United States presidential election in Montana, 1996

United States presidential election in Montana, 1996
Montana
November 5, 1996 (1996-11-05)

 
Nominee Bob Dole Bill Clinton Ross Perot
Party Republican Democratic Reform
Home state Kansas Arkansas Texas
Running mate Jack Kemp Al Gore Patrick Choate
Electoral vote 3 0 0
Popular vote 179,652 167,922 55,229
Percentage 44.11% 41.23% 13.56%


President before election

Bill Clinton
Democratic

Elected President

Bill Clinton
Democratic

The 1996 United States presidential election in Montana took place on November 5, 1996. Voters chose 3 representatives, or electors to the Electoral College, who voted for President and Vice President.

Montana voted for Senate Majority Leader Bob Dole over President Bill Clinton by a slim margin of 2.88%.[1] Billionaire businessman Ross Perot (Reform Party of the United States of America-TX) finished in third with 13.56% of the popular vote in Montana.[2]

Results

United States presidential election in Montana, 1996[3]
Party Candidate Running mate Votes Percentage Electoral votes
Republican Bob Dole Jack Kemp 179,652 44.11% 3
Democratic Bill Clinton Al Gore 167,922 41.23% 0
Reform Ross Perot Patrick Choate 55,229 13.56% 0
Libertarian Harry Browne Jo Jorgensen 2,526 0.62% 0
Natural Law (write in) John Hagelin Mike Tompkins 1,754 0.43% 0
Constitutional (write in) Howard Phillips Herbert Titus 152 0.04% 0
(write in) Charles E. Collins Rosemary Giumarra 20 0.00% 0
Independent (write in) Write-ins 5 0.00 0
Socialist (write in) Mary Cal Hollis Eric Chester 1 0.00 0

References

This article is issued from Wikipedia - version of the Friday, March 18, 2016. The text is available under the Creative Commons Attribution/Share Alike but additional terms may apply for the media files.