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1800 United States presidential election in Kentucky
From Wikipedia - Reading time: 6 min
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| Elections in Kentucky |
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 Government |
The 1800 United States presidential election in Kentucky took place between 31 October and 3 December 1800, as part of the 1800 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.[1]
Kentucky cast four electoral votes for the Democratic-Republican candidate and incumbent Vice President Thomas Jefferson over the Federalist candidate and incumbent President John Adams. The electoral votes for Vice president were cast for Jefferson's running mate Aaron Burr from New York. The state was divided into two electoral districts with two electors each, whereupon each district's voters chose the electors.[2]
Results
[edit]| 1800 United States presidential election in Kentucky[3] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Democratic-Republican | Thomas Jefferson | 119 | 100.00% | 4 | |
| Federalist | John Adams (incumbent) | 0 | 0.00% | 0 | |
| Totals | 119 | 100.00% | 4 | ||
See also
[edit]References
[edit]- ^ "A New Nation Votes". elections.lib.tufts.edu. Retrieved July 14, 2023.
- ^ "1800 Presidential General Election Results". U.S. Election Atlas. Retrieved July 14, 2023.
- ^ "1800 Presidential Election". 270towin.com. Retrieved July 14, 2023.
Licensed under CC BY-SA 3.0 | Source: https://en.wikipedia.org/wiki/1800_United_States_presidential_election_in_Kentucky1 | ↧ Download as ZWI file
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