Question

Ned currently has an account balance of $3634.51 he open the account 13 years ago with the deposit of $2564.65 if the interest compound twice a year what is the interest rate on the account

Answer 1

Answer:

1.47%

Step-by-step explanation:

Using the formula

F = P(1 + r/n)^(nt)

Where F = future value,

P = present value (or principal),

r = interest rate written as a decimal,

n = number of times interest is compounded per year,  

t = time in years

Therefore;

3634.51 = 2564.65*(1 + r/2)^(2*13)

3634.51/2564.65= (1 + t/2)^(26)

1.4172 =(1 + t/2)^(26)

Then we get the 26th root on both sides, and solve for r

we get

r = 1.47%

Answer 2

Answer:

r = 1.4%

Step-by-step explanation:

We are given that Ned has an account balance of $3634.51 and he opened the account 113 years ago with the deposit of $2564.65.

If the interest compound twice a year, we are to find the interest rate on the account.

We will use the following formula:

$F=P(1+\frac{r}{n} )^{nt}$

Substituting the given values in the above formula:

$3634.51 = 2564.65*(1+\frac{r}{2} )^{(2*3)}$

$\frac{3634.51}{2564.65} =(1+\frac{t}{2})^{26}$

$1.4172=(1+\frac{t}{2} )^{26}$

Taking 26th root on both sides to get:

r = 1.4%