Answer to Question #4212 in Abstract Algebra for Bridgette
Then the probablility which is asked by the task will be equal to the
length of the set of all
t from [0,1]
that the longer piece is at least x times as long as the shorter piece.
Let t in [0,1] be the point of cutting such that the longer piece is at least x
times as long as the shorter
Consider two cases:1) Suppose 0 <= t <= 1/2.
Then the longer piece has length 1-t, and so we have the inequality:
1-t > tx
t < 1/(x+1)
2) If 1/2 <= t <= 1,
Then the longer piece has length t, and so we have the inequality:
t > (1-t)x
Thus t belongs to either of the following intervals
[ 0 , 1/(x+1) ] or [ x/(x+1) , 1 ].
The total length of these intervals is
1/(x+1) + 1 - x/(x+1) = 1/(x+1) + 1/(x+1) = 2/(x+1)
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