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The 1869 Rhode Island gubernatorial election took place on April 7, 1869 in order to elect the governor of Rhode Island.[1] Republican candidate and incumbent governor Seth Padelford won his first one-year term as governor[2] against Democratic candidate Lyman Pierce.[3]
Party | Candidate | Votes | % | |
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Republican | Seth Padelford | 7,359 | 68.46 | |
Democratic | Lyman Pierce | 3,390 | 31.54 | |
Total votes | 10,749 | 100.00 | ||
Republican hold |
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