Tag: mathematics and statistics

Quizzes (38) Questions (955)

Questions Related to mathematics and statistics

Multiple choice mathematics and statistics set language de morgan's law for set theory complement of sets different sets de morgan's law

If $U = \left {x|x\epsilon N, x < 5\right }, A = \left {x|x\epsilon N, x\leq 2\right }$ then $A' =$ __________.

  1. $\left \{1, 2\right \}$
  2. $\left \{1, 2, 3, 4, 5\right \}$
  3. $\left \{3, 4\right \}$
  4. $\left \{3, 4, 5\right \}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$ \cup  = { x|x \in N,x < 5} $

${\rm A} = \left| {x|x \in N,x \leqslant 2} \right|$
then $A' = { 3,4} $
part $C$ is correct answer.

Multiple choice mathematics and statistics set language de morgan's law for set theory complement of sets different sets de morgan's law

Comment true or false  on the following statements
$ A\cap \left( B-C \right) =\left( A\cap B \right) -\left( A\cap C \right)$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $x\in A\cap (B - C)$

$\Rightarrow x\in A$ and $x\in (B-C)$
$\Rightarrow x\in A$ and $(x\in B \text{ and } x \notin C)$
$\Rightarrow( x\in A \text{ and } x\in B)$ and $(x\in A \text{ and } x \notin C)$
$\Rightarrow (A\cap B) - (A\cap C)$
Hence true.

Multiple choice mathematics and statistics set language de morgan's law for set theory complement of sets different sets de morgan's law

A survey on a sample of $25$ new cars being sold at a local auto dealer was conducted to see which of the three popular options - air-conditioning, radio and  power windows - were already installed.
The survey found:
$15$ had air-conditioning
$2$ had air-conditioning and power windows but no radios.
$12$ had power windows
$6$ had air-conditioning and radio but no power windows.
$11$ had radio.
$4$ had radio and power windows.
$3$ had all three options.
What is the number of cars that had none of the options?

  1. $4$
  2. $3$
  3. $1$
  4. $2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$n(A\cup R\cup P)= n(A) +n(R) +n(P)-n(A\cap R)-n(R\cap P) -n(P\cap A) + n(A\cap R \cap P)$


$2= n(A\cap P)-n(A\cap P\cap R)$

$\Rightarrow n(A\cap P) = 2+3=5$

$6= n(A\cap R)-n(A\cap P\cap R)$

$\Rightarrow n(A\cap R) = 6+3=9$

$n(A\cup R\cup P)= n(A) +n(R) +n(P)-n(A\cap R)-n(R\cap P) -n(P\cap A) + n(A\cap R \cap P)$

$=15+11+12-9-5-4+3$

So answer $= 25-23 = 2$

Multiple choice mathematics and statistics set language de morgan's law for set theory complement of sets different sets de morgan's law

With usual notations $n\left( A\cup B\cup C \right) =20,n\left( A\cap B\cap C\prime  \right) =2,n\left( B\cap C\cap A\prime  \right) =n\left( A\cap C\cap B\prime  \right) =4\quad$

$ and\quad n\left( A\cap B\cap C \right) =1$, then the number of elements belonging to exactly one of the sets is

  1. $9$
  2. $13$
  3. $14$
  4. $16$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the principle of inclusion-exclusion and Venn diagram regions: n(A only) + n(B only) + n(C only) + n(A and B only) + n(B and C only) + n(A and C only) + n(A and B and C) = 20. Given n(A and B and C) = 1, n(A and B only) = 2, n(B and C only) = 4, and n(A and C only) = 4, we solve for the sum of 'only' regions.

Multiple choice mathematics and statistics set language de morgan's law for set theory complement of sets different sets de morgan's law

Let $n(u)=700,n(A)=200,n(B)=300$
$n\left( A\cap B \right) =100,n\left( A^{\prime} \cap B^{\prime}  \right) =$

  1. $400$
  2. $600$
  3. $300$
  4. $None$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Ans. $(c). n(A\cap B)=n(A\cup B)$
$=n(u)-n(A\cup B)$
$=n(u)-\left{ n\left( A \right) +n\left( B \right) -n\left( A\cap B \right)  \right}$
$=700-\left{ 200+300-100 \right} = 300$

Multiple choice mathematics and statistics set language de morgan's law for set theory complement of sets different sets de morgan's law

A - (A - B) =$ A  \cap \, B $

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

By set theory definition, A - B = A intersect B complement. Therefore, A - (A - B) = A intersect (A intersect B complement) complement = A intersect (A complement union B) = (A intersect A complement) union (A intersect B) = empty union (A intersect B) = A intersect B.

Multiple choice mathematics and statistics set language de morgan's law for set theory complement of sets different sets de morgan's law

The value of $(A\cup B\cup C)\cap {(A\cap {B}^{c}\cap {C}^{c})}^{c}\cap {C}^{c}$

  1. $B\cap {C}^{c}$
  2. ${B}^{c}\cap {C}^{c}$
  3. $B\cap C$
  4. $A\cap B\cap C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Simplifying the expression: (A union B union C) intersect (A intersect B complement intersect C complement) complement intersect C complement. The term (A intersect B complement intersect C complement) complement is (A complement union B union C). Intersecting this with C complement leaves (A complement intersect C complement) union (B intersect C complement). Further intersection with (A union B union C) simplifies to B intersect C complement.

Multiple choice mathematics and statistics set language de morgan's law for set theory complement of sets different sets de morgan's law

Given that the universal set,$ \xi =$ {x : 1 < x < 12 and x is an integer} and the sets P = {x : x is a prime number}, Q = {x : x is a multiple of 4} and R = {2, 3, 8, 9} the elements of the set $(Q \cup R)' \cap P$ are:

  1. {2, 3}

  2. {2, 3, 5}

  3. {5, 7, 11}

  4. {1, 5, 7, 11}
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Q={4,8};$     $P={2,3,5,7,11}$;     


$Q\cup R ={2.3.4.8.9}$

$(Q\cup R)' = {5,6,7,10,11}$

$(Q\cup R)'\cap P = {5,7,11}$