United States presidential election in Nebraska, 1992

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United States presidential election in Nebraska, 1992
Nebraska
← 1988 November 3, 1992 1996 →
  43 George H.W. Bush 3x4.jpg 44 Bill Clinton 3x4.jpg RossPerotColor.jpg
Nominee George H.W. Bush Bill Clinton Ross Perot
Party Republican Democratic Independent
Home state Texas Arkansas Texas
Running mate Dan Quayle Al Gore James Stockdale
Electoral vote 5 0 0
Popular vote 344,346 217,344 174,687
Percentage 46.6% 29.4% 23.6%

NE1992.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

The 1992 United States presidential election in Nebraska took place on November 3, 1992, as part of the 1992 United States presidential election. Voters chose five representatives, or electors to the Electoral College, who voted for President and Vice President.

Nebraska was won by incumbent President George H.W. Bush (R-Texas) with 46.58% of the popular vote over Governor Bill Clinton (D-Arkansas) with 29.40%. Businessman Ross Perot (I-Texas) finished in third with 23.63% of the popular vote.[1] Clinton ultimately won the national vote, defeating both incumbent President Bush and Perot.[2]

Results[edit]

United States presidential election in Nebraska, 1992[1]
Party Candidate Votes Percentage Electoral votes
Republican George H.W. Bush (incumbent) 344,346 46.58% 5
Democratic Bill Clinton 217,344 29.40% 0
Independent Ross Perot 174,687 23.63% 0
Libertarian Andre Marrou 1,344 0.18% 0
New Alliance Party Lenora Fulani 848 0.11% 0
Natural Law Dr. John Hagelin 714 0.10% 0
Totals 739,283 100.0% 5

References[edit]

  1. ^ a b "1992 Presidential General Election Results - Nebraska". U.S. Election Atlas. Retrieved 8 June 2012. 
  2. ^ "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved 8 June 2012.