United States presidential election in West Virginia, 1992
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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% | |||||||||||||||||||||||||||||||||||||||
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| Elections in West Virginia | |||||||||
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The 1992 United States presidential election in West Virginia 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.
West Virginia was won by Governor Bill Clinton (D-Arkansas) with 48.41% of the popular vote over incumbent President George H.W. Bush (R-Texas) with 35.39%. Businessman Ross Perot (I-Texas) finished in third with 15.92% of the popular vote.[1] Clinton ultimately won the national vote, defeating incumbent President Bush.[2]
Results
| United States presidential election in West Virginia, 1992[1] | |||||
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| Party | Candidate | Votes | Percentage | Electoral votes | |
| Democratic | Bill Clinton | 331,001 | 48.41% | 5 | |
| Republican | George H.W. Bush (incumbent) | 241,974 | 35.39% | 0 | |
| Independent | Ross Perot | 108,829 | 15.92% | 0 | |
| Libertarian | Andre Marrou | 1,873 | 0.27% | 0 | |
| Totals | 683,677 | 100.0% | 5 | ||
References
- 1 2 "1992 Presidential General Election Results - West Virginia". U.S. Election Atlas. Retrieved 11 June 2012.
- ↑ "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved 11 June 2012.
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| Other 1992 elections | ||