Harper is going to invest $6,900 and leave it in an account for 12 years. Assuming the interest is compounded daily, what interest rate, to the nearest tenth of a percent,
would be required in order for Harper to end up with $11,700?

4.4%

Step-by-step explanation:

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

A=P(1+  

n

r

​  

)  

nt

Compound interest formula

A=11700\hspace{35px}P=6900\hspace{35px}t=12\hspace{35px}n=365

A=11700P=6900t=12n=365

Given values

11700=

11700=

\,\,6900\left(1+\frac{r}{365}\right)^{365(12)}

6900(1+  

365

r

​  

)  

365(12)

Plug in values

11700=

11700=

\,\,6900\left(1+\frac{r}{365}\right)^{4380}

6900(1+  

365

r

​  

)  

4380

Multiply

\frac{11700}{6900}=

6900

11700

​  

=

\,\,\frac{6900\left(1+\frac{r}{365}\right)^{4380}}{6900}

6900

6900(1+  

365

r

​  

)  

4380

​  

Divide by 6900

1.695652174=

1.695652174=

\,\,\left(1+\frac{r}{365}\right)^{4380}

(1+  

365

r

​  

)  

4380

\left(1.695652174\right)^{1/4380}=

(1.695652174)  

1/4380

=

\,\,\left[\left(1+\frac{r}{365}\right)^{4380}\right]^{1/4380}

[(1+  

365

r

​  

)  

4380

]  

1/4380

Raise both sides to 1/4380 power

1.000120571=

1.000120571=

\,\,1+\frac{r}{365}

1+  

365

r

​  

-1\phantom{=}

−1=

\,\,-1

−1

Subtract 1

0.000120571=

0.000120571=

\,\,\frac{r}{365}

365

r

​  

365\left(0.0001206\right)=

365(0.0001206)=

\,\,\left(\frac{r}{365}\right)365

(  

365

r

​  

)365

Multiply by 365

0.044008415=

0.044008415=

\,\,r

r

4.4008415\%=

4.4008415%=

\,\,r

r

4.4%  

Step-by-step explanation:

the answer is b