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McKinley 40–50% 50–60% 60–70% 70–80% 80–90% 90–100%
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The1900 United States presidential election in Rhode Island took place on November 6, 1900, as part of the1900 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,PresidentWilliam McKinley, over theDemocratic nominee, formerU.S. Representative and 1896 Democratic presidential nomineeWilliam Jennings Bryan. McKinley won Rhode Island by a margin of 24.7% in this rematch of the1896 presidential election. The return of economic prosperity and recent victory in theSpanish–American War helped McKinley to score a decisive victory.
Bryan had previous lost Rhode Island to McKinleyfour years earlier and would later lose the state again in1908 toWilliam Howard Taft.
| 1900 United States presidential election in Rhode Island[1] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Republican | William McKinley ofOhio(incumbent) | Theodore Roosevelt ofNew York | 33,784 | 59.74% | 4 | 100.00% | ||
| Democratic | William Jennings Bryan ofNebraska | Adlai Ewing Stevenson I ofIllinois | 19,812 | 35.04% | 0 | 0.00% | ||
| Prohibition | John Granville Woolley ofIllinois | Henry Brewer Metcalf ofRhode Island | 1,529 | 2.70% | 0 | 0.00% | ||
| Socialist Labor | Joseph F. Malloney ofMassachusetts | Valentine Remmel ofPennsylvania | 1,423 | 2.52% | 0 | 0.00% | ||
| Total | 56,198 | 100.00% | 4 | 100.00% | ||||
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