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(C) Let $P(t)$ be the population at time $t$. The rate of change is given by $\frac{dP}{dt} = kP$,which leads to $P(t) = P_0 e^{kt}$.Let $t=0$ corresp

Vedclass · Accepted Answer

$7950$

Vedclass · Answer

(C) Let $P(t)$ be the population at time $t$. The rate of change is given by $\frac{dP}{dt} = kP$,which leads to $P(t) = P_0 e^{kt}$.Let $t=0$ correspond to the year $1984$. Then $P_A(0) = 20,000$ and $P_B(0) = 20,000$.For town $A$ at $t=5$ (year $1989$),$P_A(5) = 20,000 e^{5k_A} = 25,000$,so $e^{5k_A} = 1.25$.For town $B$ at $t=5$ (year $1989$),$P_B(5) = 20,000 e^{5k_B} = 28,000$,so $e^{5k_B} = 1.4$.At $t=10$ (year $1994$):$P_A(10) = 20,000 (e^{5k_A})^2 = 20,000 (1.25)^2 = 20,000 	imes 1.5625 = 31,250$.$P_B(10) = 20,000 (e^{5k_B})^2 = 20,000 (1.4)^2 = 20,000 	imes 1.96 = 39,200$.The difference is $|39,200 - 31,250| = 7,950$.

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