document.write( "Question 209683: Story problems always confuse me. I have tried and tried to get this one, and I can't do it. Can some one please help me?\r
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document.write( "The 1990 life expectancy of males in a certain country was 71.3 years. In 1995, it was 74.5 years. Let E represent the life expectancy in year t and let t represent the number of years since 1990.\r
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document.write( "The linear function E(t) that fits the data is\r
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document.write( "E(t) = ___t + ___ (round to the nearest tenth)\r
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document.write( "Use the function to predict the life expectancy in males in 2006\r
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document.write( "E(16) = ____ (round to the nearest tenth) \n" );
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Algebra.Com's Answer #158520 by Alan3354(69443)![]() ![]() You can put this solution on YOUR website! The 1990 life expectancy of males in a certain country was 71.3 years. In 1995, it was 74.5 years. Let E represent the life expectancy in year t and let t represent the number of years since 1990. \n" ); document.write( "The linear function E(t) that fits the data is \n" ); document.write( "E(t) = ___t + ___ (round to the nearest tenth) \n" ); document.write( "Use the function to predict the life expectancy in males in 2006 \n" ); document.write( "E(16) = ____ (round to the nearest tenth) \n" ); document.write( "--------------- \n" ); document.write( "1990 is the starting point, t = 0 years \n" ); document.write( "In 1990, t=0 and E = 71.3 \n" ); document.write( "In 1995, t=5 and E = 74.5 \n" ); document.write( "The change in E is 74.5-71.3 = 3.2 years \n" ); document.write( "The change per year is 3.2/5 = 0.64 increase per year. \n" ); document.write( "E(t) = 71.3 + 0.64t \n" ); document.write( "----------- \n" ); document.write( "For 2006: t is 2006 - 1990 = 16 \n" ); document.write( "E(16) = 71.3 + 0.64*16 = 71.3 + 10.24 = 81.54 \n" ); document.write( "--> 81.5 years in 2006 \n" ); document.write( " \n" ); document.write( " |