1788–89 United States presidential election in Pennsylvania
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(Redirected from United States presidential election in Pennsylvania, 1789)
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Elections in Pennsylvania |
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Government |
The 1788–89 United States presidential election in Pennsylvania took place on January 7, 1789, as part of the 1788–89 United States presidential election to elect the first President. Voters chose 10 representatives, or electors to the Electoral College, who voted for President and Vice President.
Pennsylvania unanimously voted for nonpartisan candidate and commander-in-chief of the Continental Army, George Washington. The total vote is composed of 6,711 for Federalist electors, known as the Lancaster Ticket, and 672 for Anti-Federalist electors, known as the Harrisburg Ticket, all of whom were supportive of Washington.[1]
Results
[edit]1788-1789 United States presidential election in Pennsylvania[1] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Independent | George Washington | 7,383 | 100.00% | 15 | |
Totals | 7,383 | 100.00% | 15 |
See also
[edit]References
[edit]- ^ a b "A New Nation Votes". elections.lib.tufts.edu. Retrieved July 16, 2024.