1904 United States presidential election in Rhode Island

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.

1904 United States presidential election in Rhode Island

November 8, 1904

Nominee Theodore Roosevelt Alton B. Parker Party Republican Democratic Home state New York New York Running mate Charles W. Fairbanks Henry G. Davis Electoral vote 4 0 Popular vote 41,605 24,839 Percentage 60.60% 36.18%

County Results Roosevelt   50-60%   60-70%   70-80%

President before election Theodore Roosevelt Republican Elected President Theodore Roosevelt Republican

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%.

Results

1904 United States presidential election in Rhode Island[1]
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	Republican	Theodore Roosevelt of New York	Charles Warren Fairbanks of Indiana	41,605	60.60%	4	100.00%
	Democratic	Alton Brooks Parker of New York	Henry Gassaway Davis of West Virginia	24,839	36.18%	0	0.00%
	Socialist	Eugene Victor Debs of Indiana	Benjamin Hanford of New York	956	1.39%	0	0.00%
	Prohibition	Silas Comfort Swallow of Pennsylvania	George Washington Carroll of Texas	768	1.12%	0	0.00%
	Socialist Labor	Charles Hunter Corregan of New York	William Wesley Cox of Illinois	488	0.71%	0	0.00%
Total				68,656	100.00%	4	100.00%

See also

• United States presidential elections in Rhode Island

References

1. "1904 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.