1788–89 United States presidential election in Delaware
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(Redirected from United States presidential election in Delaware, 1789)
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Elections in Delaware |
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The 1788–89 United States presidential election in Delaware took place on January 7, 1789 as part of the 1788–1789 United States presidential election to elect the first President. Voters chose three representatives, or electors to the Electoral College, who voted for President and Vice President.
George Mitchell, John Baning, and Gunning Bedford Jr. served as electors. George Washington and John Jay both received three electoral votes.[1]
Results[edit]
United States presidential election in Delaware, 1789 | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Independent | George Washington | 685 | 100.00% | 3 | |
Totals | 685 | 100.00% | 3 |
See also[edit]
References[edit]
- ^ Jensen & Becker 1976, p. xxviii.
Works cited[edit]
- Jensen, Merrill; Becker, Robert, eds. (1976). The First Federal Elections 1788-1790: Congress, South Carolina, Pennsylvania, Massachusetts, New Hampshire. Vol. 1. University of Wisconsin Press. ISBN 0299066908.