1904 United States presidential election in Rhode Island explained

See main article: 1904 United States presidential election.

Election Name:1904 United States presidential election in Rhode Island
Country:Rhode Island
Type:presidential
Ongoing:no
Previous Election:1900 United States presidential election in Rhode Island
Previous Year:1900
Next Election:1908 United States presidential election in Rhode Island
Next Year:1908
Election Date:November 8, 1904
Image1:President Roosevelt - Pach Bros (cropped 3x4).jpg
Nominee1:Theodore Roosevelt
Party1:Republican Party (United States)
Home State1:New York
Running Mate1:Charles W. Fairbanks
Electoral Vote1:4
Popular Vote1:41,605
Percentage1:60.60%
Nominee2:Alton B. Parker
Party2:Democratic Party (United States)
Home State2:New York
Running Mate2:Henry G. Davis
Electoral Vote2:0
Popular Vote2:24,839
Percentage2:36.18%
Map Size:250px
President
Before Election:Theodore Roosevelt
Before Party:Republican Party (United States)
After Election:Theodore Roosevelt
After Party:Republican Party (United States)

The 1904 United States presidential election in Rhode Island took place on November 8, 1904, as part of the 1904 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

Rhode Island overwhelmingly voted for the Republican nominee, President Theodore Roosevelt, over the Democratic nominee, former Chief Judge of New York Court of Appeals Alton B. Parker. Roosevelt won Rhode Island by a margin of 24.42%.

See also