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The1904 United States presidential election in Rhode Island took place on November 8, 1904, as part of the1904 United States presidential election. Voters chose four representatives, or electors to theElectoral College, who voted forpresident andvice president.
Rhode Island overwhelmingly voted for theRepublican nominee,PresidentTheodore Roosevelt, over theDemocratic nominee, formerChief Judge of New York Court of AppealsAlton B. Parker. Roosevelt won Rhode Island by a margin of 24.42%.
| 1904 United States presidential election in Rhode Island[1] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Republican | Theodore Roosevelt ofNew York(incumbent) | Charles Warren Fairbanks ofIndiana | 41,605 | 60.60% | 4 | 100.00% | ||
| Democratic | Alton Brooks Parker ofNew York | Henry Gassaway Davis ofWest Virginia | 24,839 | 36.18% | 0 | 0.00% | ||
| Socialist | Eugene Victor Debs ofIndiana | Benjamin Hanford ofNew York | 956 | 1.39% | 0 | 0.00% | ||
| Prohibition | Silas Comfort Swallow ofPennsylvania | George Washington Carroll ofTexas | 768 | 1.12% | 0 | 0.00% | ||
| Socialist Labor | Charles Hunter Corregan ofNew York | William Wesley Cox ofIllinois | 488 | 0.71% | 0 | 0.00% | ||
| Total | 68,656 | 100.00% | 4 | 100.00% | ||||
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