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1828 United States presidential election in Rhode Island

From Wikipedia, the free encyclopedia

1828 United States presidential election in Rhode Island

← 1824 October 31 – December 2, 1828 1832 →
 
JohnQAdams.jpg
Andrew Jackson.jpg
Nominee John Quincy Adams Andrew Jackson
Party National Republican Democratic
Home state Massachusetts Tennessee
Running mate Richard Rush John C. Calhoun
Electoral vote 4 0
Popular vote 2,754 821
Percentage 77.03% 22.97%

The 1828 United States presidential election in Rhode Island took place between October 31 and December 2, 1828, as part of the 1828 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 National Republican candidate, John Quincy Adams, over the Democratic candidate, Andrew Jackson. Adams won Rhode Island by a margin of 54.06%.

With 77.03% of the popular vote, Adams' victory in Rhode Island made it his strongest state in the 1828 election.[1]

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Transcription

Results

1828 United States presidential election in Rhode Island[2]
Party Candidate Votes Percentage Electoral votes
National Republican John Quincy Adams 2,754 77.03% 4
Democratic Andrew Jackson 821 22.97% 0
Totals 3,575 100.0% 4

References

  1. ^ "1828 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. ^ "1828 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 28 February 2013.


This page was last edited on 26 January 2020, at 15:52
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