(0.75)^t\]
(Choice B)
\[R(t)=401-50t\]
B
\[R(t)=401-50t\]
(Choice C)
\[R(t)=401\cdot(0.88)^t\]
C
\[R(t)=401\cdot(0.88)^t\]
(Choice D)
\[R(t)=401-10t\]
D
\[R(t)=401-10t\]
Answers
GPT-4o mini
To determine which model best fits the data for the radius of the balloon over time after it is punctured, we can analyze the values given in the table against the equations provided in the choices.
First, we look at the values from the table:
- At \(t = 0\), \(R(0) = 401\)
- At \(t = 5\), \(R(5) = 352\)
- At \(t = 10\), \(R(10) = 302\)
- At \(t = 15\), \(R(15) = 250\)
- At \(t = 20\), \(R(20) = 199\)
- At \(t = 25\), \(R(25) = 149\)
Now let's assess each model based on the pattern of the data:
1. **Exponential Decay Models:**
- **Choice A:** \(R(t) = 401 \cdot (0.75)^t\)
- **Choice C:** \(R(t) = 401 \cdot (0.88)^t\)
These models suggest that the radius decreases by a constant percentage (relative decrease) each second. We would calculate a few values:
For choice A:
- \(R(5) = 401 \cdot (0.75)^5 \approx 401 \cdot 0.2373 \approx 95.1\) (Not accurate)
For choice C:
- \(R(5) = 401 \cdot (0.88)^5 \approx 401 \cdot 0.5277 \approx 211.0\) (Not accurate)
Both exponential models don't match the initial points closely.
2. **Linear Models:**
- **Choice B:** \(R(t) = 401 - 50t\)
- **Choice D:** \(R(t) = 401 - 10t\)
These models suggest that the radius decreases by a constant value each second. We can calculate:
For choice B:
- \(R(5) = 401 - 50 \cdot 5 = 401 - 250 = 151\)
- \(R(10) = 401 - 50 \cdot 10 = 401 - 500 = -99\) (Not valid for negative radius)
For choice D:
- \(R(5) = 401 - 10 \cdot 5 = 401 - 50 = 351\) (Very close to 352)
- \(R(10) = 401 - 10 \cdot 10 = 401 - 100 = 301\) (Very close to 302)
- \(R(15) = 401 - 10 \cdot 15 = 401 - 150 = 251\) (Very close to 250)
- \(R(20) = 401 - 10 \cdot 20 = 401 - 200 = 201\) (Very close to 199)
- \(R(25) = 401 - 10 \cdot 25 = 401 - 250 = 151\) (Very close to 149)
The linear decay model D fits the data points more accurately. Hence, the best model for \(R(t)\), the radius of the balloon \(t\) seconds after it's punctured, is:
**Choice D: \[R(t) = 401 - 10t\]**.
First, we look at the values from the table:
- At \(t = 0\), \(R(0) = 401\)
- At \(t = 5\), \(R(5) = 352\)
- At \(t = 10\), \(R(10) = 302\)
- At \(t = 15\), \(R(15) = 250\)
- At \(t = 20\), \(R(20) = 199\)
- At \(t = 25\), \(R(25) = 149\)
Now let's assess each model based on the pattern of the data:
1. **Exponential Decay Models:**
- **Choice A:** \(R(t) = 401 \cdot (0.75)^t\)
- **Choice C:** \(R(t) = 401 \cdot (0.88)^t\)
These models suggest that the radius decreases by a constant percentage (relative decrease) each second. We would calculate a few values:
For choice A:
- \(R(5) = 401 \cdot (0.75)^5 \approx 401 \cdot 0.2373 \approx 95.1\) (Not accurate)
For choice C:
- \(R(5) = 401 \cdot (0.88)^5 \approx 401 \cdot 0.5277 \approx 211.0\) (Not accurate)
Both exponential models don't match the initial points closely.
2. **Linear Models:**
- **Choice B:** \(R(t) = 401 - 50t\)
- **Choice D:** \(R(t) = 401 - 10t\)
These models suggest that the radius decreases by a constant value each second. We can calculate:
For choice B:
- \(R(5) = 401 - 50 \cdot 5 = 401 - 250 = 151\)
- \(R(10) = 401 - 50 \cdot 10 = 401 - 500 = -99\) (Not valid for negative radius)
For choice D:
- \(R(5) = 401 - 10 \cdot 5 = 401 - 50 = 351\) (Very close to 352)
- \(R(10) = 401 - 10 \cdot 10 = 401 - 100 = 301\) (Very close to 302)
- \(R(15) = 401 - 10 \cdot 15 = 401 - 150 = 251\) (Very close to 250)
- \(R(20) = 401 - 10 \cdot 20 = 401 - 200 = 201\) (Very close to 199)
- \(R(25) = 401 - 10 \cdot 25 = 401 - 250 = 151\) (Very close to 149)
The linear decay model D fits the data points more accurately. Hence, the best model for \(R(t)\), the radius of the balloon \(t\) seconds after it's punctured, is:
**Choice D: \[R(t) = 401 - 10t\]**.