1996 United States presidential election in Rhode Island
Appearance
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![]() County Results
Clinton—60-70%
Clinton—50-60% | |||||||||||||||||||||||||||||||||
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Elections in Rhode Island |
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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 .[1]
Results
United States presidential election in Rhode Island, 1996 | ||||||
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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 |