United States presidential election in Vermont, 1880
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The 1880 United States presidential election in Vermont took place on November 2, 1880, as part of the 1880 United States presidential election. Voters chose 5 representatives, or electors to the Electoral College, who voted for president and vice president.
Vermont voted for the Republican nominee, James A. Garfield, over the Democratic nominee, Winfield Scott Hancock. Garfield won Vermont by a margin of 41.66%.
With 69.81% of the popular vote, Vermont would be Garfield's strongest victory in terms of percentage in the popular vote[1].
Vice President-elect Chester Alan Arthur was born in Vermont, more specifically in the town of Fairfield.
Results
United States presidential election in Vermont, 1880[2] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Republican | James Abram Garfield of Ohio | Chester Alan Arthur of New York | 45,091 | 69.81% | 5 | 100.00% | ||
Democratic | Winfield Scott Hancock of Pennsylvania | William Hayden English of Indiana | 18,182 | 28.15% | 0 | 0.00% | ||
Greenback | James Baird Weaver of Iowa | Barzillai Jefferson Chambers of Texas | 1,212 | 1.88% | 0 | 0.00% | ||
N/A | Others | Others | 109 | 0.17% | 0 | 0.00% | ||
Total | 48,829 | 100.00% | 5 | 100.00% |
References
- ↑ "1880 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
- ↑ "1880 Presidential General Election Results - Vermont".
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