1792 United States presidential election in Massachusetts
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Elections in Massachusetts |
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The 1792 United States presidential election in Massachusetts took place between November 2 and December 5, 1792, as part of the 1792 United States presidential election. In this election, two Congressional districts chose five electors each, the remaining two districts chose three electors. Each elector chosen by majority vote of voters in Congressional district. If an insufficient number of electors are chosen by majority vote from a Congressional district, remaining electors would be appointed by the state legislature.[1]
Massachusetts unanimously voted for independent candidate and incumbent president, George Washington. The total vote is composed of 20,343 for Federalist electors, all of whom were supportive of Washington.[2]
Results
[edit]1792 United States presidential election in Massachusetts[2] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Independent | George Washington | 20,343 | 100.00% | 10 | |
Totals | 20,343 | 100.00% | 10 |
See also
[edit]References
[edit]- ^ "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.
- ^ a b Dubin, Michael J. (2002). United States Presidential Elections, 1788-1860: The Official Results by County and State. Jefferson: McFarland & Company. pp. 4–5. ISBN 9780786410170.