1788–89 United States presidential election in Delaware
(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][2]
Results
[edit]| Party | Candidate | Votes | % | ±% | |
|---|---|---|---|---|---|
| Federalist | George Mitchell | 522 | |||
| Federalist | Gunning Bedford Jr. | 163 | |||
| Federalist | John Banning | [a] | |||
| Total votes | >685 | 100.00% | |||
See also
[edit]References
[edit]- ^ Jensen & Becker 1976, p. xxviii.
- ^ a b c "City and County Statistics". Tufts University. Archived from the original on September 2, 2024.
Notes
[edit]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.
State and district results of the 1788–89 United States presidential election | ||
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