Question 151899
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1) Jay's test scores were 78,80,95, and 81. He took his last test for the quarter on Friday. What must he get to maintain a B average (between 80 and 89) 
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Since {{{B}}} average is bet. 80-89, we'll use the 80 average (minimum) for the eqn. Atleast if he gets higher than this is better right?
{{{Test=T[1]}}}, We'll do averaging,
{{{(T[1]+T[2]+T[3]+T[5])/5=80}}}
{{{(78+80+95+81+T[5])/5=80}}}, adding & cross multiply:
{{{334+T[5]=5*80}}}
{{{T[5]=400-334}}} ----> {{{T[5]=66}}} ---------> FINAL SCORE to get ave at least 80 and maintain {{{B}}} average.


2) Ernie's average after 3 tests is at least 72%. If two of his test scores were 55 & 85, what could he have scored on his 3rd test?


Same thing, {{{Averaging}}}
{{{(T[1]+T[2]+T[3])/3=72}}}
{{{(55+85+T[3])/3=72}}}, adding & cross multiply,
{{{140+T[3]=3*72}}} 
{{{T[3]=216-140}}} ----> {{{T[3]=76}}} ------> FINAL SCORE to get 72 AVERAGE.


3) If the square patio has {{{X}}} dimensions, the new dimensions now are:
{{{Length=x+6}}} ----------> increased by 6'
{{{Width=2x}}} ------------> doubled
This becomes a rectangle now since it has 2 diff dimensions.
The {{{loophole}}} with this you can get different orig. dimensions because the New Perimeter ranges from 30'-45'. Why? Simply we 'll base our computation in solving for {{{X}}} for the New Perimeter.
Okay, we'll pick the New Perimeter as {{{42ft}}}, then
{{{P=2(L+W)}}}
{{{42=2(x+6+2x)}}}
{{{42=2(3x+6)}}} -------------> {{{42=6x+12}}} ---------> {{{42-12=6x}}}
{{{cross(30)5/cross(6)=cross(6)x/cross(6)}}}
{{{x=5ft}}}
SIZE of the orig square patio is {{{5ftx5ft}}} based on the NEW PERIMETER of {{{42ft}}}


Thank you,
Jojo</font>