Jump to content

1879 Rhode Island gubernatorial election

From Wikipedia, the free encyclopedia

1879 Rhode Island gubernatorial election

← 1878 April 2, 1879 1880 →
 
Nominee Charles C. Van Zandt Thomas W. Segar
Party Republican Democratic
Popular vote 9,717 5,506
Percentage 62.09% 35.18%

County results
Zandt:      50–60%      60–70%      70–80%      80–90%

Governor before election

Charles C. Van Zandt
Republican

Elected Governor

Charles C. Van Zandt
Republican

The 1879 Rhode Island gubernatorial election was held on April 2, 1879. Incumbent Republican Charles C. Van Zandt defeated Democratic nominee Thomas W. Segar with 62.09% of the vote.

General election

[edit]

Candidates

[edit]

Major party candidates

  • Charles C. Van Zandt, Republican
  • Thomas W. Segar, Democratic

Other candidates

  • Samuel Hill, Independent

Results

[edit]
1879 Rhode Island gubernatorial election[1]
Party Candidate Votes % ±%
Republican Charles C. Van Zandt (incumbent) 9,717 62.09%
Democratic Thomas W. Segar 5,506 35.18%
Greenback Samuel Hill 318 2.03%
Majority 4,211
Turnout
Republican hold Swing

References

[edit]
  1. ^ Moore, John Leo, ed. (1994). Congressional Quarterly's Guide to U.S. elections. CQ Press. ISBN 9780871879967. Retrieved November 1, 2020.