Election in Rhode Island
1996 United States presidential election in Rhode Island Turnout 64.8%[1] 11.8 pp County Results Municipality Results Clinton
40–50%
50–60%
60–70%
70–80%
Dole
40–50%
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 four 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.[2]
As of 2020, this was the most recent presidential election in which the town of Scituate voted for a Democrat.
Results [ edit ] By county [ edit ] County Bill Clinton Democratic Bob Dole Republican Various candidates Other parties Margin Total votes cast # % # % # % # % Bristol 12,257 56.56% 6,988 32.25% 2,426 11.19% 5,269 24.31% 21,671 Kent 41,048 57.35% 19,992 27.95% 10,516 14.70% 21,056 29.40% 71,526 Newport 18,951 54.00% 11,500 32.77% 4,645 13.23% 7,451 21.23% 35,096 Providence 134,866 63.58% 49,901 23.52% 27,355 12.90% 84,965 40.06% 212,122 Washington 25,958 52.09% 16,302 32.71% 7,572 15.20% 9,656 19.38% 49,462 Totals 233,050 59.71% 104,683 26.82% 52,551 13.47% 128,367 32.89% 390,284
See also [ edit ] References [ edit ]