1886_Rhode_Island_gubernatorial_election

1886 Rhode Island gubernatorial election

1886 Rhode Island gubernatorial election

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The 1886 Rhode Island gubernatorial election was held on April 7, 1886. Incumbent Republican George P. Wetmore defeated Democratic nominee Amasa Sprague with 53.36% of the vote.

Quick Facts Nominee, Party ...

General election

Candidates

Major party candidates

Other candidates

  • George H. Slade, Prohibition

Results

More information Party, Candidate ...

Notes

  1. Some sources gives Sprague's vote as 9,994.[6][7]

References

  1. [1]
    Dubin, Michael J. (2010). United States Gubernatorial Elections, 1861-1911. Jefferson, NC: McFarland & Co. p. 14. ISBN 978-0-7864-4722-0.
  2. [2]
    Glashan, Roy R. (1979). American Governors and Gubernatorial Elections, 1775-1978. Westport, CT: Meckler Books. pp. 272–273. ISBN 0-930466-17-9.
  3. [3]
    Kallenbach, Joseph E.; Kallenbach, Jessamine S., eds. (1977). American State Governors, 1776-1976. Vol. I. Dobbs Ferry, N.Y.: Oceana Publications, Inc. p. 517. ISBN 0-379-00665-0.
  4. [4]
    McPherson, Edward, ed. (1887). The Tribune Almanac and Political Register for 1887. New York, NY: The Tribune Association. p. 79.
  5. [5]
    Manual with Rules and Orders for the use of the General Assembly of the State of Rhode Island. 1886-7. State of Rhode Island manual. Providence, R. I.: E. L. Freeman & Son, State Printers. 1886. pp. 91–92.
  6. [6]
    "RI Governor, 1886". Our Campaigns. Retrieved September 14, 2021.
  7. [7]
    Moore, John Leo, ed. (1994). Congressional Quarterly's Guide to U.S. elections. CQ Press. ISBN 9780871879967. Retrieved October 14, 2020.
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