United States presidential election in Oregon, 1988
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Main article: United States presidential election, 1988
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| County Results
Dukakis—60-70%
Dukakis—50-60%
Dukakis—<50%
Bush—<50%
Bush—50-60%
Bush—60-70%
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The 1988 United States presidential election in Oregon took place on November 8, 1988 as part of the 1988 United States presidential election. Voters chose seven representatives, or electors to the Electoral College, who voted for President and Vice President.
Oregon was won by the Democratic nominee, Massachusetts governor Michael Dukakis, over the Republican nominee, Vice President George H. W. Bush. Oregon was one of just 10 states won by Dukakis in an election won by Vice President Bush. It also marked was the first victory by a Democratic Presidential candidate in Oregon since 1964; Democrats have won every Presidential election in Oregon since.[1]
[edit] Results
| United States presidential election in Oregon, 1988[2] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Democratic | Michael Dukakis | 616,206 | 51.28% | 7 | |
| Republican | George H. W. Bush | 560,126 | 46.61% | 0 | |
| Libertarian | Ron Paul | 14,811 | 1.23% | 0 | |
| Independent | Lenora Fulani | 6,487 | 0.54% | 0 | |
| No party | Write-in | 3,974 | 0.33% | 0 | |
| No party | David Duke (write-in) | 90 | 0.01% | 0 | |
[edit] References
- ^ "Oregon Presidential Election Voting History". 270ToWin.com. Retrieved June 12, 2012.
- ^ "1988 Presidential General Election Results - Oregon". Dave Leip's U.S. Election Atlas. Retrieved June 12, 2012.
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