The formula for calculating the future value of an interest earning account is

\(\displaystyle FV = PV(1+\frac{r}{n})^{tn}\),

where

\(\displaystyle FV\)= future value,

\(\displaystyle PV\)= present value,

\(\displaystyle r\)= annual interest rate,

\(\displaystyle n\)= number of times the interest is compounded per year, and

\(\displaystyle t\)= the number of years that have passed.

The problem asks for the amount of money in the account after 5 years, with 10% interested compounded four times per year (quarterly).

Plug in the given quantities and simplify:

\(\displaystyle FV=$3500 (1+.025)^{5\cdot 4}\)

\(\displaystyle FV=3500(1.025)^{20}\)

\(\displaystyle {\color{Red} FV=\$5,735.15}\)

The formula for finding the future value of an investment is

\(\displaystyle FV=PV(1+r)^n\),

where

\(\displaystyle FV\)= future value,

\(\displaystyle PV\)= present value,

\(\displaystyle r\)= interest rate, and

\(\displaystyle n\)= number of times interest is compounded.

Plug in the given numbers and solve for the present value:

\(\displaystyle 10,000=PV(1+.05)^5\)

\(\displaystyle PV=\frac{10,000}{1.05^5}\)

\(\displaystyle {\color{Red} PV=7,835.26}\)

Initial amount = 5000

The account earns 4% compounded yearly ===> Each $1.00 will grow into $1.04.

Growth rate = 1.04

Jamie will withdraw the money after 5 years. Since the interest is compounded yearly, the number of periods is equal to the number of years the money will be in the account.

number of periods = 5

From the above information, we can calculate the amount accumulated (or final amount) after 5 years using the following formula:

final amount = initial amount * (growth rate)^{number of periods}

\(\displaystyle 5000\times(1.04)^{5}= 6083.2645\approx 6083\)

The formula for computing interest is:

Beginning Amount x ((1 + rate)^number of years) = Ending Amount After number of years

\(\displaystyle E=B(1+r)^t\)

Make sure to convert the rate from percent to number: 3% = 0.03

\(\displaystyle E=6000 (1+0.03)^{15}\)

\(\displaystyle =6000(1.5580)\)

\(\displaystyle =9347.80\)

So the answer is \(\displaystyle \$9,347.80\)

Using the equation for continuous compound interest and the given information, we get 

\(\displaystyle P = Ce^{rt}\)

     \(\displaystyle P = \text{ final principal}\)

     \(\displaystyle C = 250\)

      \(\displaystyle r = \text{ 0.06}\)

      \(\displaystyle t =3\)

\(\displaystyle P = 250e^{\left ( .06\cdot 3\right )} =250e^{.18} = 250 \cdot 1.1972 = 299\)

\(\displaystyle A = P(1+\frac{r}{n})^{nt}\)   is the compound interest formula where

P = Initial deposit = 5000

r = Interest rate = 0.12

n = Number of times interest is compounded per year = 1

t = Number of years that have passed = 5

\(\displaystyle 5000(1+.12)^5\)  \(\displaystyle =5000(1.762341683) = $8811.7084...\)

Round to the nearest cent or hundredth is \(\displaystyle \$8811.71\).

To determine Julio's account balance, we must use the interest formula given below:

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

where P is his principal (initial) investment, r is the interest rate (as a decimal), n is the number of times the interest is compounded, and t is the amount of time elapsed.

Plugging in all of our given information into the above formula - knowing that quarterly means four times a year - we get

\(\displaystyle 5000(1+\frac{0.025}{4})^{4\cdot 1}=5126.1767...\approx5126.18\)

To find the balance, B, of a continuously compounded interest account after a certain amount of time, we must use the following formula:

\(\displaystyle B=Pe^{rt}\), where P is the initial investment, r is the interest rate (as a decimal), and t is the amount of time being considered.

Plugging in all of the given information, we get

\(\displaystyle B=2000e^{0.025\cdot 2}=2102.542193\)

which rounded becomes $2102.54

The formula to find the balance, B, of a continuously compounded interest account with interest rate, r, after a certain time, t, is given by

\(\displaystyle B=Pe^{rt}\)

To solve this problem, we need to know only the initial investment (P), our final balance (three times P) and the interest rate (expressed as a decimal), 0.019.

Plugging in our known information into the formula for continuously compounded interest, we get

\(\displaystyle 3P=Pe^{0.019t}\)

We now solve for t:

\(\displaystyle 3=e^{0.019t}\)

\(\displaystyle \ln(3)=\ln(e^{0.019t})\)

Exponentiating both sides allows us to get rid of the exponential:

\(\displaystyle \ln3=0.019t\)

\(\displaystyle t=57.82\)

To determine the amount of time needed to double the initial investment - P - into a compound interest account, we simply plug in our given information into the formula:

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

where B is the balance, P is the initial investment, r is the interest rate (as a decimal), n is the number of times the interest is compounded, and t is the time elapsed.

Now, because we are doubling P, our balance B becomes two times P:

\(\displaystyle 2P=P(1+\frac{0.04}{1})^{1\cdot t}\)

Now, we can solve for P:

\(\displaystyle 2=1.04^t\)

To bring the time variable down from being an exponent, we take the logarithm of both sides (common or natural):

\(\displaystyle \ln2=t\cdot \ln1.04\)

\(\displaystyle t=17.67\)