A baseball card sold for $300 in 1979 and was sold again in 1986 for $481. Assume that the growth in the value V of the collector's item was exponenti a) Find the value k of the exponential growth rate. Assume V₁ = 300. k= 0.067 (Round to the nearest thousandth.) b) Find the exponential growth function in terms of t, where t is the number of years since 1979. V(t)= example OCEED Get more help. Clear all Check answe
A baseball card sold for $300 in 1979 and was sold again in 1986 for $481. Assume that the growth in the value V of the collector's item was exponenti a) Find the value k of the exponential growth rate. Assume V₁ = 300. k= 0.067 (Round to the nearest thousandth.) b) Find the exponential growth function in terms of t, where t is the number of years since 1979. V(t)= example OCEED Get more help. Clear all Check answe
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![## Exponential Growth of a Collector's Item
### Scenario
A baseball card sold for $300 in 1979 and was sold again in 1986 for $481. Assume that the growth in the value \( V \) of the collector's item was exponential.
### Questions:
1. **Find the value \( k \) of the exponential growth rate.**
- Initial value \( V_0 = 300 \).
- Calculated \( k \) is \( 0.067 \).
- Instruction: Round to the nearest thousandth.
2. **Find the exponential growth function in terms of \( t \), where \( t \) is the number of years since 1979.**
- Express as \( V(t) = \)
### Explanation:
The problem requires calculating the exponential growth rate from the given data and using this rate to derive a growth function over time.
### Steps for Solution:
1. **Calculate the Growth Rate \( k \):**
- Use the formula for exponential growth, which relates initial and final values with the growth rate over time:
\[ V = V_0 \times e^{kt} \]
- Substitute known values: \( V_0 = 300 \), \( V = 481 \), \( t = 7 \) (years between 1979 and 1986).
- Solve for \( k \).
2. **Derive the Exponential Growth Function:**
- Insert \( k \) into the general exponential growth formula:
\[ V(t) = 300 \times e^{0.067t} \]
This mathematical modeling of exponential growth can be applied to various scenarios in finance and economics to predict future values of items or investments based on past performance data.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F400cbf84-1293-43d3-b625-14e2acb9a968%2F5eba1627-9bf6-442b-84c4-6645d80cc8d2%2Fcgbst_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Exponential Growth of a Collector's Item
### Scenario
A baseball card sold for $300 in 1979 and was sold again in 1986 for $481. Assume that the growth in the value \( V \) of the collector's item was exponential.
### Questions:
1. **Find the value \( k \) of the exponential growth rate.**
- Initial value \( V_0 = 300 \).
- Calculated \( k \) is \( 0.067 \).
- Instruction: Round to the nearest thousandth.
2. **Find the exponential growth function in terms of \( t \), where \( t \) is the number of years since 1979.**
- Express as \( V(t) = \)
### Explanation:
The problem requires calculating the exponential growth rate from the given data and using this rate to derive a growth function over time.
### Steps for Solution:
1. **Calculate the Growth Rate \( k \):**
- Use the formula for exponential growth, which relates initial and final values with the growth rate over time:
\[ V = V_0 \times e^{kt} \]
- Substitute known values: \( V_0 = 300 \), \( V = 481 \), \( t = 7 \) (years between 1979 and 1986).
- Solve for \( k \).
2. **Derive the Exponential Growth Function:**
- Insert \( k \) into the general exponential growth formula:
\[ V(t) = 300 \times e^{0.067t} \]
This mathematical modeling of exponential growth can be applied to various scenarios in finance and economics to predict future values of items or investments based on past performance data.
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