To rent a certain meeting room, a college charges a reservation fee of $\mathrm{\$}45$ and an additional fee of $\mathrm{\$}7.30$ per hour. The math club wants to spend less than $\mathrm{\$}103.40$ on renting the meeting room.
What are the possible amounts of time for which they could rent the meeting room? Use $t$ for the number of hours the meeting room is rented, and solve your inequality for $t$.

Step 1 ：The total cost of renting the meeting room is the sum of the reservation fee and the hourly fee times the number of hours. This total cost must be less than $103.40. So, we can set up the inequality as follows: $45+7.3t<103.4$

Step 2 ：We can solve this inequality for $t$ to find the maximum number of hours the math club can rent the meeting room. The solution to the inequality is $t=8$.

Step 3 ：However, since the question asks for the possible amounts of time for which they could rent the meeting room, we need to consider all values of $t$ that are less than 8.

Step 4 ：$\boxed{{\displaystyle 0\le t<8}}$ is the final answer. The possible amounts of time for which the math club could rent the meeting room are all values of $t$ such that $0\le t<8$. In other words, the math club could rent the meeting room for any amount of time less than 8 hours.