1904 United States presidential election in Rhode Island

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1904 United States presidential election in Rhode Island
Flag of Rhode Island.svg
  1900 November 8, 1904 1908  
  President Roosevelt - Pach Bros (cropped 3x4).jpg AltonBParker.png
Nominee Theodore Roosevelt Alton B. Parker
Party Republican Democratic
Home state New York New York
Running mate Charles W. Fairbanks Henry G. Davis
Electoral vote40
Popular vote41,60524,839
Percentage60.60%36.18%

Rhode Island Presidential Election Results 1904.svg
County Results
Roosevelt
  50–60%
  60–70%
  70–80%

The 1904 United States presidential election in Rhode Island took place on November 8, 1904, as part of the 1904 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

Contents

Rhode Island overwhelmingly voted for the Republican nominee, President Theodore Roosevelt, over the Democratic nominee, former Chief Judge of New York Court of Appeals Alton B. Parker. Roosevelt won Rhode Island by a margin of 24.42%.

Results

1904 United States presidential election in Rhode Island [1]
PartyCandidateRunning matePopular voteElectoral vote
Count%Count%
Republican Theodore Roosevelt of New York (incumbent) Charles Warren Fairbanks of Indiana 41,60560.60%4100.00%
Democratic Alton Brooks Parker of New York Henry Gassaway Davis of West Virginia 24,83936.18%00.00%
Socialist Eugene Victor Debs of Indiana Benjamin Hanford of New York 9561.39%00.00%
Prohibition Silas Comfort Swallow of Pennsylvania George Washington Carroll of Texas 7681.12%00.00%
Socialist Labor Charles Hunter Corregan of New York William Wesley Cox of Illinois 4880.71%00.00%
Total68,656100.00%4100.00%

See also

References

  1. "1904 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved December 23, 2013.