1884 United States presidential election in Nebraska
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| Elections in Nebraska |
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 Government |
The 1884 United States presidential election in Nebraska took place on November 4, 1884, as part of the 1884 United States presidential election. Voters chose five representatives, or electors to the Electoral College, who voted for president and vice president.
Nebraska voted for the Republican nominee, James G. Blaine, over the Democratic nominee, Grover Cleveland. Blaine won the state by a margin of 16.78%.
With 57.31% of the popular vote, Nebraska would prove to be Blaine's fifth strongest victory in terms of percentage in the popular vote after Vermont, Minnesota, Kansas and Rhode Island.[1]
Results
| 1884 United States presidential election in Nebraska[2] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Republican | James Gillespie Blaine of Maine | John Alexander Logan of Illinois | 76,912 | 57.31% | 5 | 100.00% | ||
| Democratic | Grover Cleveland of New York | Thomas Andrews Hendricks of Indiana | 54,391 | 40.53% | 0 | 0.00% | ||
| Prohibition | John Pierce St. John of Kansas | William Daniel of Maryland | 2,899 | 2.16% | 0 | 0.00% | ||
| Total | 134,202 | 100.00% | 5 | 100.00% | ||||
References
- ^ "1884 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
- ^ "1884 Presidential General Election Results - Nebraska".
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