1840 United States presidential election in New Jersey
The 1840 United States presidential election in New Jersey took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose 8 representatives, or electors to the Electoral College, who voted for President and Vice President.
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Elections in New Jersey | ||||||||
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New Jersey voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won New Jersey by a margin of 3.59%.
Results
1840 United States presidential election in New Jersey[1] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Whig | William Henry Harrison of Ohio | John Tyler of Virginia | 33,351 | 51.74% | 8 | 100.00% | ||
Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 31,034 | 48.15% | 0 | 0.00% | ||
Liberty | James G. Birney of New York | Thomas Earle of Pennsylvania | 69 | 0.11% | 0 | 0.00% | ||
Total | 64,454 | 100.00% | 8 | 100.00% |
gollark: Most things are.
gollark: https://ieeexplore.ieee.org/ielx5/5/6021970/06021978.pdfSome actual reasonable arguments against nuclear power.
gollark: And if you fall into water, your phone will also break.
gollark: Then you would have other problems.
gollark: Perhaps we need some way to get people to not do that...
References
- "1840 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved 23 December 2013.
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