We need to flip a coin until we get two heads in a row. The smallest number of possible flips is 2, which would occur if our first two flips are both heads. This eliminates three of our answer choices, because we know the sample space must start at 2. 

This leaves us with {x : x = 2, 3, 4 . . .} and {x : x = 2, 3, 4, 5, 6}. Let's think about {x : x = 2, 3, 4, 5, 6}. What if I flip a coin 6 times and get 6 tails? Then I have to keep flipping beyond 6 flips until I get two heads in a row; therefore the answer must be {x : x = 2, 3, 4 . . .}, because we don't have an upper limit on the number of flips it will take to produce two successive heads.

$\displaystyle A \cup \overline{B}$ is the set of elements that fall either in $\displaystyle A$ or the complement of $\displaystyle B$, or both - that is, either in $\displaystyle A$ , outside of $\displaystyle B$, or both. This union is intersected with the complement of $\displaystyle C$, meaning that only the elements of the union that also fall outside of $\displaystyle C$ are considered.

"Color" in all of $\displaystyle A$ and everything outside of $\displaystyle B$ - but then, uncolor everything inside $\displaystyle C$. That makes the correct choice:

525 is a multiple of all three of the integers 3, 5, and 7:

$\displaystyle 525 \div 3 = 175$

$\displaystyle 525 \div 5 = 105$

$\displaystyle 525 \div 7 = 75$

Therefore, 525 is an element of each of sets $\displaystyle A,B,C$, and, subsequently, falls into region $\displaystyle II$, which represents $\displaystyle A \cap B \cap C$.

Since order doesn't matter here, set this up as a combination:

$\displaystyle \frac{8!}{5!3!}=\frac{8\times 7\times 6\times 5!}{5!(3\times 2)} = 8 \times7=56$

$\displaystyle 1,728 \div 3 = 576$, making 1,728 a multiple of 3, and thus, an element of $\displaystyle A$.

1,728 is not a perfect square; $\displaystyle 41^{2} = 1,681 < 1,728 < 1,764 = 42^{2}$. Thus, 1,728 is not an element of $\displaystyle B$.

1,728 is a perfect cube: $\displaystyle 1,728 = 12^{3}$. Thus, 1,728 is an element of $\displaystyle C$.

$\displaystyle 1,728 \in A \cup \overline{B} \cup C$, which is represented by the region inside circles $\displaystyle A$ and $\displaystyle C$ and outside $\displaystyle B$. This is region $\displaystyle IV$.

In order to find the median, the set needs to be written in numerical order:

$\displaystyle {6,7,13,29,41,67}$

Since $\displaystyle 13$ and $\displaystyle 29$ are both the middle numbers, taking their average will give the median of the set.

$\displaystyle \frac{(13+29)}{2}= 21$

Let $\displaystyle x$ be the number of students who don't take any of the three classes.

$\displaystyle 3+7+5+0+3+0+5+x=30$

$\displaystyle 23+x=30$

$\displaystyle x=7$

Prime numbers are numbers with no other factors than themselves and one. Two is the first prime number and the only even prime number. Other examples are 5, 7, 11, etc.

Even numbers are numbers divisible by 2. Set C includes all numbers ending in 0, 2, 4, 6, or 8.

Thus, there is one number common to both sets: 2.

The sets $\displaystyle A$ and $\displaystyle C$ do not intersect, so no senior is enrolled in both French IV and physics; the sets $\displaystyle B$ and $\displaystyle C$ do not intersect, so no senior is enrolled in both French IV and calculus. 

$\displaystyle B \subseteq A$, so every senior enrolled in calculus is also enrolled in physics; contrapositively, every senior not enrolled in physics is also not enrolled in calculus.

The correct choice is the remaining statement - every senior enrolled in physics is also enrolled in calculus - since $\displaystyle A$ is not a subset of $\displaystyle B$.

Let $\displaystyle T$ and $\displaystyle E$ be the set of all Toastmasters and Elks, respectively, and let $\displaystyle U$ be the set of all people. $\displaystyle \textup{John} \in T$ and $\displaystyle \textup{John} \notin E$, so the set to which John belongs is the shaded set in this Venn diagram:

the logical opposite of this is that John belongs to the shaded set in the diagram:

A way of saying this is  $\displaystyle \textup{John} \in E$ or $\displaystyle \textup{John} \notin T$, or, equivalently, if $\displaystyle \textup{John} \notin E$, then $\displaystyle \textup{John} \notin T$.

In plain English, if John is not an Elk, then John is not a Toastmaster.