1890 New Hampshire gubernatorial election
Appearance
| |||||||||||||||||
| |||||||||||||||||
County results Tuttle: 40–50% 50–60% Amsden: 40–50% 50–60% | |||||||||||||||||
|
Elections in New Hampshire |
---|
The 1890 New Hampshire gubernatorial election was held on November 4, 1890. Republican nominee Hiram A. Tuttle defeated Democratic nominee Charles H. Amsden with 49.26% of the vote.
General election
Candidates
Major party candidates
- Hiram A. Tuttle, Republican
- Charles H. Amsden, Democratic
Other candidates
- Josiah M. Fletcher, Prohibition
Results
Party | Candidate | Votes | % | ±% | |
---|---|---|---|---|---|
Republican | Hiram A. Tuttle | 42,479 | 49.26% | ||
Democratic | Charles H. Amsden | 42,386 | 49.15% | ||
Prohibition | Josiah M. Fletcher | 1,343 | 1.56% | ||
Majority | 93 | ||||
Turnout | |||||
Republican hold | Swing |
References
- ^ Kalb, Deborah (December 24, 2015). Guide to U.S. Elections. ISBN 9781483380353. Retrieved July 16, 2020.
Categories:
- New Hampshire gubernatorial elections
- 1890 United States gubernatorial elections
- 1890 New Hampshire elections
- November 1890 events
- 1890 in New Hampshire
- 1890s in New Hampshire
- 1890s New Hampshire elections
- 1890 elections
- 1890 elections in North America
- 1890 elections in the United States
- United States gubernatorial elections in the 1890s
- Government of New Hampshire