Answer
a) $75$
b) $P(t) = 75 \cdot 2^{t/6}$.
c) $1905$
Work Step by Step
(a) Initial Population:
We are told that 18 years after introduction the population is 600. Hence,
\[
600 = P(18) = P_0 \cdot 2^{18/6} = P_0 \cdot 2^3 = 8P_0.
\]
Solving for \(P_0\),
\[
P_0 = \frac{600}{8} = 75.
\]
\[
\boxed{75}.
\]
(b) Population \(t\) Years After Introduction:
The population model is:
\[
\boxed{P(t) = 75 \cdot 2^{t/6}}.
\]
(c) Expected Population 10 Years from Now:
Since the population is currently 18 years after introduction, 10 years from now corresponds to \(t = 18 + 10 = 28\) years after introduction. Thus,
\[
P(28) = 75 \cdot 2^{28/6} = 75 \cdot 2^{14/3}.
\]
To approximate \(2^{14/3}\), note that:
\[
2^{14/3} \approx 25.398.
\]
Therefore,
\[
P(28) \approx 75 \times 25.398 \approx 1905.
\]
Rounding to the nearest whole number, the expected population is approximately \(\boxed{1905}\) squirrels.
