United States presidential election in Delaware, 1988
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County Results
Dukakis—70-80%
Dukakis—60-70%
Dukakis—50-60%
Bush—50-60%
Bush—60-70%
Bush—70-80% | ||||||||||||||||||||||||||||||||
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The 1988 United States presidential election in Delaware took place on November 8, 1988. All 50 states and the District of Columbia, were part of the 1988 United States presidential election. Delaware voters chose 3 electors to the Electoral College, which selected the President and Vice President.
Delaware was won by incumbent United States Vice President George H. W. Bush of Texas, who was running against Massachusetts Governor Michael Dukakis. Bush ran with Indiana Senator Dan Quayle as Vice President, and Dukakis ran with Texas Senator Lloyd Bentsen.
Delaware weighed in for this election as about 1% more Republican than the national average.
Partisan background
The presidential election of 1988 was a very partisan election for Delaware, with nearly 99.5% of the electorate voting for either the Democratic or Republican parties, and only four political parties on the ballot, statewide.[1] All counties in Delaware voted in majority for Bush.
Elections in Delaware | ||||||||
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Results
United States presidential election in Delaware, 1988 | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Republican | George H. W. Bush | 139,639 | 55.88% | 3 | |
Democratic | Michael Dukakis | 108,647 | 43.48% | 0 | |
Libertarian | Ron Paul | 1,162 | 0.47% | 0 | |
New Alliance Party | Lenora Fulani | 443 | 0.18% | 0 | |
Totals | 249,891 | 100.0% | 3 |
See also
References
- ↑ "Dave Leip's Atlas of U.S. Presidential Elections". Uselectionatlas.org. Retrieved 2013-07-21.