Answer:
After 43 minutes, the bacteria will be 40 remainings.
Step-by-step explanation:
Given that:
Initial population of bacteria [tex]A_o[/tex] = 800
After time (t) = 18 mins
A = 240
Using the model:
[tex]A = A_o e^{k*t}[/tex]
[tex]240=800 e^{k*18}[/tex]
[tex]\dfrac{240}{800}= e^{18k}[/tex]
[tex]0.3= e^{18k}[/tex]
By applying natural log on both sides, we get:
In(0,3) = 18k (In e)
- 1.20397 = 18 k * (1)
18 k = -1.20397
k = -1.20397/ 18
k = - 0.06688
k ≅ -0.07
After time (t), the bacteria population (A) = 40
Now;
[tex]40 = 800 e^{k*t}[/tex]
[tex]\dfrac{40 }{800}= e^{-0.07*t}[/tex]
[tex]0.05= e^{-0.07*t}[/tex]
In (0.05) = -0.07t * In (e)
- 2.9957 = - 0.07t
t = - 2.9957/ -0.07
t = 42.795
t ≅ 43 minutes
Thus, after 43 minutes, the bacteria will be 40 remainings.
