Cody

Christian

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Christian submitted a Comment to Solution 591260

Easy, but not the weighted average. Even assuming that the first vector contains the weighting factors (for example the sizes of different sample groups) and the second vector contains the means of the sample groups, than the divisor has to be the sum of the weighting factors instead of the number.

on 4 Mar 2015

Christian received Commenter badge for Solution 591231

on 4 Mar 2015

Christian submitted a Comment to Solution 591231

Nice question. Good reminder how vector multiplication works.

on 4 Mar 2015

Christian submitted Solution 590601 to Problem 167. Pizza!

on 3 Mar 2015

Christian received Solver badge for Solution 545101

on 12 Dec 2014

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