Bacterial growthBacteria of species A and species B are kept in a single test tube, where they are fed two nutrients. Each day the test tube is supplied with 10,600 units of the first nutrient and 19,650 units of the second nutrient.
Each bacterium of species A requires 2 units of the first nutrient and 3 units of the second, and each bacterium of species B requires 1 unit of the first nutrient and 4 units of the second. What populations of each species can coexist in the test tube so that all the nutrients are consumed each day?
A = 2I + 3II (1) B = I + 4II (2) Now, Daily Requirement = 10600 I + 19650 II Thus, 2I + 3II = 10600 (3) And I + 4II = 19650 (4) Solving (3) and (4) 2(19650-4II) +3II = 10600…
300 8II +3II = 10600 39300 – 5II = 10600 5II = 28700 II = 5740 Finding I, I = 19650 4II I = 19650 4(5740) I = 3310
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