Let the town's population in 2010 be $P$. Then the population in 2020 is $1.25P$.

Bus preference:
- In 2010, the number of residents who preferred bus transportation was $0.30P$.
- In 2020, the number was $0.42(1.25P)=0.525P$.

Now compare these amounts:

So the number of residents who preferred bus transportation increased by $75\%$, which is more than $25\%$. Therefore, statement A must be true.

To check the others:
- Car: 2010 number was $0.50P$, and 2020 number was $0.38(1.25P)=0.475P$, a decrease of only $5\%$, not more than $12\%$.
- Bike: 2010 number was $0.20P$, and 2020 number was $0.20(1.25P)=0.25P$, so it increased by $25\%$, not stayed the same.
- Car or bus combined: in 2010, $80\%$ of $P$ is $0.80P$; in 2020, $80\%$ of $1.25P$ is $1.00P$, so that total increased, not stayed the same.

Because all the prisms have the same square base side length, the base area is constant. From the table, use volume $=$ base area $\times$ height. For example, when the height is $6$ cm, the volume is $216$ cm${}^3$, so the base area is $\frac{216}{6}=36$ cm${}^2$. This matches the other rows: $\frac{360}{10}=36$ and $\frac{540}{15}=36$. Since the base is a square, its side length is $\sqrt{36}=6$ cm.

Now write the surface area of a right rectangular prism with square base side length $6$ and height $h$:

We are told the surface area is $312$, so:

So the prism's height is $10$ centimeters.

First find the total number of vehicles that entered downtown during the full day: $4,800+3,600+2,700+5,100=16,200$.

The camera outage happened only during the 3 p.m. to 6 p.m. interval, which lasted 3 hours. In that interval, $5,100$ vehicles entered downtown over 3 hours, so the average rate was

vehicles per hour.

A 30-minute outage is $0.5$ hour, so the estimated number of vehicles missed during the outage was

Therefore, the number of vehicles the camera actually recorded was

So the best estimate is $15,350$.

Because $C(t)$ is quadratic and the values at $t=1$ and $t=5$ are both 18, the axis of symmetry is halfway between 1 and 5, so it is at $t=3$. The table confirms this, since $C(3)=10$ is the middle value and must be the vertex value. So the function can be written in vertex form as

Use the point $(1,18)$ to find $a$:

So the model is

Now evaluate at $t=7$:

Therefore, the correct answer is $42$.

To find the average for 5 days, first add the numbers of attendees: $18+24+21+27+30=120$. Then divide by the number of days: $\frac{120}{5}=24$. So the average number of attendees per day was $24$.