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United States presidential election in Utah, 1988

United States presidential election in Utah, 1988
Utah
← 1984 November 8, 1988 1992 →
  George H. W. Bush, Vice President of the United States, official portrait.jpg Dukakis1988rally cropped.jpg
Nominee George H. W. Bush Michael Dukakis
Party Republican Democratic
Home state Texas Massachusetts
Running mate Dan Quayle Lloyd Bentsen
Electoral vote 5 0
Popular vote 428,442 207,343
Percentage 66.2% 32.1%

United States presidential election in Utah, 1968.svg
County Results
  Dukakis—60-70%
  Bush—50-60%
  Bush—60-70%
  Bush—70-80%
  Bush—80-90%

President before election

Ronald Reagan
Republican

Elected President

George H. W. Bush
Republican


Ronald Reagan
Republican

George H. W. Bush
Republican

The 1988 United States presidential election in Utah took place on November 8, 1988. All 50 states and the District of Columbia, were part of the 1988 United States presidential election. Utah voters chose 5 electors to the Electoral College, which selected the President and Vice President.

Utah 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.

Utah weighed in for this election as 13% more Republican than the national average.

The presidential election of 1988 was a very partisan election for Utah, with over 98% of the electorate in voting for either the Republican or Democratic parties, though several other parties appeared on the ballot. Every county in Utah voted in majority for the Republican candidate, except for Carbon County, which voted primarily for Dukakis.


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