Please Help!!! :)


Ned currently has an account balance of $3,634.51. He opened the account 13 years ago with a deposit of $2,564.65. If the interest compounds twice a year, what is the interest rate on the account?

A. 4.7%

B. 5.4%

C. 2.7%

D. 1.4%



Leroy opened a savings account 18 years ago with a deposit of $2,651.43. The account has an interest rate of 2.8% compounded twice a year. How much interest has Leroy earned?


A. $4,373.68

B. $2,725.67

C. $4,358.69

D. $1,722.25

Respuesta :

abug23
I can help! Let's start with 1.
Your formula is r = n(ln(A/P)/(nt)) R being the interest rate of what your trying to find. So A= 3,634.51 P = 2,564.65 N is the number of times compounded per year.N= 2 and t is the number of years t= 13. Also while doing so use PEMDAS Let me know if you need any more help! 


Answer:

Compound interest formula is :

[tex]A=p(1+\frac{r}{n} )^{nt}[/tex]

1.

A = 3634.51

p = 2564.65

r = ?

n = 2

t = 13

Putting the values in formula, we get

[tex]3634.51=2564.65(1+\frac{r}{2} )^{2*13}[/tex]

=> [tex]3634.51=2564.65(1+\frac{r}{2} )^{26}[/tex]

=> [tex]\frac{3634.51}{2564.65}=(\frac{2+r}{2} )^{26}[/tex]

=> [tex]1.417=(\frac{2+r}{2} )^{26}[/tex]

We get [tex]\sqrt[26]{1.417}=\frac{2+r}{2}[/tex]

We get r = 0.02699 and r = -4.02699 (neglecting the negative value)

We get r = 0.027 (rounded)

And in percentage, it is 2.7%.

So, option C is the answer.

2.

p = 2651.43

r = 2.8% or 0.028

n = 2

t = 18

Putting the values in formula we get;

[tex]A=2651.43(1+\frac{0.028}{2} )^{2*18}[/tex]

=> [tex]A=2651.43(1.014)^{36}[/tex]

A = $4373.67

So, interest earned = [tex]4373.67-2651.43=1722.24[/tex] dollars

Hence, option D. $1,722.25 is the answer.