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Figuring the first "year's" population of 3-E
One in 1.82*10e12 divisions of A-1 organisms will produce a 2-B organism.
5.49450549e16 2-B will be produced per saturation cycle.
One in 1.82*10e12 is also the number of 3-E generated by 2-B organisms
rem finds populations of 3-E at the end of "year one"
countit = 0
original = 5.49450549*10^16
E3 = 0
population = 0
[start]
population = population + original
E3 = E3 + (population/(1.82*10^12))
countit = countit + 1
IF countit = 1000 then [end]
GOTO [start]
[end]
print E3
end
answer: 1.51098901*10e10 or about 1.511*10e10 3-E at end of "year one"
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