United States presidential election in Rhode Island, 1996

Main article: United States presidential election, 1996

United States presidential election in Rhode Island, 1996

November 5, 1996

Nominee Bill Clinton Bob Dole Ross Perot Party Democratic Republican Reform Home state Arkansas Kansas Texas Running mate Al Gore Jack Kemp Patrick Choate Electoral vote 4 0 0 Popular vote 233,050 104,683 43,723 Percentage 59.71% 26.82% 11.20%

180px County Results   Clinton—60-70%   Clinton—50-60%

President before election Bill Clinton Democratic Elected President Bill Clinton Democratic

The 1996 United States presidential election in Rhode Island took place on November 5, 1996 as part of the 1996 United States presidential election. Voters chose 4 representatives, or electors to the Electoral College, who voted for President and Vice President.

Rhode Island was won by President Bill Clinton (D) over Senator Bob Dole (R-KS), with Clinton winning 59.71% to 26.82% by a margin of 32.89%. Billionaire businessman Ross Perot (Reform Party of the United States of America-TX) finished in third with 11.20% of the popular vote .

Results

United States presidential election in Rhode Island, 1996
Party		Candidate	Running mate	Votes	Percentage	Electoral votes
	Democratic	Bill Clinton (incumbent)	Al Gore	233,050	59.71%	4
	Republican	Bob Dole	Jack Kemp	104,683	26.82%	0
	Reform	Ross Perot	Patrick Choate	43,723	11.20%	0
	Green	Ralph Nader	Winona LaDuke	6,040	1.55%	0
	Libertarian	Harry Browne	Jo Jorgensen	1,109	0.28%	0
	U.S. Taxpayers' Party	Howard Phillips	Herbert Titus	1,021	0.26%	0
	Natural Law	Dr. John Hagelin	Dr. V. Tompkins	435	0.11%	0
	Workers World Party	Monica Moorehead	Gloria La Riva	186	0.05%	0
	No party	Write-in		37	0.01%	0