Lets make a list assuming they have 2 cubes with 1-6 on them combos are 1,1 1,2 1,3 1,4 1,5 1,6 2,1 2,2 2,3 2,4 2,5 2,6 3,1 3,2 3,3 3,4 3,5 3,6 4,1 4,2 4,3 4,4 4,5 4,6 5,1 5,2 5,3 5,4 5,5 5,6 6,1 6,2 6,3 6,4 6,5 6,6
and out of all of them (when you add them), only 5 possible roles add up to 8 only 2 possible roles add up to 11 that means that if you rolled the max number of times, then 8 would come up 5 times, but 11 woule come up 2 times
To find the average of numbers, you add all of the numbers together and divide the total by the number of numbers. In this case, your numbers would be 95 twice, 87 five times, 73 three times, 63 four times, and 40 once. The total number of numbers is 15, and the total of all of the exam scores if you add them up is 1136. Finally, you would divide 1136 by 15, which would be 75.7333333, which would be an average of about 76%.
Point D is the intersection of three angle bisector. BE is the angle bisector of ∠B CF is the angle bisector of ∠C AG is the angle bisector of ∠A
Point D is also the intersection between three perpendicular bisector BE is the perpendicular bisector of AC AG is the perpendicular bisector of BC CF is the perpendicular bisector of AB
Hence the correct statements is statement 1 and statement 4