USA and WWE(R) Unveil 'The Sexiest Woman on Television'
2006 Diva Search Winner to Be Chosen During Live TV Special
NEW YORK, Aug. 14 -- USA Network and WWE will unveil "The
Sexiest Woman on Television" when the $250,000 WWE Diva Search winner is
announced from the Hard Rock Cafe during their live 1-hour television
event, "The Sexiest Woman on Television: The WWE Diva Search Finals," this
Wednesday, August 16, 2006 (10-11p.m. ET/ 9-10p.m. CT).
Thousands of hopefuls from across North America answered an open
casting call and WWE fans have been voting via online and SMS for their
favorite Diva on Monday Night RAW(R) (USA 9 p.m. ET/PT) and on "Friday
Night SmackDown(R)" (UPN 8 p.m. ET / 7 p.m. CT). After five weeks of
elimination rounds, USA will televise the event as the winner will be
crowned before a live audience at the Hard Rock Cafe, becoming the newest
WWE Diva and earning $250,000 for being named "The Sexiest Woman on
Fans who watched the WWE's broadcasts were asked to consider each
finalists performance each week to determine whether each contestant
possessed the following qualities of a WWE Diva (1) Glamour; (2) Beauty;
(3) Physical Fitness; (4) Poise; (5) Presence; (6) Charisma; (7)
Personality; and (8) Talent. The finalist with the fewest fan votes has
been eliminated each week. As of today, the final four contestants
competing for the title of 2006 Diva Search Winner are Jen England, Layla
El, Milena Roucka and J.T. Tinney.
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copyrights are the property of their respective owners.
SOURCE USA Network