Question

a certain type of bacteria grows slowly at a rate of 0.9% per day. if 45 bacteria are present today, find how many will be present in 60 days. use the following model:
$b(t)=45e^{0.009t}$
if needed, round your answer to 1 decimal place.
$b(t)\approx$ bacteria.
check answer

Explanation:

Step1: Substitute \(t = 60\) into the formula

Given \(B(t)=45e^{0.009t}\), when \(t = 60\), we have \(B(60)=45e^{0.009\times60}\).
First, calculate the exponent: \(0.009\times60 = 0.54\). So \(B(60)=45e^{0.54}\).

Step2: Calculate the value of \(e^{0.54}\)

We know that \(e^{x}\approx1 + x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots\). Using a calculator, \(e^{0.54}\approx1.7160079\).

Step3: Calculate \(B(60)\)

Multiply \(45\) by \(e^{0.54}\): \(B(60)=45\times1.7160079\).
\(45\times1.7160079 = 77.2203555\).

Answer:

\(77.2\) bacteria.