I dont think so. All start break shouldnt count in the argument because AL teams all have an extra batter and its stupid anyway. I also think the worst teams in the AL are the worst in baseball overall. The money is even more divided so teams like Boston or Anaheim just crush on the lower teams giving them better records. There is no dominate team in the NL but there are still good teams that can beat the best AL team in the world series. Besides STL won it all last year and they are NL.
2007-07-16 10:05:05
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answer #1
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answered by LA 2
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Yes the AL is better:
evidence
#1 AL has won 10 of the last 11 all-star games (the 11th was a tie)
#2 In 2007 the AL won interleague play 137-115 and in 2006 154-98 and has an all-time record since it began in 1997 of 1404-1300.
#3 Since 1987, the AL has won the world series 12-7
There ya go. The AL has had a run of about 10 years or so of dominance over the NL.
2007-07-16 16:10:20
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answer #2
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answered by IamCount 4
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I would say that the AL is stronger. Both divisions have its weak team. In the NL, there's Cincinnati, San Francisco, and Washington, and Pittsburgh, while in the AL there's Kansas City, Texas, Tampa Bay, and Chicago. Based on strength, the Indians, Tigers, Red Sox, and Angels are all better than the best NL team.
2007-07-16 16:02:04
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answer #3
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answered by Anonymous
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That and the past 10 years of all-star games are pointng to the strength of the AL. It's surprising though because of the NL showing in the series of late.
2007-07-16 16:26:06
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answer #4
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answered by Oz 7
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the AL is just making the NL More even
2007-07-16 16:03:37
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answer #5
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answered by Kramer 4
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The all star game was decided by one run and the American league won. The NL Cardinals are defending word series champs.
Seems pretty even to me.
2007-07-16 16:06:03
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answer #6
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answered by Willie J 5
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Yes-the hitters are definitely better...look at the AL East. They also have the DH.
2007-07-16 15:58:27
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answer #7
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answered by edwin m 2
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yup
2007-07-16 15:50:50
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answer #8
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answered by That 70s Show Dude 6
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ya
2007-07-16 15:55:41
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answer #9
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answered by Anonymous
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