document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1082951>Question 1082951</A>: <I><SPAN STYLE=\"word-break: break-all;\"> What is the percentatge error made in the computed surface are of a sphere if the error made in measuring the radius is 3%.<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #696973 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=696973\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> It would be (1.03)^3=1.093=9.3%<BR>\n" );
document.write( "check with a sphere of radius 2 and 2.06<BR>\n" );
document.write( "Volume is pi*2.06^3*(4/3) divided by pi*2^3*(4/3)=8.74/8=1.093 or 9.3%<BR>\n" );
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