United States presidential election in South Dakota, 1992

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United States presidential election in South Dakota, 1992

← 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 3 0 0
Popular vote 136,718 124,888 73,295
Percentage 40.66% 37.14% 21.80%

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

South Dakota was won by incumbent President George H.W. Bush (R-Texas) with 40.66% of the popular vote over Governor Bill Clinton (D-Arkansas) with 37.14%. Businessman Ross Perot (I-Texas) finished in third with 21.80% of the popular vote.[1] Clinton ultimately won the national vote, defeating incumbent President Bush and Perot.[2]

Results

United States presidential election in South Dakota, 1992[1]
Party Candidate Votes Percentage Electoral votes
Republican George H.W. Bush (incumbent) 136,718 40.66% 3
Democratic Bill Clinton 124,888 37.14% 0
Independent Ross Perot 73,295 21.80% 0
Libertarian Andre Marrou 814 0.24% 0
Natural Law Dr. John Hagelin 429 0.13% 0
New Alliance Party Lenora Fulani 110 0.03% 0
Totals 336,254 100.0% 3

References

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