MathematicsMedium115×since 2008Q4389The statement (p∧(∼q))∨((∼p)∧q)∨((∼p)∧(∼q))(p \wedge(\sim q)) \vee((\sim p) \wedge q) \vee((\sim p) \wedge(\sim q))(p∧(∼q))∨((∼p)∧q)∨((∼p)∧(∼q)) is equivalent to _________.A(∼p)∨(∼q)(\sim p) \vee(\sim q)(∼p)∨(∼q)Bp∨(∼q)p \vee(\sim q)p∨(∼q)Cp∨q\mathrm{p} \vee \mathrm{q}p∨qD(∼p)∨q(\sim p) \vee q(∼p)∨qCheck answerSkip