1900 United States presidential election in Rhode Island
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| Elections in Rhode Island |
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The 1900 United States presidential election in Rhode Island took place on November 6, 1900. Voters chose 4 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
| 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
- ^ "1900 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.
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