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Use the compound interest formulas \(A=P\left(1+\frac{r}{n}\right)^{n t}\) and \(A=P e^{n t}\) to solve. Round answers to the nearest cent. Find the accumulated value of an investment of \(\$ 10,000\) for 5 years at an interest rate of \(5.5 \%\) if the money is a. compounded semiannually; b. compounded quarterly; c. compounded monthly; d. compounded continuously.

Short Answer

Expert verified
The accumulated value of the $10,000 investment would be approximately \$14106.41 if compounded semiannually, \$14164.36 if compounded quarterly, \$14189.94 if compounded monthly, and \$14294.08 if compounded continuously.

Step by step solution

01

Compounded Semiannually

For an investment compounded semiannually, interest is compounded twice a year, so 'n' is 2. Plugging these values into the formula gives \(A = 10000 \left(1+\frac{0.055}{2}\right)^{2*5} = \$ 14106.41\).
02

Compounded Quarterly

If the investment is compounded quarterly, interest is compounded four times a year, so 'n' is 4. Using these values in the formula gives \(A = 10000 \left(1+\frac{0.055}{4}\right)^{4*5} = \$ 14164.36\).
03

Compounded Monthly

For monthly compounding, interest is compounded twelve times a year, so 'n' is 12. Using these values in the formula gives \(A = 10000 \left(1+\frac{0.055}{12}\right)^{12*5} = \$14189.94\).
04

Compounded Continuously

For continuously compounded interest, using the given formula \(A=Pe^{rt}\), we get \(A = 10000 e^{0.055*5} = \$ 14294.08\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Compounded Semiannually
When you're dealing with investments or savings, understanding how compound interest works is key. Let's break down what it means when your money is compounded semiannually. Basically, the bank or financial institution will calculate and add interest to your principal amount twice a year.

Putting this into a formula, you'd use the compound interest formula:
\[ A = P\left(1 + \frac{r}{n}\right)^{nt} \]
where \( A \) is the final amount, \( P \) is the principal, \( r \) is the annual interest rate in decimal form, \( n \) is the number of compounding periods per year, and \( t \) is the time in years. For semiannual compounding, \( n \) would be 2.

So, with an annual interest rate of 5.5% and an initial deposit of \(10,000 over 5 years, we plug the values into the formula as \( n = 2 \), \( r = 0.055 \), and \( t = 5 \) to find that the investment will grow to approximately \)14,106.41.

This way of compounding is relatively common for things like savings accounts and certificates of deposit (CDs). It's important to know because it affects how much money you'll earn over time.
Compounded Quarterly
If an investor chooses to have their money compounded quarterly, this means that the interest is calculated and added to the account four times a year. This can lead to more earnings compared to semiannual compounding if the interest rate is the same.

Again, we use the compound interest formula, setting \( n = 4 \) since interest will be compounded four times a year:
\[ A = P\left(1 + \frac{r}{n}\right)^{nt} \]
For our example with a \(10,000 investment at a 5.5% interest rate over 5 years, the calculation would look like this:
\[ A = 10000\left(1 + \frac{0.055}{4}\right)^{4 \times 5} \]The result is a slightly higher ending balance of \)14,164.36 compared to semiannual compounding. Quarterly compounding is a common option offered by financial institutions and can significantly impact your investment earnings over time.
Compounded Monthly
Taking it a step further, compounded monthly involves breaking down the interest calculations into twelve periods within a year. This method of compounding can lead to even more growth of an investment since interest is added more frequently.

The compound interest formula for monthly compounding is similar to the other periods, but now \( n = 12 \):
\[ A = P\left(1 + \frac{r}{n}\right)^{nt} \]
For our ongoing example of \(10,000 at a 5.5% interest rate for 5 years, the math would be:
\[ A = 10000\left(1 + \frac{0.055}{12}\right)^{12 \times 5} \]As a result, the investment grows slightly more to a value of \)14,189.94. With monthly compounding, you can see how the additional compounding periods can affect your savings, particularly over long periods.
Compounded Continuously
The concept of compounded continuously may sound a bit abstract, but it's an important theoretical premise in the world of finance. When an investment is compounded continuously, interest is theoretically added to the principal at every possible moment.

For continuous compounding, the formula you use is different:
\[ A = Pe^{rt} \]
Where \( e \) is Euler's number (approximately equal to 2.71828). With a \(10,000 investment at a 5.5% interest rate over 5 years, the compound interest formula is:
\[ A = 10000e^{0.055 \times 5} \]Which equals roughly \)14,294.08. While continuous compounding is not practical in most real-world scenarios, it's especially relevant for certain financial products like continuous compounding bonds or certain interest rate derivates. It represents the upper limit of how much compound interest can accrue and serves as a useful benchmark for comparing other compounding frequencies.

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Most popular questions from this chapter

Use a calculator with \(a\left[y^{x}\right]\) key or \(a \square\) key to solve. India is currently one of the world's fastest-growing countries. By \(2040,\) the population of India will be larger than the population of China; by \(2050,\) nearly one-third of the world's population will live in these two countries alone. The exponential function \(f(x)=574(1.026)^{x}\) models the population of India, \(f(x),\) in millions, \(x\) years after 1974 a. Substitute 0 for \(x\) and, without using a calculator, find India's population in 1974 b. Substitute 27 for \(x\) and use your calculator to find India's population, to the nearest million, in the year 2001 as modeled by this function. c. Find India's population, to the nearest million, in the year 2028 as predicted by this function. d. Find India's population, to the nearest million, in the year 2055 as predicted by this function. e. What appears to be happening to India's population every 27 years?

Exercises \(153-155\) will help you prepare for the material covered in the next section. The formula \(A=10 e^{-0.003 t}\) models the population of Hungary, \(A\), in millions, \(t\) years after 2006 a. Find Hungary's population, in millions, for 2006,2007 \(2008,\) and \(2009 .\) Round to two decimal places. b. Is Hungary's population increasing or decreasing?

Hurricanes are one of nature's most destructive forces. These low-pressure areas often have diameters of over 500 miles. The function \(f(x)=0.48 \ln (x+1)+27\) models the barometric air pressure, \(f(x),\) in inches of mercury, at a distance of \(x\) miles from the eye of a hurricane. Use this function to solve Exercises \(133-134\) Graph the function in a [0,500,50] by [27,30,1] viewing rectangle. What does the shape of the graph indicate about barometric air pressure as the distance from the eye increases?

Exercises \(153-155\) will help you prepare for the material covered in the next section. U.S. soldiers fight Russian troops who have invaded New York City. Incoming missiles from Russian submarines and warships ravage the Manhattan skyline. It's just another scenario for the multi-billion-dollar video games Call of Duty, which have sold more than 100 million games since the franchise's birth in 2003 The table shows the annual retail sales for Call of Duty video games from 2004 through 2010 . Create a scatter plot for the data. Based on the shape of the scatter plot, would a logarithmic function, an exponential function, or a linear function be the best choice for modeling the data? $$\begin{array}{cc} \hline \text { Year } & \begin{array}{c} \text { Retail Sales } \\ \text { (millions of dollars) } \end{array} \\ \hline 2004 & 56 \\ 2005 & 101 \\ 2006 & 196 \\ 2007 & 352 \\ 2008 & 436 \\ 2009 & 778 \\ 2010 & 980 \end{array}$$

From 1970 through \(2010 .\) The data are shown again in the table. Use all five data points to solve Exercises \(70-74\). $$\begin{array}{cc}\hline \begin{array}{c}x, \text { Number of Years } \\\\\text { after } 1969 \end{array} & \begin{array}{c}y, \text { U.S. Population } \\\\\text { (millions) }\end{array} \\ \hline 1(1970) & 203.3 \\\11(1980) & 226.5 \\\21(1990) & 248.7 \\\31(2000) & 281.4 \\\41(2010) & 308.7 \end{array}$$ Use your graphing utility's linear regression option to obtain a model of the form \(y=a x+b\) that fits the data. How well does the correlation coefficient, \(r,\) indicate that the model fits the data?

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