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Unformatted text preview: { 10 < T 50 <1 δ log[1(1 + ¯ P )(1v 10 )] } =1 δ log[1(1 + ¯ P )(1v 10 )]10 40 . Hence ¯ P = 1 . 467 . Alternatively, 0 . 25 = Pr { L ≥ } = Pr { ¯ a T 50(1 + ¯ P )¯ a 10 ≥ , T 50 > 10 } , which gives ¯ P = 1 . 467 . 4. We have L = v K x +1P ¨ a K x +1 = ± 1 + P d ² v K x +1P d and V ar ( L ) = ± 1 + P d ² 2 V ar ( v K x +1 ) = ± 1 + P d ² 2 ± 2 A x( A x ) 2 ² . By the equivalence principle, A x = P ¨ a x . Thus, P = A x ¨ a x = A x 1A x d . Hence, V ar ( L ) = 2 A x( A x ) 2 (1A x ) 2 . 2...
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 Summer '08
 MATTHEWTILL

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