1904 United States presidential election in Rhode Island
|Elections in Rhode Island|
The 1904 United States presidential election in Rhode Island took place on November 8, 1904. 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 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%.
|United States presidential election in Rhode Island, 1904|
|Party||Candidate||Running mate||Popular vote||Electoral vote|
|Republican||Theodore Roosevelt of New York||Charles Warren Fairbanks of Indiana||41,605||60.60%||4||100.00%|
|Democratic||Alton Brooks Parker of New York||Henry Gassaway Davis of West Virginia||24,839||36.18%||0||0.00%|
|Socialist||Eugene Victor Debs of Indiana||Benjamin Hanford of New York||956||1.39%||0||0.00%|
|Prohibition||Silas Comfort Swallow of Pennsylvania||George Washington Carroll of Texas||768||1.12%||0||0.00%|
|Socialist Labor||Charles Hunter Corregan of New York||William Wesley Cox of Illinois||488||0.71%||0||0.00%|
- "1904 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.
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