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1904 United States presidential election in Rhode Island

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Main article:1904 United States presidential election
1904 United States presidential election in Rhode Island

← 1900
November 8, 1904
1908 →
 
NomineeTheodore RooseveltAlton B. Parker
PartyRepublicanDemocratic
Home stateNew YorkNew York
Running mateCharles W. FairbanksHenry G. Davis
Electoral vote40
Popular vote41,60524,839
Percentage60.60%36.18%

County Results
Municipality Results

Roosevelt

  50–60%
  60–70%
  70–80%
  80–90%
  90–100%

Bryan

  50–60%


President before election

Theodore Roosevelt
Republican

Elected President

Theodore Roosevelt
Republican

Elections in Rhode Island
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The1904 United States presidential election in Rhode Island took place on November 8, 1904, as part of the1904 United States presidential election. Voters chose four representatives, or electors to theElectoral College, who voted forpresident andvice president.

Rhode Island overwhelmingly voted for theRepublican nominee,PresidentTheodore Roosevelt, over theDemocratic nominee, formerChief Judge of New York Court of AppealsAlton B. Parker. Roosevelt won Rhode Island by a margin of 24.42%.

Results

[edit]
1904 United States presidential election in Rhode Island[1]
PartyCandidateRunning matePopular voteElectoral vote
Count%Count%
RepublicanTheodore Roosevelt ofNew York(incumbent)Charles Warren Fairbanks ofIndiana41,60560.60%4100.00%
DemocraticAlton Brooks Parker ofNew YorkHenry Gassaway Davis ofWest Virginia24,83936.18%00.00%
SocialistEugene Victor Debs ofIndianaBenjamin Hanford ofNew York9561.39%00.00%
ProhibitionSilas Comfort Swallow ofPennsylvaniaGeorge Washington Carroll ofTexas7681.12%00.00%
Socialist LaborCharles Hunter Corregan ofNew YorkWilliam Wesley Cox ofIllinois4880.71%00.00%
Total68,656100.00%4100.00%

See also

[edit]

References

[edit]
  1. ^"1904 Presidential General Election Results - Rhode Island". U.S. Election Atlas. RetrievedDecember 23, 2013.
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State and district results of the1904 United States presidential election
Electoral map, 1904 election


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