Choice A is the correct answer. Find the carrying capacity.

1. As $t \to \infty$: $e^{-0.4t} \to 0$.
2. Limit: $E(\infty) = \frac{100}{1 + 0} = 100\%$.
3. Result: Maximum adoption is 100%.

💡 Strategic Tip: In $\frac{K}{1 + Ae^{-rt}}$, the numerator $K$ is always the limit.

Choice B is incorrect because 50% would be if $K = 50$. Choice C is incorrect because this is the initial adoption. Choice D is incorrect because 19 is the coefficient $A$, not the limit.