No 1+2+3+4+5+6+7+8+9+10=54.
So it can't and there will be one ball left over.
If there is a group of 3 coloured balls, then any groups of 2 balls selected from it will be considered as a combination, whereas the different arrangements of every combination will be considered as a permutation
You need to add them all together in each group then divide by the amour of numbers. That will give you the average.
2 groups
if you take a vector (= group of numbers) and you divide it by a scalar (=one number) then you get a vector (=group of numbers)
The different group names for the keyword "5" include integers, whole numbers, natural numbers, and positive numbers.
The way to pinpoint the heaviest ball is as follows:1 - Split the balls into three groups of three balls each.2 - First use of the scale: Weigh any two groups against each other.3 - If the groups weigh the same then the heavier ball is in the third group, otherwise it is in the group that weighs more.4 - We now know that the heavy ball is one of three.5 - Second use of the scale: Using the group of balls that we know contains the heavy ball, weigh any two balls against each other.6 - If one of the balls is heavier then we have our answer. If the balls weigh the same then the third ball is the heavy one and we again have our answer.
There are three different groups modes - No groups, Separate Groups, Visible Groups. In 'Separate groups' mode, each group can only see their own group - other groups are invisible. In 'Visible groups' mode, each group works in their own group, but can also see other groups.
A four-group series of numbers separated by periods typically represents an IP address. Each group of numbers in an IP address can range from 0 to 255, and the groups represent different parts of the address that help identify devices on a network.
1- Take ou one ball and slit the other six into two groups of three balls 2- put the two groups in the balance ( first use ) and if they are the same weight then the ball you took out in the first place is the heavier if not and one group is heavier than the other then the heavier ball is in that group so you can exclude the other group and the ball you took out at first. 3- If one group is heavier than the other then you take one ball from that group ( it has 3 balss remember? ) and you weight the other two remaining balls ( second and last use of the balance ) if the two balls have the same weight then the ball you took out is the heavier one if not look to the balance to see which of the balls is lower and there is your answer. Please, tell me if you understood : )
There are two different types of groups in the experiment, a control group and a experimental group
There are three possibilities:-- 7 equal groups, with 3 in each group-- 3 equal groups, with 7 in each group-- 21 equal groups, with 1 in each group
Separate the balls in 3 separate groups of 3 balls each. Get two groups of 3 and place it on either side of the balance. If the balance is the same then the probability is that the heavy ball is in the other group off the balance. Or if the ball was on the group in the balance it would be noticed right away also. After recognizing the heavier group, you take 2 balls from the remaining group and you weight them, if it's the same, then the remaining ball must be the heavier one, if not then you would automatically see it in the scale.