1876 United States presidential election in Nebraska
(Redirected from United States presidential election in Nebraska, 1876)
The 1876 United States presidential election in Nebraska took place on November 7, 1876, as part of the 1876 United States presidential election. Voters chose three representatives, or electors to the Electoral College, who voted for president and vice president.
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Nebraska voted for the Republican nominee, Ohio Governor Rutherford B. Hayes, over the Democratic nominee, New York Governor Samuel J. Tilden by a margin of 29.4%.
With 64.70% of the popular vote, Nebraska would be Hayes' second strongest victory in terms of percentage in the popular vote after Vermont.[1]
Results
edit1876 United States presidential election in Nebraska[2] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Republican | Rutherford B. Hayes of Ohio | William A. Wheeler of New York | 31,915 | 64.70% | 3 | 100.00% | ||
Democratic | Samuel J. Tilden of New York | Thomas A. Hendricks of Indiana | 17,413 | 35.30% | 0 | 0.00% | ||
Total | 49,328 | 100.00% | 3 | 100.00% |
See also
editReferences
edit- ^ "1876 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
- ^ "1876 Presidential General Election Results - Nebraska".