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Suppose that an airline uses a seat width of 16.516.5 in. assume men have hip breadths that are normally distributed with a mean of 14.414.4 in. and a standard deviation of 0.90.9 in. complete parts (a) through (c) below. (a) find the probability that if an individual man is randomly selected, his hip breadth will be greater than 16.516.5 in. the probability is nothing. (round to four decimal places as needed.)

In the figure below, ∠dec ≅ ∠dce, ∠b ≅ ∠f, and segment df is congruent to segment bd. point c is the point of intersection between segment ag and segment bd, while point e is the point of intersection between segment ag and segment df. the figure shows a polygon comprised of three triangles, abc, dec, and gfe. prove δabc ≅ δgfe.

Prove (a) cosh2(x) − sinh2(x) = 1 and (b) 1 − tanh 2(x) = sech 2(x). solution (a) cosh2(x) − sinh2(x) = ex + e−x 2 2 − 2 = e2x + 2 + e−2x 4 − = 4 = . (b) we start with the identity proved in part (a): cosh2(x) − sinh2(x) = 1. if we divide both sides by cosh2(x), we get 1 − sinh2(x) cosh2(x) = 1 or 1 − tanh 2(x) = .