\(\displaystyle A=P(1+\frac{r}{n})^{nt}\)

where \(\displaystyle P\) is the principal, \(\displaystyle n\) is the number of times per year interest is compounded, \(\displaystyle t\) is the time in years, and \(\displaystyle r\) is the interest rate.

\(\displaystyle A=5000(1+\frac{0.06}{12})^{(12)(5)}=6744\)

Recall the formula for compound interest:

\(\displaystyle Amount=P(1+r/n)^{nt}\), where n is the number of periods per year, r is the annual interest rate, and t is the number of years.

Plug in the values given in the question:

\(\displaystyle 400,000(1+0.07/2)^{2\cdot 4}=400,000(1.035)^{8}=\)

\(\displaystyle 400,000(1.3168)=526,720\)

For compound interest, the amount Nick owes is

\(\displaystyle P(1+i)^t\)

where \(\displaystyle P\) is the principal, or starting amount of the loan ($1000), \(\displaystyle i\) is the interest rate per year (30% = .3). and \(\displaystyle t\) is the time that has passed since Nick took out the loan. (2)

We have

\(\displaystyle 1000(1+.3)^2 =1000(1.3)^2 = 1000(1.3)(1.3)=1000(1.69)=1690\)

Hence our answer is $1690.

We will make use of the formula \(\displaystyle A = P(1+i)^t\) where \(\displaystyle A\) is the accumulated amount, \(\displaystyle P\) is the starting amount, \(\displaystyle i\) is the rate of interest, and \(\displaystyle t\) is the time in year the money is invested.

At the beginning, Casey starts with $1000, at an interest rate of 10% (or .1) and saves his money for 2 years. So after 2 years, he has

\(\displaystyle 1000(1+.1)^2 = 1000(1.1)^2 = 1000(1.21) = 1210\) dollars in his account.

After he sees the $1210, he deposits $100 more, and then waits one more year.

Now \(\displaystyle P\) becomes 1310. And after this 3rd year, Casey has

\(\displaystyle 1310(1.1)^1 = 1441\) in his account.

The accumulated amount at the end of 3 years will be \(\displaystyle $10,000 x (1.1)^3\).

It is easier to find the correct answer by using the following approach:

Calculate the amount accumulated at the end of each year. (Note that the interest is compounded, so use the amount accumulated at the end of the previous year to calculate the interest for the next year.)

At the end of year 1

\(\displaystyle Amount\:accumulated= 10000 x (1+0.1) = 10000+1000=11,000\)

At the end of year 2

\(\displaystyle Amount\:accumulated= 11000 x (1+0.1) = 11000+1100=12,100\)

At the end of year 3

\(\displaystyle Amount\:accumulated= 12100 x (1+0.1) = 12100+1210=13,310\)

This problem simply ask for the amount of interest received one year after having invested this money. We could, for this problem, either use the simple interest formula or the compounded interest formula, which is \(\displaystyle P\left(1+\frac{r}{c}\right)^{n\cdot c}-P\), where \(\displaystyle P\) is the principal,  \(\displaystyle r\) the rate, \(\displaystyle c\) the number of compounding periods and \(\displaystyle n\) the number of years for which we invest. We substract \(\displaystyle P\) because we only want the amount of interest, not the total value at the end of \(\displaystyle n\) periods. Note that  \(\displaystyle r\) is positive for increases and negative for decreases.

Applying this formula we get, 

\(\displaystyle 45,000\left(1+\frac{0.06}{1}\right)^{1\cdot 1}-45,000\), giving us \(\displaystyle 2700\).

We apply the compound interest rate formula

\(\displaystyle A=P(1+r)^t\)

where P=principal, r=rate, and t=time.

Pluggin in our values we get

 \(\displaystyle 45,000(1+0.06)^{2}\), or \(\displaystyle 50,562\).

The way we should treat this problem is as an equation and plug in what we know in the compounded interest formula, as follows: \(\displaystyle 100,000(1+r)^{2}=132,250\). By manipulating the terms we get: \(\displaystyle (1+r)^{2}= 1.3225\). Now, since it would be too complicated to solve this quadratic equation, we should just try with the values in the answers. For example let's try with the square of \(\displaystyle 1.12\), which turns out to be \(\displaystyle 1.2544\), therefore it is too small and we should look for a larger rate. Let's try until we find the right answer. Remember when you test answers to find the right answer, make sure you go slow so you don't have to test twice in case you would make an error.

The formula for compound interest is 

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

where \(\displaystyle A\) is the current, or accrued, value of the investment, \(\displaystyle P\) is the initial amount invested, or principal, \(\displaystyle r\) is the annual rate expressed as a decimal, \(\displaystyle n\) is the number of periods per year, and \(\displaystyle t\) is the number of years.

In this scenario, 

\(\displaystyle A = 6,400 , r=0.036, n = 12, t = 10\),

so the equation becomes

\(\displaystyle 6,400 = P \left \(1+ \frac{0.036}{12} \right \)^{12 \cdot 10}\)

\(\displaystyle 6,400 = P \left \(1+0.003 \right \)^{12 0}\)

\(\displaystyle 6,400 = P \left \(1 .003 \right \)^{12 0}\)

\(\displaystyle P = \frac{6,400}{\left \(1 .003 \right \)^{12 0}}\)

The formula for continuously compounded interest is 

\(\displaystyle A = P e^{rt}\)

where \(\displaystyle A\) is the current, or accrued, value of the investment, \(\displaystyle P\) is the initial amount invested, or principal, \(\displaystyle r\) is the annual rate expressed as a decimal, and \(\displaystyle t\) is the number of years.

In this scenario, 

\(\displaystyle P = 4,000 , r=0.04 , t = 18\),

so

\(\displaystyle A = 4,000 e^{0.04 \cdot 18} = 4,000 e^{0.72}\)