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(Solved): The population \( P \) (in thousands) of a country can be modeled by \[ P=-14.75 t^{2}+788.5 t+117, ...




The population \( P \) (in thousands) of a country can be modeled by
\[
P=-14.75 t^{2}+788.5 t+117,228
\]
where \( t \) is ti
The population \( P \) (in thousands) of a country can be modeled by \[ P=-14.75 t^{2}+788.5 t+117,228 \] where \( t \) is time in years, with \( t=0 \) corresponding to 1980 . (a) Evaluate \( P \) for \( t=0,10,15,20 \), and 25. \( \begin{array}{ll}P(0)= & \text { thousand people } \\ P(10)= & \text { thousand people } \\ P(15)= & \text { thousand people } \\ P(20)= & \text { thousand people } \\ P(25)= & \text { thousand people }\end{array} \) Explain these values. The is (b) Determine the population growth rate, \( d P / d t \). \[ \frac{d P}{d t}= \] (c) Evaluate \( d P / d t \) for the same values as in part (a). \[ \begin{aligned} P^{\prime}(0)= & \text { thousand people per year } \\ P^{\prime}(10)= & \text { thousand people per year } \\ P^{\prime}(15)= & \text { thousand people per year } \\ P^{\prime}(20)= & \text { thousand people per year } \\ P^{\prime}(25)= & \text { thousand people per year } \end{aligned} \] Explain your results. The is


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given P=P(t)=?14.75t2+788.5t+117,228……….(i) t=0 corresponding to 1980. (a) put t=0 in place of t in (i), we get P(0)=?14.75×02+788.5×0+117,228=117,228
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