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Elections in Arkansas |
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The 1992 United States presidential election in Arkansas took place on November 3, 1992, as part of the 1992 United States presidential election. State voters chose six representatives, or electors to the Electoral College, who voted for president and vice president.
Arkansas was won by incumbent Governor Bill Clinton (D) with 53.21% of the popular vote over incumbent President George H. W. Bush (R-Texas) with 35.48%. Businessman Ross Perot (I-Texas) finished in third, with 10.43% of the popular vote.[1] Clinton thus won his home state by a wide margin of 17.73%, becoming the first Democratic candidate to win the state since Jimmy Carter in 1976. Arkansas and Washington, D.C., which Clinton also won, were the only contests in 1992 in which any candidate received an absolute majority of the popular vote.
Clinton carried all but five counties (Benton, Crawford, Pope, Searcy, and Sebastian) in the state. As of the 2020 presidential election[update], this is the last election in which Baxter County, Carroll County, Newton County, Boone County, and Polk County have voted for a Democratic presidential candidate.[2] This is also the most recent time that Arkansas has voted more Democratic than Maryland.[3][4]
Arkansas would be one of only three states, along with the District of Columbia, where Clinton's raw vote total exceeded that of Bush and Perot combined, the others being Maryland and New York. This is also the most recent election in which Arkansas trended more Democratic relative to the nation. Clinton's 505,823 votes is the most votes received by a Democratic presidential candidate in the state's history.
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