Question

If 44000 dollars is invested at an interest rate of 9 percent per year, find the value of the investment at the end of 5 years for the following compounding methods, to the nearest cent.

(a) Annual: $

(b) Semiannual: $

(c) Monthly: $

(d) Daily: $

Answer 1

The value of a $44,000 investment at a 9% interest rate compounded annually, semiannually, monthly, and daily after 5 years is $68,233.60, $69,042.56, $69,715.87, and $69,836.58, respectively.

Step-by-step explanation:

The value of an investment compounded at different intervals can be calculated using the compound interest formula:

A = P(1 + ​r/n)^{nt}

Where:

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

To solve the problem, we need to calculate the future value of a $44,000 investment at a 9% interest rate for each compounding method:

1. Annual Compounding:
   A = 44000(1 + 0.09/1)^(1*5) = $68,233.60
2. Semiannual Compounding:
   A = 44000(1 + 0.09/2)^(2*5) = $69,042.56
3. Monthly Compounding:
   A = 44000(1 + 0.09/12)^(12*5) = $69,715.87
4. Daily Compounding:
   A = 44000(1 + 0.09/365)^(365*5) = $69,836.58

All values are rounded to the nearest cent as required.

Answer 2

(a) Annual: $67,699.45

(b) Semiannual: $68,330.65

(c) Monthly: $68,889.97

(d) Daily: $69,001.91

Step-by-step explanation:

To determine the value of a $44,000 investment at the end of 5 years with an annual interest rate of 9%, we can use the compound interest formula.

`\boxed{\begin{array}{l}\underline{\textsf{Compound Interest Formula}}\\\\A=P\left(1+(r)/(n)\right)^(nt)\\\\\textsf{where:}\\\phantom{ww}\bullet\;\;\textsf{$A$ is the final amount.}\\\phantom{ww}\bullet\;\;\textsf{$P$ is the principal amount.}\\\phantom{ww}\bullet\;\;\textsf{$r$ is the interest rate (in decimal form).}\\\phantom{ww}\bullet\;\;\textsf{$n$ is the number of times interest is applied per year.}\\\phantom{ww}\bullet\;\;\textsf{$t$ is the time (in years).}\end{array}}`

In this case:

• P = $44,000
• r = 9% = 0.09
• t = 5 years

Substitute these values into the compound interest formula:

`A=44000\left(1+(0.09)/(n)\right)^(5n)`

(a) To find the value of the investment (A) when the interest is compounded annually, substitute n = 1 into the equation:

`A=44000\left(1+(0.09)/(1)\right)^(5\cdot 1)`

`A=44000\left(1.09\right)^(5)`

`A=67699.4540156`

`A=\$67\:699.45`

(b) To find the value of the investment (A) when the interest is compounded semi-annually, substitute n = 2 into the equation:

`A=44000\left(1+(0.09)/(2)\right)^(5\cdot 2)`

`A=44000\left(1.045\right)^(10)`

`A=68330.654556...`

`A=\$68\:330.65`

(c) To find the value of the investment (A) when the interest is compounded monthly, substitute n = 12 into the equation:

`A=44000\left(1+(0.09)/(12)\right)^(5\cdot 12)`

`A=44000\left(1.0075\right)^(60)`

`A=68889.9651854...`

`A=\$68\:889.97`

(d) To find the value of the investment (A) when the interest is compounded daily, substitute n = 365 into the equation:

`A=44000\left(1+(0.09)/(365)\right)^(5\cdot 365)`

`A=44000\left(1.0002465753...\right)^(1825)`

`A=69001.9084965...`

`A=\$69\:001.91`