Eli invested $37,000 in an account paying an interest rate of 6 3 4 6 4 3 ​ % compounded quarterly. Xavier invested $37,000 in an account paying an interest rate of 6 1 4 6 4 1 ​ % compounded annually. After 9 years, how much more money would Eli have in his account than Xavier, to the nearest dollar?

3732 dollars

Step-by-step explanation:

$3732

Step-by-step explanation:

Rate 1: 6  

4

3

​  

%=6+3/4=

\,\,6.75\%\rightarrow 0.0675

6.75%→0.0675

\text{Rate 2: }6\tfrac{1}{4}\%=6+1/4=

Rate 2: 6  

4

1

​  

%=6+1/4=

\,\,6.25\%\rightarrow 0.0625

6.25%→0.0625

\text{Calculate Final Amount for Eli}

Calculate Final Amount for Eli

\overline{\phantom{\text{Calculate Final Amount for Eli}}}

Calculate Final Amount for Eli

\text{Compounded Quarterly:}

Compounded Quarterly:

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

A=P(1+  

n

r

​  

)  

nt

Compound interest formula

P=37000\hspace{35px}r=0.0675\hspace{35px}t=9\hspace{35px}n=4

P=37000r=0.0675t=9n=4

Given values

A=37000\left(1+\frac{0.0675}{4}\right)^{4(9)}

A=37000(1+  

4

0.0675

​  

)  

4(9)

Plug in values

A=37000(1.016875)^{36}

A=37000(1.016875)  

36

Simplify

A=67582.5002

A=67582.5002

Use calculator

\text{Calculate Final Amount for Xavier}

Calculate Final Amount for Xavier

\overline{\phantom{\text{Calculate Final Amount for Xavier}}}

Calculate Final Amount for Xavier

\text{Compounded Annually:}

Compounded Annually:

A=P(1+r)^t

A=P(1+r)  

t

P=37000\hspace{35px}r=0.0625\hspace{35px}t=9

P=37000r=0.0625t=9

Given values

A=37000(1+0.0625)^{9}

A=37000(1+0.0625)  

9

Plug in values

A=37000(1.0625)^{9}

A=37000(1.0625)  

9

Add

A=63850.1869

A=63850.1869

Use calculator

\text{How much more money Eli has:}

How much more money Eli has:

67582.5002-63850.1869

67582.5002−63850.1869

3732.3133

3732

the answer is d

step-by-step explanation: