| 1 | A | B | C | D | E |
| --- | --- | --- | --- | --- | --- |
| 2 | 1894 United States House of Representatives election in Wyoming - Results |  |  |  |  |
| 3 | Party | Party | Candidate | Votes | % |
| 4 | - | Republican | Frank Wheeler Mondell | 10,068 | 52.64% |
| 5 | - | Democratic | Henry A. Coffeen | 6,152 | 32.17% |
| 6 | - | People's | Shakespeare E. Sealey | 2,906 | 15.19% |
| 7 | Total votes | Total votes | Total votes | 19,126 | 100% |

Question: What was the outcome of the 1894 United States House of Representatives election in Wyoming?
Answer: Republican Frank Wheeler Mondell defeated Democrat Henry A. Coffeen and Populist Shakespeare E. Sealey, receiving 52.64% of the vote, making Coffeen the second Representative from Wyoming to lose the election.