SOLUTION: the average (arithmetic mean) of 10 exam grades is 87.if the highest and lowest grades are removed from the set, the average of the remaining 8 exam grades is 90. what was the aver

Question 867768: the average (arithmetic mean) of 10 exam grades is 87.if the highest and lowest grades are removed from the set, the average of the remaining 8 exam grades is 90. what was the average of the 2 exam grades that were removed?

circle A has a circumference of 4pi, and circle b has a circumference of 8pi. if the circles intersect at two points , what is a possible distance from the center of circle A to the center of circle B?
Found 2 solutions by ankor@dixie-net.com, solver91311:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
the average (arithmetic mean) of 10 exam grades is 87.if the highest and lowest grades are removed from the set, the average of the remaining 8 exam grades is 90. what was the average of the 2 exam grades that were removed?
Let h = the high grade
let L = the lowest grade
the other 8 average 90
the equation
= 87
multiply both sides by 10
L + 720 + h = 870
L + h = 870 - 720
L + h = 150
therefore: = 75 is the average of the 2 removed grades
:
circle A has a circumference of 4pi, and circle b has a circumference of 8pi. if the circles intersect at two points , what is a possible distance from the center of circle A to the center of circle B?
Find the radius of Circle A
=
2r = r
r = 2
Find the radius of Circle b
=
r = 4
:
Place A's center on the circumference of b, the centers would be 4 units apart
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

If the average of 10 scores is 87, then the sum of all 10 scores must be 10 times 87. Likewise, if the average of 8 of those scores is 90, the sum of those 8 scores must be 8 times 90. The sum of the two scores that were removed must be the sum of the 10 scores minus the sum of the 8 scores.

John

My calculator said it, I believe it, that settles it

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