Pick a random number (evenly distributed) between 0 and 1. Continue picking random numbers as long as they keep decreasing; stop picking when you obtain a number that is greater than the previous one you picked. What is the expected number of numbers you pick?

e = 2.718

Step-by-step explanation:

- There're "n!" ways of arranging "n" numbers, supposing that there're n picks, then the first (n−1) picks are in descending order, there are (n−1) ways of choosing the first (n−1) numbers and thus the probability of picking just n numbers is:

                                      (n-1) / n!

- The expected value (E) would be:

                   E = ∑ n*(n-1)/n!

                      = ∑ n*(n-1)/n*(n-1)*(n-2)!

                      = ∑ (n = 2 to infinity) [ 1 / (n-2)! ] = ∑ (n = 0 to infinity) [ 1 / (n)! ]

                      = e   (Maclaurin series approximation)

8(-10x+3.5y-7)

8×-10x + 8×3.5y + 8×-7

-80x + 28y - 56

therefore

-80x+28y-56 is the correct answer