Question 892267
If all grades count the same (for example, if the grades are for quizzes given equal value for the average), after a zero, {{{n}}} 75%(c) grades would bring the average percentage grade to
{{{(75*n+0)/(n+1)=75n/(n+1)}}}%
for a 70.00%, we need {{{75n/(n+1)=70}}} .
Solving that equation:
{{{75n/(n+1)=70}}} --> {{{75n=70(n+1)}}} --> {{{75n=70n+70}}} --> {{{75n-70n=70}}} --> {{{5n=70}}} --> {{{n=70/5}}} --> {{{highlight(n=14)}}} .
In practice, {{{highlight(13)}}} 75%(c) grades would do,
because a 69.50% would be rounded to 70%, and the equation would be
{{{75n/(n+1)=69.5}}} --> {{{75n=69.5(n+1)}}} --> {{{75n=69.5n+69.5}}} --> {{{75n-69.5n=69.5}}} --> {{{5.5n=69.5}}} --> {{{n=69.5/5.5}}} --> {{{n=about 12.64}}} .