Q: What is the sum of squared deviations for 2 2 0 5 1 4 1 3 0 0 1 4 4 0 1 4 3 4 2 1 0?

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Given a set of n scores, the variance is sum of the squared deviation divided by n or n-1. We divide by n for the population and n-1 for the sample.

The sum of 0 and 1 is 1. 0+1=1 If you have 1, and you add 0(which is nothing), then you will still have 1.

Because the letters can represent any sum not necessarily its only on a triangle where that happens the hypotenuse squared equals the sum of the other two sides squared added together like if a was 1 and b was two c could be three so 1+4 doesn't equal 9

No, unless "a" happens to be equal to 0, or to 1.

It is impossible to answer the question. In the specified isosceles triangle, x can have any value between 0 and 8. Therefore 0 < x2 < 64. There are infinitely many numbers between 0 and 1 (leave alone 0 and 64) and their sum is infinite.

Related questions

Given a set of n scores, the variance is sum of the squared deviation divided by n or n-1. We divide by n for the population and n-1 for the sample.

Usually the sum of squared deviations from the mean is divided by n-1, where n is the number of observations in the sample.

0

The sum of 0 and 1 is 1. 0+1=1 If you have 1, and you add 0(which is nothing), then you will still have 1.

x^2 + x -x^2 +1 =x+1

No. Standard deviation is the square root of the mean of the squared deviations from the mean. Also, if the mean of the data is determined by the same process as the deviation from the mean, then you loose one degree of freedom, and the divisor in the calculation should be N-1, instead of just N.

Because the letters can represent any sum not necessarily its only on a triangle where that happens the hypotenuse squared equals the sum of the other two sides squared added together like if a was 1 and b was two c could be three so 1+4 doesn't equal 9

The following will sum all integers from 1-100. Adjust according to your needs.int sum = 0;for (int i = 1; i

No, unless "a" happens to be equal to 0, or to 1.

int i, sum; /* example using while */ sum = 0; i = 1; while (i <= 100) { sum += i; ++i; } /* example using for */ sum = 0; for (i = 1; i <= 100; ++i) sum += i;

There are no two whole numbers which when squared sum to 62. There are infinitely may pairs of irrational numbers that when squared sum to 62, eg 1 and √61, 2 and √58

-1, 0, 1