sat-dog

Choice C is correct. It's given that the function $P$ models the population of the city $t$ years after $2005$. Since there are $12$ months in a year, $18$ months is equivalent to $\frac{18}{12}$ years. Therefore, the expression $\frac{18}{12}x$ can represent the number of years in $x$ $18$-month periods. Substituting $\frac{18}{12}x$ for $t$ in the given equation yields $P\left(\frac{18}{12}x\right)=290{\left(1.04\right)}^{\left(\frac{4}{6}\right)\left(\frac{18}{12}x\right)}$, which is equivalent to $P\left(\frac{18}{12}x\right)=290{\left(1.04\right)}^x$. Therefore, for each $18$-month period, the predicted population of the city is $1.04$ times, or $104\%$ of, the previous population. This means that the population is predicted to increase by $4\%$ every $18$ months.

Choice A is incorrect and may result from conceptual or calculation errors. 

Choice B is incorrect. Each year, the predicted population of the city is $1.04$ times the previous year's predicted population, which is not the same as an increase of $1.04\%$.

Choice D is incorrect and may result from conceptual or calculation errors.