document.write( "Question 1030969: the age of a machine X in years is, is related to the probability of breakdown, y by the formula x=3+Ln(y/1-y)
\n" ); document.write( "Determine the probability of a breakdown for 1 3 and 10 years.\r
\n" ); document.write( "\n" ); document.write( "x = 3 + l n ( y 1 − y )
\n" ); document.write( "x=3+ln(y1−y)
\n" ); document.write( "3 + l n ( y 1 − y )
\n" ); document.write( "= 1 3+ln(y1−y)
\n" ); document.write( "=1 3 + l n ( y 1 − y )
\n" ); document.write( "= 3 3+ln(y1−y)=3
\n" ); document.write( "3 + l n ( y 1 − y ) = 10
\n" ); document.write( "I'm lost, lost lost!\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
Algebra.Com's Answer #645709 by Alan3354(69443)\"\" \"About 
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the age of a machine X in years is, is related to the probability of breakdown, y by the formula x=3+Ln(y/1-y)
\n" ); document.write( "Determine the probability of a breakdown for 1 3 and 10 years.
\n" ); document.write( "--------------
\n" ); document.write( "Solve for y:
\n" ); document.write( "x = 3 + ln(y/1-y)
\n" ); document.write( "x - 3 = ln(y/1-y)
\n" ); document.write( "y/(1-y) = e^(x-3)
\n" ); document.write( "y/(y-1) = -e^(x-3)
\n" ); document.write( "y = -e^(x-3)(y-1) = -e^(x-3)*y + e^(x-3)
\n" ); document.write( "y + e^(x-3)*y = e^(x-3)
\n" ); document.write( "y*(e^(x-3) + 1) = e^(x-3)
\n" ); document.write( "\"y+=+e%5E%28x-3%29%2F%28e%5E%28x-3%29+%2B+1%29\"\r
\n" ); document.write( "\n" ); document.write( "-----
\n" ); document.write( "x = 1 year
\n" ); document.write( "y = e^(-2)/(e^(-2)+1)
\n" ); document.write( "y =~ 0.1192
\n" ); document.write( "--------------
\n" ); document.write( "x = 3 years
\n" ); document.write( "y = e^0/(e^0 + 1)
\n" ); document.write( "y = 0.5
\n" ); document.write( "----------------
\n" ); document.write( "x = 10 years
\n" ); document.write( "y = e^(7)/(e^(7)+1)
\n" ); document.write( "y =~ 0.9991\r
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