document.write( "Question 392523: In 1992, the life expetancy of males in a certain country was 70.8 years. In 1998, it was 74.5 years. Let E represent the life expentancy in year t and let t represent the number of years since 1992.\r
\n" ); document.write( "\n" ); document.write( "The linear function E(t) that fits the data is:\r
\n" ); document.write( "\n" ); document.write( "Et=☐t + ☐
\n" ); document.write( "(round to the nearest tenth)\r
\n" ); document.write( "\n" ); document.write( "Use the function to predict the life expectancy of males in 2009.\r
\n" ); document.write( "\n" ); document.write( "E(17)=☐
\n" ); document.write( "(round to the nearest tenth)
\n" ); document.write( "

Algebra.Com's Answer #278605 by mananth(16946)\"\" \"About 
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In 1992, the life expetancy of males in a certain country was 70.8 years. In 1998, it was 74.5 years. Let E represent the life expentancy in year t and let t represent the number of years since 1992.\r
\n" ); document.write( "\n" ); document.write( "The linear function E(t) that fits the data is:\r
\n" ); document.write( "\n" ); document.write( "1992 to 1998 = 6 years difference
\n" ); document.write( "increase in life expectancy = 74.5-70.8=3.7
\n" ); document.write( "3.7 in 6 years\r
\n" ); document.write( "\n" ); document.write( "so per year increas = 3.7/6 = 0.62
\n" ); document.write( "E(t) = 0.6(t)+ 70.8\r
\n" ); document.write( "\n" ); document.write( "Use the function to predict the life expectancy of males in 2009.\r
\n" ); document.write( "\n" ); document.write( "E(17)=0.6*17+70.8
\n" ); document.write( "E(17)= 81
\n" ); document.write( "...
\n" ); document.write( "m.ananth@hotmail.ca
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