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1890 Idaho gubernatorial election

From Wikipedia, the free encyclopedia

1890 Idaho gubernatorial election

October 1, 1890
1892 →
 
NomineeGeorge L. ShoupBenjamin Wilson
PartyRepublicanDemocratic
Popular vote10,2627,948
Percentage56.35%43.65%

County results
Shoup:     50–60%     60–70%
Wilson:     50–60%

Governor before election

George L. Shoup (Territorial)
Republican

Elected Governor

George L. Shoup
Republican

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The1890 Idaho gubernatorial election was held on October 1, 1890, in order to elect the firstgovernor of Idaho uponIdaho acquiring statehood in July 1890. IncumbentRepublican governor of theIdaho TerritoryGeorge L. Shoup defeatedDemocratic nomineeBenjamin Wilson.[1]

General election

[edit]

On election day, October 1, 1890,Republican nomineeGeorge L. Shoup won the election by a margin of 2,314 votes against his opponentDemocratic nomineeBenjamin Wilson, thereby retaining Republican control over the new office of governor. Shoup was sworn in as the first governor of the new state ofIdaho on December 8, 1890.[2]

Results

[edit]
Idaho gubernatorial election, 1890
PartyCandidateVotes%
RepublicanGeorge L. Shoup (incumbent[a])10,26256.35%
DemocraticBenjamin Wilson7,94843.65%
Total votes18,210100.00%
Republicanhold

Notes

[edit]
  1. ^Incumbent governor ofIdaho Territory.

References

[edit]
  1. ^"George L. Shoup". National Governors Association. Retrieved2023-05-12.
  2. ^"ID Governor". ourcampaigns.com. September 26, 2005. Retrieved2023-05-12.
Retrieved from "https://en.wikipedia.org/w/index.php?title=1890_Idaho_gubernatorial_election&oldid=1325728211"
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