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(A) For $t \le 1 \ hour$,the population grows as $N(t) = N_0 \exp(t)$. Thus,$\frac{N_0}{N} = \exp(-t)$. This is a decreasing exponential function.For

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(A) For $t \le 1 \ hour$,the population grows as $N(t) = N_0 \exp(t)$. Thus,$\frac{N_0}{N} = \exp(-t)$. This is a decreasing exponential function.For $t > 1 \ hour$,the population follows $\frac{dN}{dt} = -5N^2$. Integrating this from $t=1$ to $t$,we get $\int_{N_1}^{N} \frac{dN}{N^2} = \int_{1}^{t} -5 dt$,where $N_1 = N_0 \exp(1)$.This gives $[-\frac{1}{N}]_{N_1}^{N} = -5(t-1)$,so $\frac{1}{N} - \frac{1}{N_1} = 5(t-1)$.Multiplying by $N_0$,we get $\frac{N_0}{N} = \frac{N_0}{N_1} + 5N_0(t-1) = \exp(-1) + 5N_0(t-1)$.This is a linear function with a positive slope for $t > 1$. Therefore,the plot of $\frac{N_0}{N}$ vs. $t$ decreases until $t=1$ and then increases linearly.

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