Let’s analyze each statement based on the given population formula $P(t) = 5.5^t$.

(i) The average rate of change between 0 and 1 is 4.5

The average rate of change of $P(t)$ between two points $t = a$ and $t = b$ is given by the formula:

$\frac{P(b) - P(a)}{b - a}$

Substituting $a = 0$ and $b = 1$, we calculate:

$\text{Average rate of change} = \frac{P(1) - P(0)}{1 - 0} = \frac{5.5^1 - 5.5^0}{1} = \frac{5.5 - 1}{1} = 4.5$

Thus, statement (i) is true.

(ii) $P'(t) = \lim_{h \to 0} \frac{5.5^h \cdot (5.5^t - 1)}{h}$

The derivative $P'(t)$ of the function $P(t) = 5.5^t$ can be calculated using the chain rule and exponential differentiation. The derivative of $a^t$ is $a^t \ln(a)$, so:

$P'(t) = 5.5^t \ln(5.5)$

To verify if the given limit expression correctly represents this, we analyze it:

$P'(t) = \lim_{h \to 0} \frac{5.5^h \cdot (5.5^t - 1)}{h}$

This expression does not directly match the standard limit form for derivatives. Instead, it seems to represent something different (possibly involving a different approach to the derivative), and it does not yield the correct derivative formula for $5.5^t$. Hence, statement (ii) is false.

(iii) $P'(0)$ is greater than the average rate of change of $P$ between 0 and 1

We calculated the average rate of change between 0 and 1 to be 4.5.

Next, we calculate $P'(0)$:

$P'(t) = 5.5^t \ln(5.5)$

Substituting $t = 0$:

$P'(0) = 5.5^0 \ln(5.5) = 1 \cdot \ln(5.5) \approx 1.7047$

Clearly, $P'(0) \approx 1.7047$, which is less than the average rate of change of 4.5. Thus, statement (iii) is false.

Conclusion:

Only statement (i) is true. Thus, the correct answer is:

$\boxed{\text{(C) (i) only}}$

Would you like more details or have any other questions?

Here are 5 related questions to expand on this topic:

1. What is the derivative of an exponential function in general?
2. How is the average rate of change related to the instantaneous rate of change?
3. What are some typical applications of average rate of change in population models?
4. Can the limit definition of a derivative be applied to non-exponential functions?
5. How does exponential growth differ from linear growth?

Tip: To calculate the derivative of $a^t$, always remember to multiply by $\ln(a)$, where $a$ is the base of the exponential function.