document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=75801>Question 75801</A>: <I><SPAN STYLE=\"word-break: break-all;\"> each stroke of a vacuum pump removes one third of the air ramianing in a container. wut percent of the original quanity of the air ramains in the container after 10 strokes, to the nearest percentage?\r<BR>\n" );
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document.write( "we r workin with geometric sequence </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #54451 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley75><B>checkley75(3666)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley75><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=54451\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> SINCE EACH STROKE REMOVES 1/3 OF THE AIR THEN THE REMAINING AIR IS 2/3 WHICH IS AGAIN REDUCED BY 1/3 WITH EACH SUCCESSIVE STROBE LEAVING 2/3 OF THE REMAING AIR, ETC. SO AFTER 10 STROKES WE HAVE LEFT:<BR>\n" );
document.write( "2/3*2/3*2/3*2/3*2/3*2/3*2/3*2/3*2/3*2/3=1,024/59,049=.0173415=1.7% OF THE AIR IS LEFT.<BR>\n" );
document.write( "OR<BR>\n" );
document.write( "(2/3)^10=.0173415=1.7%  </SPAN>\n" );
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