Question 1086775
<font color="black" face="times" size="3">Exam Scores = {67,84,7,x} where x is a real number and {{{0 <= x <= 100}}}
Sum of the scores = S = 67+84+78+x = 229+x
sample size = n = 4
Average = A = (sum of scores)/(sample size) = S/n = (229+x)/4


We want the average to be a B or higher. I'm assuming the grade "B" corresponds to a score of 80, which you could argue is a B- but I'm going to ignore the plus/minus for the letter grades. 


So,
A > 80
(229+x)/4 > 80
4*(229+x)/4 > 4*80
229+x > 320
229+x-229 > 320-229
x > 91
So you need to get a score of 91 to get an overall average of 80
If you score higher than 91, then the average will be larger than 80


If you place a B grade at something like 85, then 
A > 80
(229+x)/4 > 85
4*(229+x)/4 > 4*85
229+x > 340
229+x-229 > 340-229
x > 111
which is impossible to accomplish (x maxes out at 100)


You're probably wondering: "What is the highest average possible?". That happens when x = 100. Let's plug that into the equation below to find A
A = (229+x)/4
A = (229+100)/4
A = 82.25


So the best average possible is 82.25% or 82% if you round to the nearest whole percent. </font>