Answer
a) $\$820$
b) $\$8.20$
c) $c(p)= \frac{4.55 p+365.00}{p}$
d)The total cost person to attend an event is $\$6.98$.
e) Domain: $1\leq p\leq 250$
Range: $6.01\leq C(p)\leq 369.55$
Work Step by Step
Given $$C(p)=4.55 p+365.00.
$$ a) Find $C(100)$ when $p= 100$.
\begin{equation}
\begin{aligned}
C(100)&= 4.55 \cdot 100+365.00\\
&=820.
\end{aligned}
\end{equation} The total cost for $100$ people to attend an event is $\$820$.
b) Find $C(100)/100$ when $p= 100$: \begin{equation}
\begin{aligned}
C(100)/100&= \frac{820}{100}= 8.2.
\end{aligned}
\end{equation} The total cost per person to attend an event is $\$8.20$.
c) Let $c(p)= C(p)/p$ the cost per person to attend an event. Then, we may write:
\begin{equation}
\begin{aligned}
c(p)&= \frac{4.55 p+365.00}{p}.
\end{aligned}
\end{equation} d) Find $c(150)$ when $p= 150$.
\begin{equation}
\begin{aligned}
c(150)&= \frac{4.55 \cdot 150+365.00}{150}= 6.98.
\end{aligned}
\end{equation} The total cost person to attend an event is $\$6.98$.
e) Take minimum and maximum number of people to attend the even to be $p= 1$ and $p= 250$ to set the range.
\begin{equation}
\begin{aligned}
c(1)&= \frac{4.55 \cdot 1+365.00}{1}= 369.55\\
c(250)&= \frac{4.55 \cdot 250+365.00}{250}= 6.01.
\end{aligned}
\end{equation} A reasonable range and domain of the cost function would be:
\begin{equation}
\begin{aligned}
\textbf{Domain:}&\quad 1\leq p\leq 250\\
\textbf{Range:}&\quad 6.01\leq C(p)\leq 369.55\\
\end{aligned}
\end{equation}
