Beyonce',Brooke Valentine,Cassie,Lindsay Lohan,Nicole Richie,Paris Hilton,or Trina.
Just wanted to see what others thought.
2006-11-03 13:52:15 · 17 answers · asked by heyeverybody 1 in Entertainment & Music ➔ Polls & Surveys
someone who is neither anorexic or obese.
2006-11-03 13:55:25 · answer #1 · answered by owned by a siberian husky 4 · 1⤊ 0⤋
Beyonce
2006-11-03 14:24:40 · answer #2 · answered by Katherine 3 · 0⤊ 0⤋
Beyonce I love having T&A why give it up to be an ugly girl like Nicole, Paris, Lindsay or Trina??
2006-11-03 13:56:54 · answer #3 · answered by Anonymous · 1⤊ 0⤋
Jessica Alba
2006-11-03 13:54:35 · answer #4 · answered by Angel Girl 7 · 0⤊ 0⤋
Beyonce, but lets not forget how good Janet Jackson looks. Shes in shape and looks good for 40!
2006-11-03 14:07:13 · answer #5 · answered by MixedQtee 2 · 0⤊ 0⤋
Liv Tyler or Jessica Alba.
2006-11-03 13:58:09 · answer #6 · answered by shea_8705 5 · 0⤊ 0⤋
Lol well if your American of African descent everybody will choose to have a body like Beyonce and I want her body and all but I love my own body!
2006-11-03 13:56:18 · answer #7 · answered by Careatha 3 · 1⤊ 0⤋
Somehow....I don't think I'd look good with a pair of breasts. If anything, I look more like my daughter. And she is WAY better looking than any of the names that you picked.
But hey, that's just from a parent's point of view.
Even if your kid looks like Leatherface, they are still the most beautiful things in the world.
2006-11-03 14:00:19 · answer #8 · answered by Mr. Maul 4 · 1⤊ 0⤋
none of them because:
beyonce' is too curvy
brooke-?
cassie-thin
lindsay-she's ok, but she looked better when she wasn't thin
nicole-i'm not aneroxic. she's too damnmed thin
paris-i don't want a slutty body
trina-?
i prefer jessica alba's. not too thin or curvy. just perfectly toned.
2006-11-03 13:56:58 · answer #9 · answered by ALESANA 7 · 0⤊ 0⤋
I gotta go with Jessica Alba.
2006-11-03 13:56:04 · answer #10 · answered by ♫ Insane_Princess ♪ 5 · 1⤊ 0⤋
Give me Beyonce, she's hot.
2006-11-03 13:58:21 · answer #11 · answered by *COCO* 6 · 0⤊ 0⤋