P t = 260 1.04 6 4 t The function P models the population, in thousands,

Choice A is correct. It’s given that the function $P$ models the population, in thousands, of a certain city $t$ years after $2003$. The value of the base of the given exponential function, $1.04$, corresponds to an increase of $4\%$ for every increase of $1$ in the exponent, $\left(\frac{6}{4}\right)t$. If the exponent is equal to $0$, then $\left(\frac{6}{4}\right)t=0$. Multiplying both sides of this equation by $\left(\frac{4}{6}\right)$ yields $t=0$. If the exponent is equal to $1$, then $\left(\frac{6}{4}\right)t=1$. Multiplying both sides of this equation by $\left(\frac{4}{6}\right)$ yields $t=\frac{4}{6}$, or $t=\frac{2}{3}$. Therefore, the population is predicted to increase by $4\%$ every $\frac{2}{3}$ of a year. It’s given that the population is predicted to increase by $4\%$ every $n$ months. Since there are $12$ months in a year, $\frac{2}{3}$ of a year is equivalent to $\left(\frac{2}{3}\right)\left(12\right)$, or $8$, months. Therefore, the value of $n$ is $8$.

Choice B is incorrect. This is the number of months in which the population is predicted to increase by $4\%$ according to the model $P\left(t\right)=260{\left(1.04\right)}^t$, not $P\left(t\right)=260{\left(1.04\right)}^{\left(\frac{6}{4}\right)t}$.

Choice C is incorrect. This is the number of months in which the population is predicted to increase by $4\%$ according to the model $P\left(t\right)=260{\left(1.04\right)}^{\left(\frac{4}{6}\right)t}$, not $P\left(t\right)=260{\left(1.04\right)}^{\left(\frac{6}{4}\right)t}$.

Choice D is incorrect. This is the number of months in which the population is predicted to increase by $4\%$ according to the model $P\left(t\right)=260{\left(1.04\right)}^{\left(\frac{1}{6}\right)t}$, not $P\left(t\right)=260{\left(1.04\right)}^{\left(\frac{6}{4}\right)t}$.