1900 United States presidential election in Rhode Island

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1900 United States presidential election in Rhode Island

← 1896 November 6, 1900 1904 →
  William McKinley by Courtney Art Studio, 1896.jpg WilliamJBryan1902.png
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 33,784 19,812
Percentage 59.74% 35.04%

President before election

William McKinley
Republican

Elected President

William McKinley
Republican

The 1900 United States presidential election in Rhode Island took place on November 6, 1900. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

Rhode Island 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 Rhode Island by a margin of 24.7% 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[edit]

1900 United States presidential election in Rhode Island[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican William McKinley of Ohio Theodore Roosevelt of New York 33,784 59.74% 4 100.00%
Democratic William Jennings Bryan of Nebraska Adlai Ewing Stevenson I of Illinois 19,812 35.04% 0 0.00%
Prohibition John Granville Woolley of Illinois Henry Brewer Metcalf of Rhode Island 1,529 2.70% 0 0.00%
Socialist Labor Joseph F. Malloney of Massachusetts Valentine Remmel of Pennsylvania 1,423 2.52% 0 0.00%
Total 56,198 100.00% 4 100.00%

References[edit]

  1. ^ "1900 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.