# Î 4 ë 4 ½ e b 3 ov4 o k³ ì4¹ ¹ ¹ ¹¹ ¹ó¹³ì¹ó¹³ ì4¹ ¹ ¹

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î4ë4=½Eb)°±3 ¸ov¸|4¸ ¸ o¸ k¸(¸³ =ì±4¹ ¹ ‚¹ »¹!¹|= ¹‚Ó¹³ì±=¹‚Ó¹³ì±4¹ ¹ ‚¹ »¹!¹³=ûÒ´ü³P·P¸Òü=½F¶½c) X=al menos una a la alza, Y=tres a la alza, notemos que°±W³ = 1 − °±l³,conl =ninguna a la alza°±X|W³ =°±X ∩ W³°±W³=61059481 − °±l³=61059481 −4103928=52915.-Notemos que para aceptar a todas es necesario que al menos 2 no lleguen por lo tanto°±¸SNµÀ¸Æ ¸ ÀUw¸ê³ = °±¸o ÇN*Uê 2 *U ooNpÔN*³ = 1 − °±ooNpÔN ¸ oU Çáê 1³ = 1 −\$. 2+ 51±. 2. 8³%0.9998717.-i)°±² ∩ ´³ = °±²³°±´|²³ = 1/8ii) Si fuese subconjunto no sucedería que:°±² ∩ ´³ ≠ °±²³
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Chapter 3 / Exercise 38
Single Variable Calculus: Early Transcendentals
Stewart
Expert Verified
iii) Como²·∩ ´·=Ω− ² ∪ ´, entoncesP±A¹|B¹³ =º±Û»∩Ü»³ì±Ü»³=½¼ì±Û∪Ü³½¼ì±Ü³=½¼±ì±Û³º¼±½¾³¼±½|¾³¼ì±Û∩Ü³½¼¼±½¾³¼±½|¾³=3/4iv) Como² ∩ ´·= ² − ² ∩ ´, entoncesP±A|B³ + P±A|B¹³ =½4+º±¿³¼º±¿À³¼±½¾³¼±½|¾³= 3/4v) Claramente² ∩ ´y²·∩ ´forman una partición de´, por lo tantoP±A|B³ +P±A¹|B³ =º±Û∩Ü³ºì±Û»∩Ü³ì±Ü³=ì±Ü³ì±Ü³= 119.-a)îº¶ë½FFb)4F½FFc)4Fº¶îºî½FFd).Fî.=e)4F4Fº¶îf)4Fº¶¶º¶î½FF24.-1/7, 2/7 y 4/725.-a)Y°±² ∪ ´³ = °±²³ + °±´³ − °±²´³ = °±²³ + °±´³ = 0.4
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Term
Fall
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Estados Unidos, Desviaci n t pica, Media aritm tica, Variable aleatoria
##### We have textbook solutions for you!
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Chapter 3 / Exercise 38
Single Variable Calculus: Early Transcendentals
Stewart
Expert Verified