SOLUTION: The logistic growth function f(t) =53,000/1+1059.0e^-1.5t models the number of people who have become ill with a particular infection t weeks after its initial outbreak in a parti

Question 1089550: The logistic growth function f(t) =53,000/1+1059.0e^-1.5t
models the number of people who have become ill with a particular infection t weeks after its initial outbreak in a particular community. What is the limiting size of the population that becomes ill?

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
i believe the formula is:

f(T) = 53000 / (1 + 1059 * e ^ (-1.5 * T))

the parentheses are necessary to ensure a correct interpretation of the problem.

if i understood the formula correctly, then e^(-1.5 * T) is the same as 1 / e^(1.5 * T) and the formula would look like this:

f(T) = 53000 / (1 + 1059 / (e ^ (1.5 * T))

as T approaches infinity, then 1059 / (e ^ (1.5 * T)) approaches 0 and the formula becomes:

f(T) = 53000 / 1 which is equal to 53000.

therefore, the limiting size of the population that becomes ill is 53000.

i checked this out graphically and determined that it is correct.

T represents the number of weeks since the onset.

in the graph, y represents f(T) and x represents T.

the graph looks like this:

$$$

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