1880 United States presidential election in Florida
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Turnout | 19.15% of the total population 5.77 pp[1] | |||||||||||||||||||||||||
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County Results
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Elections in Florida |
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Government |
The 1880 United States presidential election in Florida took place on November 2, 1880, as part of the 1880 United States presidential election. Florida voters chose four representatives, or electors, to the Electoral College, who voted for president and vice president.[2]
Florida was won by General Winfield Scott Hancock (D–Pennsylvania), running with former Representative William Hayden English, with 54.17% of the popular vote, against Representative James Garfield (R-Ohio), running with the 10th chairman of the New York State Republican Executive Committee Chester A. Arthur, with 41.05% of the vote.[2]
Results
United States presidential election in Florida, 1880[2] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Democratic | Winfield Scott Hancock of Pennsylvania | William Hayden English of Indiana | 27,964 | 54.17% | 4 | 100.00% | ||
Republican | James Garfield of Ohio | Chester A. Arthur of New York | 23,654 | 45.83% | 0 | 0.00% | ||
Total | 51,618 | 100.00% | 4 | 100.00% |
See also
- 1880 United States House of Representatives elections in Florida
- 1880 Florida gubernatorial election
References
Categories:
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- Short description matches Wikidata
- Use mdy dates from September 2023
- Elections using electoral votes
- 1880 United States presidential election by state
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- 1880 Florida elections
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