1880 United States presidential election in Rhode Island
The 1880 United States presidential election in Rhode Island took place on November 2, 1880, as part of the 1880 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.
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Elections in Rhode Island |
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Rhode Island voted for the Republican nominee, James A. Garfield, over the Democratic nominee, Winfield Scott Hancock. Garfield won the state by a margin of 25.37%.
With 62.24% of the popular vote, Rhode Island would be Garfield's fourth strongest victory in terms of percentage in the popular vote after Vermont, Nebraska and Minnesota.[1]
Results
1880 United States presidential election in Rhode Island[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 | 18,195 | 62.24% | 4 | 100.00% | ||
Democratic | Winfield Scott Hancock of Pennsylvania | William Hayden English of Indiana | 10,779 | 36.87% | 0 | 0.00% | ||
Greenback | James Baird Weaver of Iowa | Barzillai Jefferson Chambers of Texas | 236 | 0.81% | 0 | 0.00% | ||
Prohibition | Neal Dow of Maine | Henry Adams Thompson of Ohio | 20 | 0.07% | 0 | 0.00% | ||
Anti-Masonic | John Wolcott Phelps of Vermont | Samuel Clarke Pomeroy of Kansas | 4 | 0.01% | 0 | 0.00% | ||
N/A | Others | Others | 1 | 0.01% | 0 | 0.00% | ||
Total | 29,235 | 100.00% | 4 | 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 - Rhode Island". Dave Leip's Atlas of U.S. Presidential Elections.
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