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.
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 County Results
Blaine 50-60% 60-70%
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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
editReferences
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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