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1874 Rhode Island gubernatorial election

From Wikipedia, the free encyclopedia

1874 Rhode Island gubernatorial election

← 1873 1 April 1874 1875 →
 
Nominee Henry Howard Lyman Pierce
Party Republican Democratic
Popular vote 12,335 1,589
Percentage 87.48% 11.27%

County results
Howard:      60–70%      80–90%      >90%

Governor before election

Henry Howard
Republican

Elected Governor

Henry Howard
Republican

The 1874 Rhode Island gubernatorial election was held on 1 April 1874 in order to elect the governor of Rhode Island. Incumbent Republican governor Henry Howard won re-election against Democratic nominee Lyman Pierce.[1]

General election

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On election day, 1 April 1874, incumbent Republican governor Henry Howard won re-election by a margin of 10,746 votes against his opponent Democratic nominee Lyman Pierce, thereby retaining Republican control over the office of governor. Howard was sworn in for his second term on 5 May 1874.[2]

Results

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Rhode Island gubernatorial election, 1874
Party Candidate Votes %
Republican Henry Howard (incumbent) 12,335 87.48
Democratic Lyman Pierce 1,589 11.27
Scattering 177 1.25
Total votes 14,101 100.00
Republican hold

References

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  1. ^ "Henry Howard". National Governors Association. Retrieved 8 April 2024.
  2. ^ "RI Governor". ourcampaigns.com. 6 October 2005. Retrieved 8 April 2024.