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Answer :
To determine the equation for [tex]\( P(t) \)[/tex], the balance of the Plant Bank account [tex]\( t \)[/tex] years after a deposit is made, we need to identify the correct formula based on compound interest.
From the given options, it seems we are dealing with compound interest where:
- A principal amount of [tex]$10,000 is deposited.
- The balance is compounded periodically, often monthly in such problems.
Here, we look at the form of a compound interest formula:
\[ A = P\left(1 + \frac{r}{n}\right)^{n \cdot t} \]
Where:
- \( A \) is the amount of money accumulated after \( n \) years, including interest.
- \( P \) is the principal amount (initial deposit), here it is $[/tex]10,000.
- [tex]\( r \)[/tex] is the annual interest rate (as a decimal).
- [tex]\( n \)[/tex] is the number of times that interest is compounded per year.
- [tex]\( t \)[/tex] is the time the money is invested for in years.
Now let's evaluate the choices provided:
1. [tex]\( D(t) = 10000\left(1 + \frac{0.2}{12}\right)^{12 \cdot t} \)[/tex]
- This one uses an annual interest rate of 20% (0.2), compounded monthly.
- The [tex]\( \frac{0.2}{12} \)[/tex] represents the monthly interest rate.
- [tex]\( 12 \cdot t \)[/tex] represents the total number of compounding periods over [tex]\( t \)[/tex] years.
2. [tex]\( D(t) = 10000\left(1 + \frac{0.02}{12}\right)^{12 \cdot t} \)[/tex]
- This uses an annual interest rate of 2% (0.02), also compounded monthly.
3. [tex]\( D(t) = 10000(1 + 0.02)^t \)[/tex]
- This is a simple yearly compounding with an annual interest rate of 2%.
4. [tex]\( D(t) = 10000(1 + 0.2)^t \)[/tex]
- This is a simple yearly compounding with an annual interest rate of 20%.
5. [tex]\( D(t) = 10000(0.2) t \)[/tex]
- This seems to be a linear growth, possibly indicating simple interest at a 20% rate rather than compound interest.
6. [tex]\( D(t) = 10000(0.02) t \)[/tex]
- Similarly, this is a linear model for simple interest at 2%.
Considering the context is likely compound interest with monthly compounding (a common real-world scenario), the first option:
[tex]\[ D(t) = 10000\left(1 + \frac{0.2}{12}\right)^{12 \cdot t} \][/tex]
This option correctly models compound interest with monthly compounding at an annual rate of 20%. Therefore, this is the most appropriate equation for [tex]\( P(t) \)[/tex], the Plant Bank balance after [tex]\( t \)[/tex] years.
From the given options, it seems we are dealing with compound interest where:
- A principal amount of [tex]$10,000 is deposited.
- The balance is compounded periodically, often monthly in such problems.
Here, we look at the form of a compound interest formula:
\[ A = P\left(1 + \frac{r}{n}\right)^{n \cdot t} \]
Where:
- \( A \) is the amount of money accumulated after \( n \) years, including interest.
- \( P \) is the principal amount (initial deposit), here it is $[/tex]10,000.
- [tex]\( r \)[/tex] is the annual interest rate (as a decimal).
- [tex]\( n \)[/tex] is the number of times that interest is compounded per year.
- [tex]\( t \)[/tex] is the time the money is invested for in years.
Now let's evaluate the choices provided:
1. [tex]\( D(t) = 10000\left(1 + \frac{0.2}{12}\right)^{12 \cdot t} \)[/tex]
- This one uses an annual interest rate of 20% (0.2), compounded monthly.
- The [tex]\( \frac{0.2}{12} \)[/tex] represents the monthly interest rate.
- [tex]\( 12 \cdot t \)[/tex] represents the total number of compounding periods over [tex]\( t \)[/tex] years.
2. [tex]\( D(t) = 10000\left(1 + \frac{0.02}{12}\right)^{12 \cdot t} \)[/tex]
- This uses an annual interest rate of 2% (0.02), also compounded monthly.
3. [tex]\( D(t) = 10000(1 + 0.02)^t \)[/tex]
- This is a simple yearly compounding with an annual interest rate of 2%.
4. [tex]\( D(t) = 10000(1 + 0.2)^t \)[/tex]
- This is a simple yearly compounding with an annual interest rate of 20%.
5. [tex]\( D(t) = 10000(0.2) t \)[/tex]
- This seems to be a linear growth, possibly indicating simple interest at a 20% rate rather than compound interest.
6. [tex]\( D(t) = 10000(0.02) t \)[/tex]
- Similarly, this is a linear model for simple interest at 2%.
Considering the context is likely compound interest with monthly compounding (a common real-world scenario), the first option:
[tex]\[ D(t) = 10000\left(1 + \frac{0.2}{12}\right)^{12 \cdot t} \][/tex]
This option correctly models compound interest with monthly compounding at an annual rate of 20%. Therefore, this is the most appropriate equation for [tex]\( P(t) \)[/tex], the Plant Bank balance after [tex]\( t \)[/tex] years.
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