Question 1027952
it appears that the formula is:


p(t) = d/(1+ke^-ct)


that set of parentheses is very important because it changes the normal order of operations.


using the normal order of operations without parentheses, you would get:


p(t) = (d/1) + (ke^-ct)


that doesn't make a lot of sense because you would be getting no growth at all.


i'm assuming the formula is p(t) = d/(1+ke^-ct)


under that assumption, the population will grow to 1200.


when d = 1200 and k = 11 and c = .2, you get:


when t = 0, p(0) = 1200 / (1 + 11 * e^(-.2*0) which becomes p(0) = 1200 / (1 + 11 * e^0) which becomes p(0) = 1200 / (1 + 11) which becomes p(0) = 1200 / 12 which becomes p(0) = 100


the initial number of fish is therefore 100.


when t = 10, p(10) = 1200 / (1 + 11 * e^(-.2*10) which becomes p(10) = 1200 / (1 + 11 * e^-2) which becomes p(10) = 1200 / (1 + 1.488688116) which becomes p(10) = 1200 / (2.488688116) which becomes p(10) = 482 rounded to the nearest integer.


when t = 20, p(20) = 1200 / (1 + 11 * e^(-.2*20) which becomes p(20) = 1200 / (1 + 11 * e^-4) which becomes p(20) = 1200 / (1 + .2014720278) which becomes p(20) = 1200 / (1.2014720278) which becomes p(20) = 999 rounded to the nearest integer.


when t = 30, p(30) = 1200 / (1 + 11 * e^(-.2*30) which becomes p(30) = 1200 / (1 + 11 * e^(-6) which becomes p(20) = 1200 / (1 + .0024787522) which becomes p(30) = 1200 / 1.0024787522) which becomes p(30) = 1197 rounded to the nearest integer.


it is starting to become clear that, as t gets larger, e^(-.2*t) becomes smaller.  


this leads to the conclusion as t approaches infinity, e^(-.2*t) will approach 0.


p(infinity) will therefore becomes 1200 / (1 + 0) which becomes 1200 / 1 which becomes 1200.


the number of fish in the pond will saturate at 1200, given that d = 1200 and k = 11 and c = .2 in the formula of p(t) = d / (1 + e^(-ct))