Question

A culture started with 5000 bacteria. After 2 hours, it grew to 5500 bacteria. Predict how many bacteria will be present after 10 hours

Answer 1

Answer:8,056

Explanation:

Answer 2

`\bf \stackrel{\textit{started, t = 0, A = 5000}}{\qquad \textit{Amount for Exponential Growth}} \\\\ A=P(1 + r)^t\qquad \begin{cases} A=\textit{accumulated amount}\to &5000\\ P=\textit{initial amount}\\ r=rate\to r\%\to (r)/(100)\\ t=\textit{elapsed time}\to &0\\ \end{cases} \\\\\\ 5000=P(1+r)^0\implies \boxed{5000=P}\\\\ -------------------------------`


`\bf \stackrel{\textit{2 hours later, t = 2, A = 5500}}{\qquad \textit{Amount for Exponential Growth}} \\\\ A=P(1 + r)^t\qquad \begin{cases} A=\textit{accumulated amount}\to &5500\\ P=\textit{initial amount}\to &5000\\ r=rate\to r\%\to (r)/(100)\\ t=\textit{elapsed time}\to &2\\ \end{cases}`


`\bf 5500=5000(1+r)^2\implies \cfrac{5500}{5000}=(1+r)^2\implies \cfrac{11}{10}=(1+r)^2 \\\\\\ \sqrt{\cfrac{11}{10}}=1+r\implies \sqrt{\cfrac{11}{10}}-1=r\implies 0.0488\approx r\\\\ -------------------------------\\\\ \boxed{A\approx 5000(1+0.0488)^t}\qquad \qquad \stackrel{\textit{after 10 hours, t = 10}}{A\approx 5000(1.0488)^(10)} \\\\\\ A\approx 8051.87`