Let's analyze the given problem step-by-step to determine the rate of change of the number of bacteria every second.
1. We start with the function modeling the number of bacteria over time:
[tex]\[
B(t) = 8500 \cdot \left(\frac{8}{27}\right)^{\frac{t}{3}}
\][/tex]
2. We need to identify the factor by which the bacteria population is multiplied every second. This involves extracting the rate factor from the exponent in the function.
3. Notice that [tex]\( t \)[/tex] is in seconds, and the exponent on [tex]\(\left(\frac{8}{27}\right)\)[/tex] is [tex]\(\frac{t}{3}\)[/tex]. To find the factor by which the bacteria population changes every second, we need to raise [tex]\(\left(\frac{8}{27}\right)\)[/tex] to the power of [tex]\(\frac{1}{3}\)[/tex], because:
[tex]\[
\left(\frac{8}{27}\right)^{\frac{t}{3}} = \left[\left(\frac{8}{27}\right)^{\frac{1}{3}}\right]^t
\][/tex]
This transformation allows us to see the per-second multiplication factor clearly.
4. Therefore, the factor we need is:
[tex]\[
\left(\frac{8}{27}\right)^{\frac{1}{3}}
\][/tex]
5. Evaluating this expression (and rounding to two decimal places), we get:
[tex]\[
\left(\frac{8}{27}\right)^{\frac{1}{3}} \approx 0.67
\][/tex]
Hence, every second, the number of bacteria is multiplied by a factor of [tex]\( \boxed{0.67} \)[/tex].
