Answer :
Sure, let's solve question 11 first.
Given:
- Principal [tex]\( P = \$9000 \)[/tex]
- Annual interest rate [tex]\( r = 2.45\% = 0.0245 \)[/tex]
- Time duration [tex]\( t = 30 \)[/tex] years
We have two different compounding methods for comparison:
1. Semiannual Compounding:
With semiannual compounding, interest is compounded twice per year.
- Compounding frequency [tex]\( n = 2 \)[/tex] times per year.
- The formula for compound interest is:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Substituting the given values:
[tex]\[ A_{semiannual} = 9000 \left(1 + \frac{0.0245}{2}\right)^{2 \times 30} \][/tex]
[tex]\[ A_{semiannual} \approx \$18,685.71 \][/tex]
2. Continuous Compounding:
With continuous compounding, interest is compounded at every instant.
- The formula for continuous compounding is:
[tex]\[ A = Pe^{rt} \][/tex]
Substituting the given values:
[tex]\[ A_{continuous} = 9000 \cdot e^{0.0245 \times 30} \][/tex]
[tex]\[ A_{continuous} \approx \$18,769.34 \][/tex]
Now, we calculate the difference between the two account balances:
[tex]\[ \text{Difference} = A_{continuous} - A_{semiannual} \][/tex]
[tex]\[ \text{Difference} \approx 18769.34 - 18685.71 \][/tex]
[tex]\[ \text{Difference} \approx \$83.63 \][/tex]
So, the difference in the account balance after 30 years when compounded continuously instead of semiannually is \[tex]$83.63. Therefore, the correct answer for Question 11 is: \[ \boxed{\$[/tex]83.63}
\]
Let me now proceed to Question 12.
Given:
[tex]\[ A = 2.400 \left( 1 + \frac{0.031}{A} \right)^4 \][/tex]
To interpret this equation, let's break it down:
- [tex]\( A \)[/tex] represents the final amount of money in the savings account.
- [tex]\( 2.400 \)[/tex] likely represents the principal amount or the initial deposit.
- [tex]\( 0.031 \)[/tex] is the interest rate applied per compounding period.
- The term [tex]\( (1 + \frac{0.031}{A}) \)[/tex] inside the expression adjusts the interest rate relative to [tex]\( A \)[/tex], which can be further analyzed based on the context of the savings account.
- The exponent [tex]\( 4 \)[/tex] indicates that the interest is compounded four times over the period being considered.
Therefore, in the context of this problem, the value [tex]\( 0.031 \)[/tex] represents the nominal interest rate that is being applied in each compounding period, adjusted in relation to the amount [tex]\( A \)[/tex].
Thus, for Question 12, the value [tex]\( 0.031 \)[/tex] represents the nominal interest rate applied to the principal.
Given:
- Principal [tex]\( P = \$9000 \)[/tex]
- Annual interest rate [tex]\( r = 2.45\% = 0.0245 \)[/tex]
- Time duration [tex]\( t = 30 \)[/tex] years
We have two different compounding methods for comparison:
1. Semiannual Compounding:
With semiannual compounding, interest is compounded twice per year.
- Compounding frequency [tex]\( n = 2 \)[/tex] times per year.
- The formula for compound interest is:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Substituting the given values:
[tex]\[ A_{semiannual} = 9000 \left(1 + \frac{0.0245}{2}\right)^{2 \times 30} \][/tex]
[tex]\[ A_{semiannual} \approx \$18,685.71 \][/tex]
2. Continuous Compounding:
With continuous compounding, interest is compounded at every instant.
- The formula for continuous compounding is:
[tex]\[ A = Pe^{rt} \][/tex]
Substituting the given values:
[tex]\[ A_{continuous} = 9000 \cdot e^{0.0245 \times 30} \][/tex]
[tex]\[ A_{continuous} \approx \$18,769.34 \][/tex]
Now, we calculate the difference between the two account balances:
[tex]\[ \text{Difference} = A_{continuous} - A_{semiannual} \][/tex]
[tex]\[ \text{Difference} \approx 18769.34 - 18685.71 \][/tex]
[tex]\[ \text{Difference} \approx \$83.63 \][/tex]
So, the difference in the account balance after 30 years when compounded continuously instead of semiannually is \[tex]$83.63. Therefore, the correct answer for Question 11 is: \[ \boxed{\$[/tex]83.63}
\]
Let me now proceed to Question 12.
Given:
[tex]\[ A = 2.400 \left( 1 + \frac{0.031}{A} \right)^4 \][/tex]
To interpret this equation, let's break it down:
- [tex]\( A \)[/tex] represents the final amount of money in the savings account.
- [tex]\( 2.400 \)[/tex] likely represents the principal amount or the initial deposit.
- [tex]\( 0.031 \)[/tex] is the interest rate applied per compounding period.
- The term [tex]\( (1 + \frac{0.031}{A}) \)[/tex] inside the expression adjusts the interest rate relative to [tex]\( A \)[/tex], which can be further analyzed based on the context of the savings account.
- The exponent [tex]\( 4 \)[/tex] indicates that the interest is compounded four times over the period being considered.
Therefore, in the context of this problem, the value [tex]\( 0.031 \)[/tex] represents the nominal interest rate that is being applied in each compounding period, adjusted in relation to the amount [tex]\( A \)[/tex].
Thus, for Question 12, the value [tex]\( 0.031 \)[/tex] represents the nominal interest rate applied to the principal.
