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The table shows population statistics for the ages of best actor and best supporting actor winners at an awards ceremony. the distributions of the ages are approximately bell-shaped. compare the z-scores for the actors in the following situation. best actor best supporting actor muequals42.0 muequals49.0 sigmaequals7.3 sigmaequals15 in a particular year, the best actor was 59 years old and the best supporting actor was 45 years old. determine the z-scores for each. best actor: z equals best supporting actor: z equals (round to two decimal places as needed.) interpret the z-scores. the best actor was (more than 2 standard deviations above more than 1 standard deviation above less than 1 standard deviation above less than 2 standard deviations below) the mean, which (is not, is) unusual. the best supporting actor was (less than 1 standard deviation below more than 1 standard deviation above more than 2 standard deviations below more than 1 standard deviation below) the mean, which (is is not) unusual.

Use the frequency distribution, which shows the number of american voters (in millions) according to age, to find the probability that a voter chosen at random is in the 18 to 20 years old age range. ages frequency 18 to 20 5.9 21 to 24 7.7 25 to 34 20.4 35 to 44 25.1 45 to 64 54.4 65 and over 27.7 the probability that a voter chosen at random is in the 18 to 20 years old age range is nothing. (round to three decimal places as needed.)

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) = .