United States presidential election in Rhode Island, 1996

United States presidential election in Rhode Island, 1996
Rhode Island
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 59.7% 26.8% 11.2%

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 .[1]

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

References

  1. ↑ http://uselectionatlas.org/RESULTS/index.html

See also

This article is issued from Wikipedia - version of the 11/5/2016. The text is available under the Creative Commons Attribution/Share Alike but additional terms may apply for the media files.