Given $n(U) = 20$,$n(A) = 12$,$n(B) = 9$,$n(A \cap B) = 4$,where $U$ is the universal set,$A$ and $B$ are subsets of $U$,then $n((A \cup B)^C) = $

If the sets $A$ and $B$ are defined as $A = \{ (x, y) : y = \frac{1}{x}, x \in R, x \neq 0 \}$ and $B = \{ (x, y) : y = -x, x \in R \}$,then:

Let $U = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \}$,$A = \{ 1, 2, 5 \}$,and $B = \{ 6, 7 \}$. Then $A \cap B'$ is:

Let $n(U) = 700, n(A) = 200, n(B) = 300$ and $n(A \cap B) = 100,$ then $n(A^c \cap B^c) = $

Suppose $A_1, A_2, A_3, \dots, A_{30}$ are $30$ sets each having $5$ elements and $B_1, B_2, \dots, B_n$ are $n$ sets each with $3$ elements. Let $\bigcup_{i=1}^{30} A_i = \bigcup_{j=1}^n B_j = S$ and each element of $S$ belongs to exactly $10$ of the $A_i$'s and exactly $9$ of the $B_j$'s. Then $n$ is equal to:

$A$ survey shows that $63\%$ of the Americans like cheese whereas $76\%$ like apples. If $x\%$ of the Americans like both cheese and apples,then