document.write( "Question 1073211: A large company rates its employees every six months as either satisfactory or unsatisfactory. Employees who had
\n" ); document.write( "any kind of previous work experience are more likely to be rated as satisfactory than employees without previous
\n" ); document.write( "work experience. The percents of experienced and inexperienced workers and their breakdown into satisfactory
\n" ); document.write( "and unsatisfactory ratings are shown in the diagram below.\r
\n" ); document.write( "\n" ); document.write( "-Employees with work experience:85%
\n" ); document.write( "*90% satisfactory
\n" ); document.write( "*10% unsatisfactory \r
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\n" ); document.write( "\n" ); document.write( "-Employees with no previous experience:15%
\n" ); document.write( "*70% satisfactory
\n" ); document.write( "*30% unsatisfactory\r
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\n" ); document.write( "\n" ); document.write( "What is the probability that a randomly selected employee will be one who had previous work experience and is
\n" ); document.write( "rated unsatisfactory? Round your answer to the nearest whole percent.
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Algebra.Com's Answer #688050 by jorel1380(3719)\"\" \"About 
You can put this solution on YOUR website!
The randomly selected employee had previous work experience (85%), and is rated unsatisfactory (10%), so:
\n" ); document.write( ".85 x .1=.085, or 8.5% chance of being randomly selected. ☺☺☺☺
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