We read your photo as:

Given set P = 2, 4, 6. Which of the following is a subset of P? A. 2, 2, 4 B. 2, 3, 4 C. 2, 4, 6

Mathematics

ScanSolve AI Answer

Answer

C

Step-by-step solution

5 steps

Here are the solutions to the problems.

:
Let EE be the set of students who take Economics and HH be the set of students who take Home Management.
Given:
Total students n(U)=45n(U) = 45
Students taking Economics n(E)=25n(E) = 25
Students taking Home Management n(H)=21n(H) = 21
Students taking neither subject n(E∪H)′=3n(E \cup H)' = 3

  1. Find the number of students who take at least one subject.

    n(E∪H)=n(U)−n(E∪H)′n(E \cup H) = n(U) - n(E \cup H)' n(E∪H)=45−3=42n(E \cup H) = 45 - 3 = 42
  2. Use the Principle of Inclusion-Exclusion to find the number of students who take both subjects.

    n(E∪H)=n(E)+n(H)−n(E∩H)n(E \cup H) = n(E) + n(H) - n(E \cap H) 42=25+21−n(E∩H)42 = 25 + 21 - n(E \cap H) 42=46−n(E∩H)42 = 46 - n(E \cap H) n(E∩H)=46−42=4n(E \cap H) = 46 - 42 = 4
  3. Find the number of students who take Economics only.

    n(Economics only)=n(E)−n(E∩H)n(\text{Economics only}) = n(E) - n(E \cap H) n(Economics only)=25−4=21n(\text{Economics only}) = 25 - 4 = 21

    The correct option is C.
    The number of students who take Economics only is C. 21\boxed{\text{C. } 21}.

    :
    Given set P={2,4,6}P = \{2, 4, 6\}.
    A set AA is a subset of PP if every element in AA is also an element in PP.

    • A. {2,2,4}\{2, 2, 4\} is equivalent to {2,4}\{2, 4\}. Both 2 and 4 are in PP. So, {2,2,4}\{2, 2, 4\} is a subset of PP.
    • B. {2,3,4}\{2, 3, 4\}. The element 3 is in this set, but 3 is not an element of PP. Therefore, {2,3,4}\{2, 3, 4\} is not a subset of PP.
    • C. {2,4,6}\{2, 4, 6\}. All elements (2, 4, 6) are in PP. So, {2,4,6}\{2, 4, 6\} is a subset of PP.
    • D. ∅\emptyset. The empty set is a subset of every set. So, ∅\emptyset is a subset of PP. The correct option is B. The set that is not a subset of PP is B. {2,3,4}\boxed{\text{B. } \{2, 3, 4\}}.

    :
    We need to find the equivalent expression for (M′∩N)′(M' \cap N)'.
    Using De Morgan's Laws, which state that (A∩B)′=A′∪B′(A \cap B)' = A' \cup B' and (A∪B)′=A′∩B′(A \cup B)' = A' \cap B'.
    Also, (A′)′=A(A')' = A.

    Apply De Morgan's Law to (M′∩N)′(M' \cap N)':

    (M′∩N)′=(M′)′∪N′(M' \cap N)' = (M')' \cup N' (M′∩N)′=M∪N′(M' \cap N)' = M \cup N'

    The correct option is C.
    (M′∩N)′(M' \cap N)' is equivalent to C. (M∪N′)\boxed{\text{C. } (M \cup N')}.

    :
    Let SS be the set of teachers who teach pure Science and TT be the set of teachers who teach Social Science.
    Given:
    Total teachers n(U)=65n(U) = 65
    Teachers teaching pure Science n(S)=45n(S) = 45
    Teachers teaching Social Science n(T)=32n(T) = 32
    Teachers who do not teach any subject n(S∪T)′=7n(S \cup T)' = 7

  4. Find the number of teachers who teach at least one subject.

    n(S∪T)=n(U)−n(S∪T)′n(S \cup T) = n(U) - n(S \cup T)' n(S∪T)=65−7=58n(S \cup T) = 65 - 7 = 58
  5. Use the Principle of Inclusion-Exclusion to find the number of teachers who teach both subjects.

    n(S∪T)=n(S)+n(T)−n(S∩T)n(S \cup T) = n(S) + n(T) - n(S \cap T) 58=45+32−n(S∩T)58 = 45 + 32 - n(S \cap T) 58=77−n(S∩T)58 = 77 - n(S \cap T) n(S∩T)=77−58=19n(S \cap T) = 77 - 58 = 19

    The correct option is D.
    The number of teachers who teach both subjects is D. 19\boxed{\text{D. } 19}.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp