2006-08-14 16:20:06 · 12 answers · asked by Nuts Are Us 3 in Entertainment & Music ➔ Celebrities
Howard Stern is offensive to a portion of the population, but Bill O'Really is offensive to EVERYONE.
Besides, if Howard Stern made a factual statement on the air and later found out that what he said was wrong, I'm sure he'd correct himself on air. Bill O'Really made the same erroneous factual statement twice and has never apologized for the mistake. In fact, he tried to erase it from his transcript.
2006-08-14 16:31:02 · answer #1 · answered by krissydahs93 4 · 0⤊ 0⤋
Howard Stern. Does he knows how to bathe?
Stern is a total wimp & a loser.
Bill O'Reilly is a pompous so & so, but at least he talks in
plain English, Stern talks like he has a mouthful of marbles.
It's the best of two poor choices.
2006-08-14 16:34:24 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Bill O'Reilly. Howard Stern is a performer who expresses outrageous beliefs as part of an act. Bill O'Reilly expresses outrageous beliefs because he thinks they are true.
2006-08-14 16:27:51 · answer #3 · answered by charmingchatty 4 · 0⤊ 0⤋
DEFINETLY Howard Stern. Bill O' Reilly may be an idiot in some people's eyes, but Howard Stern is just gross.
2006-08-14 16:24:16 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Howard Stern. At least Bill keeps it "clean" even if everyone doesn't agree with him.
2006-08-14 16:25:18 · answer #5 · answered by Gwen 5 · 0⤊ 0⤋
I would say Howard Stern cause the guy is totally gross and vulgar.
2006-08-14 17:07:36 · answer #6 · answered by mystic rain 7 · 0⤊ 0⤋
definitely bill, Howard is an entertainer trying to have fun and make a little dough along the way
2006-08-14 16:27:31 · answer #7 · answered by Anonymous · 0⤊ 0⤋
Yes
lol
I cant stand either but imo Bill is far worse
2006-08-14 16:23:34 · answer #8 · answered by starikotasukinomiko 6 · 0⤊ 0⤋
Bill O'Reilly. He is just so arrogant.
2006-08-14 16:38:55 · answer #9 · answered by Anonymous · 0⤊ 0⤋
Howard is great, Bill is an idiot.
2006-08-14 16:27:54 · answer #10 · answered by AshleyC 2 · 0⤊ 0⤋