1884 United States presidential election in Rhode Island
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 County Results
Blaine 50-60% 60-70%
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| Elections in Rhode Island |
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The 1884 United States presidential election in Rhode Island took place on November 4, 1884, as part of the 1884 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.
Rhode Island voted for the Republican nominee, James G. Blaine, over the Democratic nominee, Grover Cleveland. Blaine won the state by a margin of 20.26%.
With 58.07% of the popular vote, Rhode Island would prove to be Blaine's fourth strongest victory in terms of percentage in the popular vote after Vermont, Minnesota and Kansas.[1]
Results
[edit]| 1884 United States presidential election in Rhode Island[2] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Republican | James Gillespie Blaine of Maine | John Alexander Logan of Illinois | 19,030 | 58.07% | 4 | 100.00% | ||
| Democratic | Grover Cleveland of New York | Thomas Andrews Hendricks of Indiana | 12,391 | 37.81% | 0 | 0.00% | ||
| Prohibition | John Pierce St. John of Kansas | William Daniel of Maryland | 928 | 2.83% | 0 | 0.00% | ||
| Greenback | Benjamin Franklin Butler of Massachusetts | Absolom Madden West of Mississippi | 423 | 1.29% | 0 | 0.00% | ||
| Total | 32,771 | 100.00% | 4 | 100.00% | ||||
See also
[edit]References
[edit]- ^ "1884 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
- ^ "1884 Presidential General Election Results - Rhode Island".
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