document.write( "Question 982906: a bricklayer measured the length of his wall to be 4.10m, if the actual length is 4.25m, what is his percentage error \n" ); document.write( "
Algebra.Com's Answer #603733 by Fombitz(32388) You can put this solution on YOUR website! Error=((Actual-Measured)/Actual)*100\r \n" ); document.write( "\n" ); document.write( ". \n" ); document.write( ". \n" ); document.write( ". \n" ); document.write( " \n" ); document.write( "3.5% \n" ); document.write( " \n" ); document.write( " |