document.write( "Question 364089: The following are based on a survey taken by a consumer research firm.x= the number of televisions in household and % percentages U.S.households.
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\n" ); document.write( "%=3% 11% 28% 39% 12% and 7%
\n" ); document.write( "what is the probability that a household selected at random has less than three televisions? \n" ); document.write( "

Algebra.Com's Answer #259706 by ewatrrr(24785) $\"\"$ $\"About$

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document.write( "Hi, 
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document.write( "(1)3%, (2)11%, (3) 28%, (4)39% , (5)12% and (>5)7% 
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document.write( "P( < 3 televisions) = .03 + .11 = .14  Or  14 % \n" );
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