Solution
In 2011 The temperature in Darrtown is
[tex]65^{\circ}F[/tex]The temperature increased by 3% in 2012
The temperature will be
[tex](1+\frac{3}{100})\times65=(1.03)(65)=66.95^{\circ}F[/tex]The temperature decreased by 4.5% in 2013
The temperature will be
[tex]\begin{gathered} (1-\frac{4.5}{100})\times66.95=0.955\times66.95=63.93725 \\ \\ (1-\frac{4.5}{100})\times66.95=64^{\circ}F\text{ (to the nearest whole number)} \end{gathered}[/tex]Therefore, the temperature in 2013 is 64 degrees Fareheint
Option B
