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Elections in Kentucky |
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Government |
The 1808 United States presidential election in Kentucky took place between 4 November and 7 December 1808, as part of the 1808 United States presidential election. Voters chose eight representatives, or electors to the Electoral College, who voted for President and Vice President. [1]
Kentucky cast seven electoral votes, as one elector didn't vote, for the Democratic-Republican candidate James Madison over the Federalist candidate Charles C. Pinckney. The electoral votes for Vice president were cast for Madison's running mate George Clinton from New York. The state was divided into two electoral districts with four electors each, whereupon each district's voters chose the electors. [2]
1808 United States presidential election in Kentucky [3] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Democratic-Republican | James Madison | 2,679 | 98.02% | 7 | |
Federalist | Charles C. Pinckney | 54 | 1.98% | 0 | |
None | Not Cast | – | – | 1 | |
Totals | 2,733 | 100.0% | 8 | ||
The 1792 United States presidential election was the second quadrennial presidential election. It was held from Friday, November 2, to Wednesday, December 5, 1792. Incumbent President George Washington was elected to a second term by a unanimous vote in the electoral college, while John Adams was re-elected as vice president. Washington was essentially unopposed, but Adams faced a competitive re-election against Governor George Clinton of New York.
The 1808 United States presidential election was the sixth quadrennial presidential election, held from Friday, November 4, to Wednesday, December 7, 1808. The Democratic-Republican candidate James Madison defeated Federalist candidate Charles Cotesworth Pinckney decisively.
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