document.write( "Question 902491: Jim s second test score was 8 points higher than hisfirst score. His third score was 88. He had a B average (between 80 and 89, inconclusive) for the three tests. What can you conclude about his first test score?\r
\n" ); document.write( "\n" ); document.write( "I am honestly unsure of where to start on this one- it is so different from the other ones I did.. I think that maybe it will be that \r
\n" ); document.write( "\n" ); document.write( "Score 1= x
\n" ); document.write( "Score 2= x+8
\n" ); document.write( "Score 3=88\r
\n" ); document.write( "\n" ); document.write( "after that I don't know what to do because of the average aspect involved- it is confusing me even though I am maybe making it more difficult than it is. \n" ); document.write( "

Algebra.Com's Answer #547291 by richwmiller(17219) $\"\"$ $\"About$

You can put this solution on YOUR website!
add them up and divide by 3
\n" ); document.write( "(x+x+8+88)=2x+96
\n" ); document.write( "(2x+96)/3=89
\n" ); document.write( "2x+96=267
\n" ); document.write( "2x=171
\n" ); document.write( "x=85.5
\n" ); document.write( "(2x+96)/3=80
\n" ); document.write( "2x+96=240
\n" ); document.write( "2x=144
\n" ); document.write( "x=72
\n" ); document.write( "x is between 72 and 85.5
\n" ); document.write( "x+8 is between 80 and 93.5
\n" ); document.write( "(72+80+88)/3=240/3=80
\n" ); document.write( "(85.5+93.5+88)/3=267/3=89\r
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