P t = 290 1.04 4 6 t The function P models the population, in thousands, of a certain city

Question

P t = 290 1.04 4 6 t The function P models the population, in thousands, of a certain city t years after 2005 . According to the model, the population is predicted to increase by n % every 18 months. What is the value of n ?

Accepted Answer

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 $StartFraction 18 Over 12 EndFraction$ years. Therefore, the expression $StartFraction 18 Over 12 EndFraction x$ can represent the number of years in $x$ $18$ -month periods. Substituting $StartFraction 18 Over 12 EndFraction x$ for $t$ in the given equation yields $P (\frac{18}{12} x) = 290 {(1.04)}^{(\frac{4}{6}) (\frac{18}{12} x)}$ , which is equivalent to $P (\frac{18}{12} x) = 290 {(1.04)}^{x}$ . Therefore, for each $18$ -month period, the predicted population of the city is $1.04$ times, or $104 percent sign$ of, the previous population. This means that the population is predicted to increase by $4 percent sign$ 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 $1.04 percent sign$ .

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

Question p-t-29

P t = 290 1.04 4 6 t The function P models the population, in thousands,

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