The United States House of Representatives has 435 seats. A political party has majority control of the House of Representatives if 218 seats belong to the party. In the second session of the 109th Congress, the Republican Party controlled 229 seats and the Democratic Party controlled 201 seats. One seat was held by an independent representative and 4 seats were vacant. Assuming the Democratic Party kept control of the 201 seats they had, what was the minimum whole percent of the remaining seats, the Democratic Party needed to win in the November 7, 2006 election to have at least 218 seats in the House of Representatives?
2006-11-15 08:44:27 · 5 answers · asked by blondefloridagrl35 1 in Science & Mathematics ➔ Mathematics
wow, you thought of all these on your own.
2006-11-15 08:46:47 · answer #1 · answered by Anonymous · 0⤊ 1⤋
Total Seats in the House = 435
Repl Seats = 229
Dem Seats = 201
Indep Seats= 001
Vacnt Seats= 004
A Majority = 218 Seats
218 Seats - 201 Seats = 17 Seats Required to gain Majority
Percentage of gain needed is 17 / 435 = 0.0390804 decimal
Multiplied by 100 for percentage = 3.908 %
Made into a whole percent = 4 %
Answer = 4%
Regards,
Zah
2006-11-15 16:57:14 · answer #2 · answered by zahbudar 6 · 0⤊ 0⤋
In an election, all seats in the House are up fro grabs. Given the assumption of losing no seats, the Democrats needed
(218 - 201)/(435 - 201) =
17/234 =
7.26% (round up for control to 8%)
2006-11-15 17:15:11 · answer #3 · answered by Helmut 7 · 0⤊ 0⤋
need 218
have 201
gain 17
17/201=9%
2006-11-15 16:48:33 · answer #4 · answered by yupchagee 7 · 0⤊ 0⤋
i will go with c, true
2006-11-15 16:51:54 · answer #5 · answered by theinfamousjae 2 · 0⤊ 1⤋