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1840 United States presidential election in Rhode Island

From Wikipedia, the free encyclopedia

1840 United States presidential election in Rhode Island

← 1836 October 30 - December 2, 1840 1844 →
 
William Henry Harrison (cropped).jpg
Van Buren.jpg
Nominee William Henry Harrison Martin Van Buren
Party Whig Democratic
Home state Ohio New York
Running mate John Tyler none
Electoral vote 4 0
Popular vote 5,278 3,301
Percentage 61.22% 38.29%

President before election

Martin Van Buren
Democratic

Elected President

William Henry Harrison
Whig

The 1840 United States presidential election in Rhode Island took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose 4 representatives, or electors to the Electoral College, who voted for President and Vice President.

Rhode Island voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Rhode Island by a margin of 22.93%.

With 61.22% of the popular vote, Rhode Island would be Harrison's third strongest state in the 1840 election after Kentucky and Vermont.[1]

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Transcription

Results

1840 United States presidential election in Rhode Island[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Whig William Henry Harrison of Ohio John Tyler of Virginia 5,278 61.22% 4 100.00%
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 3,301 38.29% 0 0.00%
Liberty James G. Birney of New York Thomas Earle of Pennsylvania 42 0.49% 0 0.00%
Total 8,621 100.00% 4 100.00%

References

  1. ^ "1840 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. ^ "1840 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.


This page was last edited on 9 February 2020, at 14:37
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