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Solve the Problem 10,000,00010,000,000 Cells Per Milliliter in a 6-Day Period r\mathrm { r }

Question 163

Multiple Choice

Solve the problem.
-Suppose the algae growth in Black Oak Lake increased from 100 cells per milliliter to approximately 10,000,00010,000,000 cells per milliliter in a 6-day period. The specific growth rate r\mathrm { r } is defined by r=logN2log1 N1x2x1\mathrm { r } = \frac { \log \mathrm { N } _ { 2 } - \log _ { 1 } \mathrm {~N} _ { 1 } } { \mathrm { x } _ { 2 } - \mathrm { x } _ { 1 } } , where N1\mathrm { N } _ { 1 } is the algae concentration at time x1x _ { 1 } and N2N _ { 2 } is the algae concentration at time x2x _ { 2 } . In this situation, what is the specific growth rate of algae? Round your results to the nearest hundredth.


A) 1,666,650.001,666,650.00
B) 1,666,650.00- 1,666,650.00
C) 0.830.83
D) 0.83- 0.83

Correct Answer:

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