Which of these methods will add the value 10.5 to times?

times.Insert (10.5);times.Add ("10.5");times.Add (10.5);times.Insert (0, 10.5);

Excellent! Insert () needs two arguments inside the parentheses, and Add () needs a double-type value to work.

Oh noes! Insert () needs two arguments inside the parentheses, and Add () needs a double-type value to work.