Around robin tounament of n contestants is a set of matches where each of the n 2 pairs of contestants play each other exactly once, with the outcome of play being that one contestant wins and one loses. for an integer k < n, let’s consider whether it is possible that the tournament outcome is such that for every set of k player, there is a player that beat each member of that set. show that if n k · 1 − 1 2 kn−k < 1 then such an outcome is possible. you may assume that the probability that any contestant beats another is 50/50

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step-by-step explanation:

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step-by-step explanation:

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt