Let $a$ be the number of adult volunteers on Saturday and $s$ be the number of student volunteers on Saturday. From the total number of volunteers, $a+s=18$. Since each adult worked 4 hours and each student worked 2 hours, the total hours on Saturday give $4a+2s=54$. Substitute $s=18-a$ into the hours equation: $4a+2(18-a)=54$. Simplify: $4a+36-2a=54$, so $2a+36=54$, and therefore $2a=18$. This gives $a=9$. So there were 9 adult volunteers on Saturday.

The table shows that the average mass increases by $6$ grams every $2$ weeks, so the rate of change is $\frac{6}{2}=3$ grams per week. From 8 weeks to 11 weeks is 3 more weeks, so the mass should increase by $3\cdot 3=9$ grams. Since the mass at 8 weeks is $32$ grams, the predicted mass at 11 weeks is $32+9=41$ grams. Therefore, the correct answer is $41$.

To find the median, order the rainfall amounts from least to greatest: $2.7,3.2,3.5,4.1,5.0$. Since there are 5 values, the median is the middle value, which is $3.5$. Because $3.5<4$, the statement in choice A must be true.

Check the other statements:
- Mean: $\frac{3.2+4.1+2.7+5.0+3.5}{5}=\frac{18.5}{5}=3.7$, which is not greater than 4.
- Months greater than 4 inches: only May and July, so $2$ out of $5$, which is not more than half.
- Range: $5.0-2.7=2.3$, which is not less than 2.

The amount of fencing needed is the perimeter of each rectangle, found with $P=2l+2w$.

Flower Bed 1: $2(12)+2(5)=24+10=34$

Flower Bed 2: $2(10)+2(6)=20+12=32$

Flower Bed 3: $2(9)+2(7)=18+14=32$

Flower Bed 4: $2(8)+2(8)=16+16=32$

The greatest perimeter is $34$ feet, so Flower Bed 1 requires the greatest amount of fencing.

The correct answer is $y=3x$. From the table, divide parking fees by visitors for any row: $\frac{360}{120}=3$, $\frac{450}{150}=3$, and $\frac{540}{180}=3$. The ratio is constant, so the amount collected is proportional to the number of visitors. That means the model has the form $y=kx$, where $k=3$. Therefore, the equation is $y=3x$.