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$$\frac{a+b+c+d+e+f}{6}=100\\ a=600-b-c-d-e-f$$ To make $a$ as small as possible, you should make $b,c,d,e$ and $f$ as large as possible, which would be $101$. This makes $a=95$.


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In your case, $| I |$ stands for the number of rows in the bicluster and similarly, $| J | = $ number of columns. More generally, the notation $|A|$ (usually) means the number of elements in a set $A$.


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It is a Weighted Mean. http://en.wikipedia.org/wiki/Weighted_arithmetic_mean Short, concise, but I must have this line. :)


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Set up your equation knowing $(price\space per\space pound)\times(pounds)=(total\space price)$ $$(1.56)\times(20+x)=(20\times 1.80)+(x\times 1.44)$$ \begin{array}{|c|c|c|} \hline \text{Price per pound}& \text{Pounds} & \text{Total Price}\\ \hline \text{1.80} & 20 & \text{36}\\ \hline \text{1.44} & \text{x}& 1.44\times x\\ \hline ...


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There is a way to express them in matrix form, but it makes less sense than intuition from statistics. Let $$ A_m := \begin{pmatrix}1/m \\ 1/m \\ \vdots \\ 1/m\end{pmatrix} \ \text{(an}\ 1×m\ \text{column vector)} $$ for any $m∈{\mathbb N}$. Then: $$\bar{X}_i = X\,A_n;\quad \text{Likewise, } \bar{X} = \bar{X}_i^{\mathsf T}\,A_r = A_r^{\mathsf T}\,X\,A_n\,. ...


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Here are some expressions, although not necessarily easier to memorize! Let $\mathbf{1}_{n}$ denote the column vector of all ones and length $n$, and $\mathbf{I}$ the $n\times n$ identity matrix. Then, $$ \hat{v}^{S} = \text{Tr}\left( \mathbf{X}^{T}\mathbf{X} \left(\mathbf{I}_{n}- \frac{1}{n}\mathbf{1}_{n}\mathbf{1}_{n}^{T}\right)\right), $$ where ...


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Mean of a continuous function with a discrete set of values is usually approximated with rectangle method, midpoint method or trapezoid method. The best way is usually trapezoid, which is calculated as : $$\frac{1}{x_n-x_0}\sum_{k=0}^{n-1}\frac{x_{k+1}-x_k}{2}(f(x_k)+f(x_{k+1}))$$


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Let $L_t$ be the number of targets unshot after $t$ turns. We have $L_0=10$ and $0 \le L_t \le L_{t-1}$. If $X_t$ is the number of targets of the $10$ original targets aimed for in turn $t$ then we also have $L_{t-1}-X_t \le L_t \le 10-X_t$. Then using the hypergeometric distribution $$\Pr(L_t=a|L_{t-1}=b, X_t=x) = \frac{{b \choose b-a}{10-b \choose ...



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