document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=831003>Question 831003</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 60,73,90,75,82,80 a.if the mode of all Eight scores is 73 find Peters average .<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 #501126 by </B><A HREF=/tutors/aboutme.mpl?userid=JulietG><B>JulietG(1812)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=JulietG><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=JulietG><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=501126\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The problem is that you've only provided 6 scores.  That the mode (most often recurring number) is 73 tells us that at least one of the two missing scores is a 73.  We'll have to assume the other missing score is 73, but that *would* change the average if it's not a 73 (for instance, it could be 100 and wouldn't change the mode, but the mean would go up).<BR>\n" );
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document.write( "Assuming that the two missing scores are 73, we then have the following scores:<BR>\n" );
document.write( "60, 73, 73, 73, 90, 85, 82, 80<BR>\n" );
document.write( "Add them up and then divide by 8<BR>\n" );
document.write( "616 / 8 = 77<BR>\n" );
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document.write( "Mode (most often recurring) = 73<BR>\n" );
document.write( "Mean = 77 <<-- </SPAN>\n" );
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