Using the principle of inclusion-exclusion, we can calculate the number of students who play exactly two sports: \[ n({exactly two}) = n(A \cap B) + n(B \cap C) + n(A \cap C) - 3n(A \cap B \cap C) \] From the problem: - \( n(A \cup B \cup C) = 100 \) - \( n(A) = 60 \), \( n(B) = 55 \), \( n(C) = 70 \) - \( n(A \cap B \cap C) = 20 \) Thus, the number of students who play exactly two sports is \( 45 \).
Thus, the correct answer is (B).
Let \(f:R→R\) be defined by \(f(x)=\){\(2x+3,x≤5 3x+α,x>5\) .Then the value of \(α\) so that f is continuous on \(R\) is
Kepler's second law (law of areas) of planetary motion leads to law of conservation of