United States presidential election in Vermont, 1900

United States presidential election in Vermont, 1900
Vermont
November 6, 1900

 
Nominee William McKinley William Jennings Bryan
Party Republican Democratic
Home state Ohio Nebraska
Running mate Theodore Roosevelt Adlai E. Stevenson
Electoral vote 4 0
Popular vote 42,569 12,849
Percentage 75.73% 22.86%

President before election

William McKinley
Republican

Elected President

William McKinley
Republican

The 1900 United States presidential election in Vermont took place on November 6, 1900 throughout 45 states. Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.

Vermont overwhelmingly voted for the Republican nominee, President William McKinley, over the Democratic nominee, former U.S. Representative and 1896 Democratic presidential nominee William Jennings Bryan. McKinley won Vermont by a landslide margin of 52.87% in this rematch of the 1896 presidential election. The return of economic prosperity and recent victory in the Spanish–American War helped McKinley to score a decisive victory.

Results

United States presidential election in Vermont, 1900[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican William McKinley of Ohio Theodore Roosevelt of New York 42,569 75.73% 4 100.00%
Democratic William Jennings Bryan of Nebraska Adlai Ewing Stevenson I of Illinois 12,849 22.86% 0 0.00%
Prohibition John Granville Woolley of Illinois Henry Brewer Metcalf of Rhode Island 383 0.68% 0 0.00%
People's Wharton Barker of Pennsylvania Ignatius Loyola Donnelly of Minnesota 367 0.65% 0 0.00%
N/A Others Others 30 0.06% 0 0.00%
Total 56,198 100.00% 4 100.00%

References

  1. "1900 Presidential General Election Results - Vermont". U.S. Election Atlas. Retrieved 23 December 2013.
This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.