1896 Rhode Island gubernatorial election
The 1896 Rhode Island gubernatorial election was held on April 1, 1896. Incumbent Republican Charles W. Lippitt defeated Democratic nominee George L. Littlefield with 56.40% of the vote.
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Elections in Rhode Island | ||||||
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General election
Candidates
Major party candidates
- Charles W. Lippitt, Republican
- George L. Littlefield, Democratic
Other candidates
- Thomas H. Peabody, Prohibition
- Edward W. Theinert, Socialist Labor
- Henry A. Burlingame, People's
Results
Party | Candidate | Votes | % | ± | |
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Republican | Charles W. Lippitt | 28,472 | 56.40% | ||
Democratic | George L. Littlefield | 17,061 | 33.79% | ||
Prohibition | Thomas H. Peabody | 2,950 | 5.84% | ||
Socialist Labor | Edward W. Theinert | 1,272 | 2.52% | ||
People's | Henry A. Burlingame | 730 | 1.45% | ||
Majority | 11,411 | ||||
Turnout | |||||
Republican hold | Swing |
References
- "Congressional Quarterly's Guide to U.S. elections". Books.google.com. Retrieved July 18, 2020.
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