3x - 2x + 5 - 6 = 9 + t + x
x - 1 = 9 + t + x
-1 = 9 + t
-10 = t
Side of square = 48/4 = 12 in
Side of equilateral triangle = 48/3 = 16 in
So
side of triangle / side of square = 16 /12 = 4 / 3
Let the number of original yellow marbles = Y
Let the number of original blue marbles be = B
And we know that B/Y = 4/3 ⇒ B = (4/3)Y ⇒ Y = (3/4)B
And we have that
B + 5 7
_____ = ___
Y - 3 3
Subbing for Y we have that
_________ = ___ cross-multiply
(3/4)B - 3 3
3(B + 5) = 7 [ (3/4)B - 3 ] simplify
3B + 15 = (21/4)B - 21 add 21 to both sidws, subtract 3B from both sides
36 = (21/4 - 3)B
36 = ( 21/4 - 12/4) B
36 = (9/4)B multiply both sides by 4/9
16 = B = original number of blue marbles
Excellent, Cal !!!!!
We have this data set
0,0,0,0,0, 1,1,1,1,1,1,1,1,1,1,1,1,1,1, 2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,3,4,4, 5,5,5,5, 6,6, 7
There are 57 data values....the median occurs at the 58/2 = 29th data value = 2
If the mean of the six grades is 87, there were 6 * 87 = 522 total points
If the mean of 6 grades is 89, there were a total of 455 points
Then the dropped grade must have been 522 - 455 = 77
100 ( .50) (.70) = $35
This question is missing some info.....
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THX, HELPMEEEEEEEEEEEEE !!!!!
