A population of a bacteria colony doubles every 3 hours. If there are currently 500 bacteria, how many will there be after 12 hours?

Write the following recurring decimal as a simplified commor fraction. \( \mathbf{0 , 7 4} \)

Resolver: a) \( \frac{5}{8} \cdot\left(\frac{1}{4}\right)^{-1}+\frac{9}{5} \cdot \sqrt{\frac{27}{9} \cdot 3}-\frac{7}{10}+0,8= \)

e) \( \sqrt[2]{\left(\frac{49}{9}\right)^{11} \cdot\left(\frac{49}{9}\right)^{-4}:\left(\frac{98}{18}\right)^{6}: \frac{196}{36}+[3-(0, \hat{6}-1)]^{2}}= \)

Hisoblang: \( \sqrt{66^{2}-44^{2}-22^{2}} \) \( \begin{array}{llll}\text { A) } 22 & \text { B) } 44 & \text { C) } 0 & \text { D) } 33\end{array} \)

The doubling period of a bacterial population is 15 minutes. At time \( t=80 \) minutes, the bacterial population was 60000 . What was the initial population at time \( t=0 \) ? Find the size of the bacterial population after 4 hours.