$\displaystyle \overline{A}$ is the complement of $\displaystyle A$, the set of all elements in the universal set not in $\displaystyle A$. In the diagram, it is represented by all elements not inside the circle that represents $\displaystyle A$. This is the set:

$\displaystyle \overline{A}= \left \{ 1,6,7,8,11,12,15,16\right \}$

$\displaystyle \overline{A} \cup \overline{B}$ is the union of the sets $\displaystyle \overline{A}$ and $\displaystyle \overline{B}$ - that is, the set of all elements in either $\displaystyle \overline{A}$ or $\displaystyle \overline{B}$. $\displaystyle \overline{A}$ is the complement of set $\displaystyle A$ - that is, the set of elements in $\displaystyle U$ not in $\displaystyle A$; $\displaystyle \overline{B}$ is defined similarly. Therefore, we need all of the elements either outside of $\displaystyle A$ or outside of $\displaystyle B$. The excluded elements will be those inside both sets, which, by the diagram, can be seen to be 3, 5, 13, and 14. Therefore,

 $\displaystyle \overline{A} \cup \overline{B} = \left \{ 1,2,4,6,7,8,9,10,11,12,15,16\right \}$.

The region in which 2,431 appears depends on the sets of which 2,431 is an element, which in turn depends on which of 7, 11, and 13 divides it evenly:

$\displaystyle 2,431 \div 7 = 347 \textrm{ R 2}$

$\displaystyle 2,431 \div 11 = 221$

$\displaystyle 2,431 \div 13 = 187$

2,431 is a multiple of 11 and 13, but not 7, so 2,431 is in $\displaystyle B$ and $\displaystyle C$, but not $\displaystyle A$. We look for a number among the choices that is in $\displaystyle B$ and $\displaystyle C$, but not $\displaystyle A$ - that is, a number divisible by 11 and 13 but not 7.

$\displaystyle 2,772 \div 7 = 396$, so 2,772 is divisible by 7. We can eliminate it.

$\displaystyle 2,184 \div 7 = 312$, so 2,184 is divisible by 7. We can eliminate it.

$\displaystyle 3,081 \div 11 = 280 \textrm{ R }1$, so 3,081 is not divisible by 11, and we can eliminate it.

$\displaystyle 2,409 \div 13 = 185 \textrm{ R }4$, so 2,409 is not divisible by 13, and we can eliminate it.

However:

$\displaystyle 2,145 \div 11 =195$

$\displaystyle 2,145 \div 13 = 165$

$\displaystyle 2,145 \div 7 = 306 \textrm{ R } 4$

2,145 is divisible by 11 and 13 but not 7, so this is the correct choice.

Cathy is in the Honor Society, meaning that she is in set $\displaystyle A$; she turned 18 after election day, so she is not in set $\displaystyle B$; she is taking a French course, so she is in set $\displaystyle C$. She is in set $\displaystyle A \cap B' \cap C$, represented by region 2.

The shaded region is $\displaystyle \left (A \cap C \right ) \cup \left (B \cap C \right )$, which will comprise all of the numbers that are multiples of 10 and of either or both 8 or 9.

$\displaystyle A \cap C$ will comprise multiples of 10 and 8 - that is, multiples of $\displaystyle LCM (8,10)= 40$. Since $\displaystyle 1,000 \div 40 = 25$, 

$\displaystyle c \left (A \cap C \right ) = 25$

$\displaystyle B \cap C$ will comprise multiples of 10 and 9 - that is, multiples of $\displaystyle LCM (9,10)= 90$. Since $\displaystyle 1,000 \div 90 \approx 11.1$, 

$\displaystyle c \left (B \cap C \right ) = 11$

To find $\displaystyle c\left [ \left (A \cap C \right ) \cup \left (B \cap C \right ) \right ]$, we first find $\displaystyle c\left [ \left (A \cap C \right ) \cap \left (B \cap C \right ) \right ]$.

$\displaystyle \left (A \cap C \right ) \cap \left (B \cap C \right ) = A \cap B \cap C$, the set of numbers that are multiples of 8, 9, and 10. $\displaystyle LCM (8,9,10)= 360$, so we look for multiples of 360, of which there are two under 1,000 (360 and 720). 

$\displaystyle c\left [ \left (A \cap C \right ) \cup \left (B \cap C \right ) \right ] = 2$

$\displaystyle c\left [ \left (A \cap C \right ) \cup \left (B \cap C \right ) \right ] = c \left (A \cap C \right )+ c \left (B \cap C \right )-c\left [ \left (A \cap C \right ) \cap \left (B \cap C \right ) \right ]$

$\displaystyle = 25+11-2 = 34$

The shaded region is inside $\displaystyle B$ and $\displaystyle C$, and outside of $\displaystyle A$, meaning that the shaded set represents

$\displaystyle A' \cap (B\cap C)$

Examine $\displaystyle B\cap C$ first. Each number in $\displaystyle B\cap C$ must be a multiple of 7 and 8; since the two are relatively prime, each number is a multiple of 56.

$\displaystyle 1,000 \div 56 \approx 17.9$

so seventeen elements are in $\displaystyle B \cap C$.

We eliminate all elements in $\displaystyle A$ - that is, all elements that also multiples of 6. These elements are 168, 336, 504, 672, 840 - a total of five. 

This leaves twelve elements in $\displaystyle A' \cap (B\cap C)$.

The shaded region is inside set $\displaystyle A$, so we are looking for a President who was born after 1850; this eliminates Lincoln.

The region is outside of $\displaystyle C$, so we want a President who served fewer than eight years. This eliminates Wilson and Reagan.

The region is outside of $\displaystyle B$, so we want the state of the President's birth to be fully or partly east of the Mississippi River. This eliminates Nixon.

The correct response is Harding.

The symbol $\displaystyle \cup$ stands for the union between two sets.  Therefore, $\displaystyle A\cup B$ means the set of all numbers that are in either A or B.  Looking at our choices, the only number that isn't in either A, B, or both is 23.

The notation $\displaystyle \cup$ stands for "union," which refers to everything that is in either set. $\displaystyle \textup{Calculus} \cup\textup{Spanish}$ refers to the group of students taking either Calculus or Spanish (everyone on this diagram except those taking only Biology). From the diagram, Patrick and Ashley are the only students taking neither Calculus nor Spanish, so Patrick is the correct answer.

The notation $\displaystyle A \cup C$ stands for "A union C," which refers to everything that is in either set $\displaystyle A$ or set $\displaystyle C$.

$\displaystyle A\cup C= \left \{ 12, 2, 3, 4, 11, 5 \right \}$

When we add the numbers together, we get:

$\displaystyle 12 + 2 + 3 + 4 + 11 + 5 = 37$