The volume of a cylindrical basin is given by $V=\pi r^{2}h$, where $r$ is the radius and $h$ is the water depth.

Compute each fountain's volume in terms of $\pi$:

- Fountain 1: $V=\pi (4{)}^2(2)=32\pi$
- Fountain 2: $V=\pi (5{)}^2(1.5)=\pi (25)(1.5)=37.5\pi$
- Fountain 3: $V=\pi (3{)}^2(3)=27\pi$
- Fountain 4: $V=\pi (6{)}^2(1)=36\pi$

Compare the values: $32\pi$, $37.5\pi$, $27\pi$, and $36\pi$. The greatest is $37.5\pi$, so Fountain 2 contains the greatest volume of water.

To find the overall percent of rides paid with a mobile app, use a weighted average based on the total number of rides each month, not just the four percentages.

First, find the total number of rides each month:
- Month 1: $1,200\times 20=24,000$
- Month 2: $1,350\times 19=25,650$
- Month 3: $1,500\times 22=33,000$
- Month 4: $1,650\times 21=34,650$

Next, find the number of mobile-app rides each month:
- Month 1: $0.25(24,000)=6,000$
- Month 2: $0.30(25,650)=7,695$
- Month 3: $0.34(33,000)=11,220$
- Month 4: $0.40(34,650)=13,860$

Now add the totals:
- Total rides: $24,000+25,650+33,000+34,650=117,300$
- Total mobile-app rides: $6,000+7,695+11,220+13,860=38,775$

Finally, compute the overall percent:

which is about $32.3\%$.

Therefore, the correct answer is $\boxed{32.3\%}$.

Since the function is $y=a{x}^2+bx+c$, use the three points to write a system.

From $(-1,6)$:

From $(2,3)$:

From $(4,11)$:

Now eliminate $c$.

Subtract the first equation from the second:

Subtract the second equation from the third:

Use $a+b=-1$, so $b=-1-a$. Substitute into $6a+b=4$:

Then

Use $a-b+c=6$:

Therefore,

So the correct answer is $2$.

At each speed, gallons used for a 120-mile trip equals $\frac{120}{m}$, where $m$ is the miles per gallon from the table. So the four gallon amounts are

The average of these four values is

Factor out 120:

This matches choice B.

Because the points lie on a straight line, the temperature changes at a constant rate. First find the rate of change using any two points, such as $(2,68)$ and $(5,56)$:

So the temperature decreases by $4^{\circ }\text{C}$ each hour. Starting from $(9,40)$, the sample must cool from $40^{\circ }\text{C}$ to $32^{\circ }\text{C}$, which is a drop of $8^{\circ }\text{C}$. At $4^{\circ }\text{C}$ per hour, that takes

hours. Therefore, the sample reaches $32^{\circ }\text{C}$ at

hours. The correct answer is $11$.