United States Presidential Election In Missouri, 1992
United States presidential election in Missouri, 1992
Missouri
1988 November 3, 1992 1996 ->
  44 Bill Clinton 3x4.jpg 43 George H.W. Bush 3x4.jpg RossPerotColor.jpg
Nominee Bill Clinton George H.W. Bush Ross Perot
Party Democratic Republican Independent
Home state Arkansas Texas Texas
Running mate Al Gore Dan Quayle James Stockdale
Electoral vote 11 0 0
Popular vote 1,053,873 811,159 518,741
Percentage 44.1% 33.9% 21.7%

MO1992.jpg
County Results
  Clinton-->70%
  Clinton--60-70%
  Clinton--50-60%
  Clinton--40-50%
  Bush--40-50%
  Bush--50-60%
  Bush--60-70%
  Bush-->70%
  Perot--40-50%

President before election

George H. W. Bush
Republican

Elected President

Bill Clinton
Democratic

United States presidential election in Missouri, 1992 [1]
Party Candidate Votes Percentage Electoral votes
Democratic Bill Clinton 1,053,873 44.07% 11
Republican George H.W. Bush 811,159 33.92% 0
Independent Ross Perot 518,741 21.69% 0
Libertarian Andre Marrou 7,497 0.31% 0
Totals 2,384,522 100.0% 11

This article describes the United States presidential election, 1992, in Missouri. Since 1904, Missouri has voted for the eventual winner of the election in a presidential election, with the exceptions of the 1956, 2008, and 2012 elections.

References


  This article uses material from the Wikipedia page available here. It is released under the Creative Commons Attribution-Share-Alike License 3.0.


United_States_presidential_election_in_Missouri,_1992



 

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