document.write( "Question 377863: In 1920 a record for a race was 45.3 seconds. In 1940 it was 44.7 seconds.\r
\n" );
document.write( "\n" );
document.write( "R(t)= the Records of the race and let t = the number of years since 1920.\r
\n" );
document.write( "\n" );
document.write( "1.Find a linear function that fits the Data
\n" );
document.write( "2. Use the function to predict the race 2003, and 2006
\n" );
document.write( "3. Find the year when the record will be 42.5 seconds.\r
\n" );
document.write( "\n" );
document.write( " \n" );
document.write( "
Algebra.Com's Answer #268486 by mananth(16946)![]() ![]() You can put this solution on YOUR website! In 1920 a record for a race was 45.3 seconds. \n" ); document.write( "In 1940 it was 44.7 seconds. \n" ); document.write( "decrease per year = (45.3-44.7) /20 \n" ); document.write( "=0.6/20 \n" ); document.write( "=0.03 per year. \n" ); document.write( "R(t) = 45.3-0.03t \n" ); document.write( "...... \n" ); document.write( "In 2003 \n" ); document.write( "------------- \n" ); document.write( "R(83)= 45.5-0.03*83 \n" ); document.write( "R(83)=45.5 -2.49 \n" ); document.write( "R(83)=42.81 \n" ); document.write( "In 2003 the record will be 42.81 seconds. \n" ); document.write( "plug t=86 for 2006 \n" ); document.write( ".. \n" ); document.write( "3.) 42.5 = 45.5 -0.03t \n" ); document.write( "0.03t=45.5-42.5 \n" ); document.write( "0.03 t = 3 \n" ); document.write( "t = 3/0.03 \n" ); document.write( "t=100 \n" ); document.write( "Add 100 years to 1920= 2020 \n" ); document.write( "In 2020 the recoed will be 42.5 seconds \n" ); document.write( "m.ananth@hotmail.ca\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |