Suppose $2,100.00 dollars is invested in an account for 11 years. Find the value of the investment at the end of 11 years if the interest is:

(a) 4.4% compounded annually:  
(b) 4.4% compounded semiannually:  
(c) 4.4% compounded quarterly:  
(d) 4.4% compounded monthly:  
(e) 4.4% compounded daily (*ignore leap years*):  

To solve these problems, we can use the compound interest formula:

\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]

Where:
- $ A $ is the amount of money accumulated after $ t $ years, including interest.
- $ P $ is the principal amount (the initial investment).
- $ r $ is the annual interest rate (decimal form).
- $ n $ is the number of times the interest is compounded per year.
- $ t $ is the time the money is invested for (in years).

In this case:
- $ P = 2100 $
- $ r = 4.4\% = 0.044 $
- $ t = 11 $ years

### (a) **Compounded Annually**:
When interest is compounded annually, $ n = 1 $.

\[
A = 2100 \left(1 + \frac{0.044}{1}\right)^{1 \times 11} = 2100 \left(1.044\right)^{11}
\]

### (b) **Compounded Semiannually**:
When interest is compounded semiannually, $ n = 2 $.

\[
A = 2100 \left(1 + \frac{0.044}{2}\right)^{2 \times 11} = 2100 \left(1 + 0.022\right)^{22}
\]

### (c) **Compounded Quarterly**:
When interest is compounded quarterly, $ n = 4 $.

\[
A = 2100 \left(1 + \frac{0.044}{4}\right)^{4 \times 11} = 2100 \left(1 + 0.011\right)^{44}
\]

### (d) **Compounded Monthly**:
When interest is compounded monthly, $ n = 12 $.

\[
A = 2100 \left(1 + \frac{0.044}{12}\right)^{12 \times 11} = 2100 \left(1 + 0.0036667\right)^{132}
\]

### (e) **Compounded Daily**:
When interest is compounded daily, $ n = 365 $.

\[
A = 2100 \left(1 + \frac{0.044}{365}\right)^{365 \times 11} = 2100 \left(1 + 0.00012055\right)^{4015}
\]

Now, let's calculate these values.Here are the values of the investment at the end of 11 years with different compounding frequencies:

- (a) Compounded Annually: **\$3,372.29**
- (b) Compounded Semiannually: **\$3,389.53**
- (c) Compounded Quarterly: **\$3,398.37**
- (d) Compounded Monthly: **\$3,404.34**
- (e) Compounded Daily: **\$3,407.26**

Would you like more details on how the formula works or have any other questions?

Here are 5 related questions:
1. How does the frequency of compounding affect the total interest earned?
2. What happens to the final amount if the interest rate increases to 5%?
3. How would the investment grow if it were compounded continuously?
4. Can we calculate the interest earned after 5 years instead of 11?
5. How does the principal amount affect the total investment growth?

**Tip**: The more frequently the interest is compounded, the larger the investment grows, though the differences become smaller as compounding frequency increases.

Suppose $2,100.00 dollars is invested in an account for 11 years. Find the value of the investment at the end of 11 years if the interest is: (a) 4.4% compounded annually; (b) 4.4% compounded semiannually; (c) 4.4% compounded quarterly; (d) 4.4% compounded monthly; (e) 4.4% compounded daily (ignore leap years).

Learn how to calculate the future value of a $2,100 investment with a 4.4% interest rate compounded annually, semiannually, quarterly, monthly, and daily. This detailed solution uses the compound interest formula to show how different compounding frequencies affect investment growth over 11 years. Suitable for Grades 10-12.

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