azsun

Yes, like as what tertre said:

Here's a similar problem, How many ways are there to arrange the word, "Waterfall"

First, count how many letters, and see how many letters in the word repeat, then divide.

Chuck can play in the three quarters of the circle with radius 3 that does not have any area inside the shed.

The area of that is $\frac {3}{4} \times9 \pi = \frac{27\pi} {4}$.

One meter of Chuck's rope can extend below the shed on the bottom right corner with one meter left. So, Chuck has a quarter circle with radius $1$ he can play in.

The total is $\frac{27\pi} {4} + \frac{\pi}{4}= \frac{28\pi}{4} = \boxed{7\pi}$ .

(a): Median is the middle score, when arranged in order. So, 140

Ok, 60 do both drama and sports, so 85-60=25 only take drama, and 200-60=140 on do sports. Adding, 140, 25, and 40, we get 205- do everything. But, the question is asking, what is the probability, that a student chosen at random is in neither.  This is just, 330-205=125. So, 125/330*100=37.87

The answer is $\boxed{41}$

This is a case of the flagpole problem which can be solved with the formula $ab/(a + b)$where a and b are the lengths of the sides

Therefore,  the length of $EF$  is $60$  

another way to do this w/o the formula is graphing it

Great solution, both hectictar and CPhill!

Um, 2 users above posted the solution. No need to say that. 

Slight mistake, CPhill?

It will always be $2^n-(2^{n-1}+2^{n-2} \cdots +2^1+2^0)=1$.Thus, the minimum value is $\boxed{1}$

Let's see if I can solve this:

Since $\triangle ABC$ and $\triangle PQR$ are equilateral, then $\angle ABC=\angle ACB=\angle RPQ=60^\circ$.

Therefore, $\angle YBP = 180^\circ-65^\circ-60^\circ=55^\circ$  and $\angle YPB = 180^\circ-75^\circ-60^\circ=45^\circ$ .

In $\triangle BYP$, we have $\angle BYP = 180^\circ - \angle YBP - \angle YPB = 180^\circ - 55^\circ-45^\circ=80^\circ$.

Since $\angle XYC = \angle BYP$ , then $\angle XYC=80^\circ$.

In $\triangle CXY$, we have $\angle CXY = 180^\circ - 60^\circ - 80^\circ = 40^\circ$.

So our final answer is $\boxed{40}$ degrees.