Someone please help me will give BRAILIEST!!!!

Mathematics

2 answers:

Naily [24]3 years ago

6 0

This is the number of combinations of 2 from 6 =
6!
-------
2! 4!

an easy way to work this out is
6*5
----- = 30 /2 = 15
2 * 1

Answer is 15

romanna [79]3 years ago

6 0

Answer:

A 2 person team can be chosen in one of fifteen ways.

Explanation:

The question is not precise because if you treated it literally the answer would be 3. First you choose one team, 4 people are left in the group, the second team takes another 2 people and the remaining create the third team.

But I assume that the question is like "In how many ways a 2 people team can be chosen from 6 people?"

Such question has an answer 15 because first member is chosen from 6 people (so there are 6 possibilities), the second person is chosen from remaining five people so the number is <span>6⋅5=30</span>, but you have to divide the result by 2 because 2 people can be chosen in 2 ways but they still form the same team. It does not matter if you choose Ann first then John or the other way John first then Ann they form the same team.

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Suppose that you need to create a list of n values that have a specific known mean. Some of the n values can be freely selected.

Leno4ka [110]

As the problem indicates, degrees of freedom are the number of values that can be independently selected before it is necessary to choose specific values to arrive at the desired result.

The average, on the other hand, results from the sum of a list of values divided by the amount of values in the summed list.

Assume that the mean sought is $x$ and consider that the list is composed of a single element $a$ , in that case no random number can be selected, since the mean $x$ must correspond to that number.

If the list were composed of two elements $a$ and $b$ , one of the two values could be chosen randomly, and according to the chosen value the second should be the one whose sum with the previous one results in $2x$ , this given that the formula of the average $\sum\limits_{i=1}^n \frac{ x_{i}}{n}$ .

With three values $a$ , $b$ and $c$ , it is possible to select two freely, since the thirteen must be the one that balances the sum of $a+b$ , that is $(a + b) + c = 3x$ .

Thus, in general, with n values, it is possible to select $n-1$ values freely whose sum must be balanced by the last value so that the whole sum is $nx$ .