Solve the Problem $10,000,000$ Cells Per Milliliter in a 6-Day Period $\mathrm { r }$

question 18

Multiple Choice

Solve the problem.
-Suppose the algae growth in Black Oak Lake increased from 100 cells per milliliter to approximately $10,000,000$ cells per milliliter in a 6-day period. The specific growth rate $\mathrm { r }$ is defined by $\mathrm { r } = \frac { \log \mathrm { N } _ { 2 } - \log _ { 1 } \mathrm {~N} _ { 1 } } { \mathrm { x } _ { 2 } - \mathrm { x } _ { 1 } }$ , where $\mathrm { N } _ { 1 }$ is the algae concentration at time $x _ { 1 }$ and $N _ { 2 }$ is the algae concentration at time $x _ { 2 }$ . In this situation, what is the specific growth rate of algae? Round your results to the nearest hundredth.

• A) \(1,666,650.00\)
• B) \(- 1,666,650.00\)
• C) \(0.83\)
• D) \(- 0.83\)