Respuesta :
[tex]\bf ~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$4000\\ P=\textit{original amount deposited}\dotfill & \$2000\\ r=rate\to r\%\to \frac{r}{100}\\ t=years\dotfill &10 \end{cases} \\\\\\ 4000=2000(1+r\cdot 10)\implies \cfrac{4000}{2000}=10r+1\implies 2=10r+1 \\\\\\ 1=10r\implies \cfrac{1}{10}=r\implies 0.1=r\implies \stackrel{\textit{converting to percentage}}{0.1\cdot 100\implies }~~\stackrel{\%}{10}=r[/tex]
Answer:
The answer is approx 7.1%
Step-by-step explanation:
The compound interest formula is :
[tex]A=p(1+r/n)^{nt}[/tex]
where, p = 2000
A = 4000
n = 1 (we will assume)
t = 10
r = ?
Now putting the values in formula we get
[tex]4000=2000(1+r/1)^{10}[/tex]
or
[tex]2=(1+r)^{10}[/tex]
Switching sides:
[tex](1+r)^{10}=2[/tex]
[tex]1+r=\sqrt[10]{2}[/tex]
Solving this we get, r=0.07177 and r=-2.07177(neglect the negative)
So, we have r=0.0717 ≈ 0.071 (taking only up to 3 decimal places)
And in percentage, this is 7.1%(approx)
