[tex]\bf \qquad \textit{Amount for Exponential change}\\\\
A=P(1\pm r)^t\qquad
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{starting amount}\to &110\\
r=rate\to 4.8\%\to \frac{4.8}{100}\to &0.048\\
t=\textit{elapsed period}\to &4\\
\end{cases}
\\\\\\
A=110(1+0.048)^4[/tex]
the rate is positive, because is a rate of increase, from 2005 to 2009, is 4 years, 2005 being t = 0