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):

$\displaystyle \ln2=t\cdot \ln1.04$

$\displaystyle t=17.67$