Question

A particular type of bacteria is found to be capable of doubling in number about every 49.3 minutes. The number N of bacteria present after t minutes could be modeled by N(t)=N↓0e^0.014t. Suppose that N↓0=300,000 is the initial number of bacteria per milliliter. Complete parts (a) and (b) below.(a) Approximate the number of bacteria per milliliter after 3 hours.

Answer 1

Given the equation :

`N(t)=N_0\cdot e^(0.014t)`
`N_0=300,000`

We need to find the number of bacteria after 3 hours

so, t = 3 * 60 = 180 minutes

Substitute with t = 180 at the given equation

`\begin{gathered} N(180)=300,000\cdot e^(0.014\cdot180)=300,000\cdot e^(2.52)=300,000\cdot12.4286 \\ N(180)=3,728,579 \end{gathered}`

So, after 3 hours , the number of bacteria = 3,728,579

to the nearest thousand it will be : 3,729,000