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United States presidential election in Nebraska, 1880

From Wikipedia, the free encyclopedia

United States presidential election in Nebraska, 1880

← 1876 November 2, 1880 1884 →
 
JamesGarfield.png
WinfieldScottHancock.png
Nominee James A. Garfield Winfield S. Hancock
Party Republican Democratic
Home state Ohio Pennsylvania
Running mate Chester A. Arthur William H. English
Electoral vote 3 0
Popular vote 54,979 28,523
Percentage 62.87% 32.62%

President before election

Rutherford B. Hayes
Republican

Elected President

James A. Garfield
Republican

The 1880 United States presidential election in Nebraska took place on November 2, 1880, as part of the 1880 United States presidential election. Voters chose three representatives, or electors to the Electoral College, who voted for president and vice president.

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 30.25%.

With 62.87% of the popular vote, Rhode Island would be Garfield's second strongest victory in terms of percentage in the popular vote after Vermont[1].

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Transcription

Results

United States presidential election in Nebraska, 1880[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican James Abram Garfield of Ohio Chester Alan Arthur of New York 54,979 62.87% 3 100.00%
Democratic Winfield Scott Hancock of Pennsylvania William Hayden English of Indiana 28,523 32.62% 0 0.00%
Greenback James Baird Weaver of Iowa Barzillai Jefferson Chambers of Texas 3,950 4.52% 0 0.00%
Total 87,452 100.00% 3 100.00%

References

  1. ^ "1880 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. ^ "1880 Presidential General Election Results - Nebraska".


This page was last edited on 12 November 2018, at 19:01
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