1788–89 United States presidential election in Massachusetts
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Elections in Massachusetts |
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The 1788–89 United States presidential election in Massachusetts took place on January 7, 1789, as part of the 1788–1789 United States presidential election to elect the first President. Massachusetts was entitled to 10 electors, with two being appointed by the state legislature and the rest being chosen by state legislature from the two most popular candidates in each U.S. House district. Each elector voted once for President and again for Vice President.[1]
Massachusetts unanimously voted for independent candidate and commander-in-chief of the Continental Army, George Washington. The total vote is composed of 17,740 for Federalist electors, all of whom were supportive of Washington.
Results
1788-1789 United States presidential election in Massachusetts | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Independent | George Washington | 17,740 | 100.00% | 10 | |
Totals | 17,740 | 100.00% | 10 |
See also
References
- ^ "The Electoral Count for the Presidential Election of 1789". The Papers of George Washington. Archived from the original on September 14, 2013. Retrieved May 4, 2005.