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16

We know that $(x-y)^2+(x+y)^2=2(x^2+y^2).$ So, if $a=x+y$ and $b=x-y$, then $c^2=x^2+y^2.$ Now see Formulas for generating Pythagorean triples.


3

In 1976 Gallagher proved, under the assumption of a uniform version of the Hardy-Littlewood $k$-tuples conjecture, that for any fixed $\lambda>0$ and integer $k$ $$\#\{\text{ integers } x\leq X\ :\ \pi(x+\lambda \log x)-\pi(x)=k\}\sim e^{-\lambda}\frac{\lambda^k}{k!}X,$$ that is it follows a Poisson distribution. Since the waiting times for a Poisson ...


2

I thought Hardy-Littlewood might come into it. Here is some numerical data following Erics great answer: x-axis: N y-axis: Geometric mean of the first 10000 prime gaps following $10^N$ divided by $\ln 10^N$. $e^{-\gamma} \approx 0.56146$.


2

You need to endow your infinite set with a measure such that the whole space has measure $1$ and then integrate (and hope that your function is measurable to begin with). For finite sets, the obvious choice of measure is the counting measure divided by the number of elements; this gives the usual average. Even weighted averages can be achieved by playing ...


2

Total men age is $32\cdot48.5+23=1575$ Total women age is $59\cdot39.2-47=2265.8$ Average age is therefore $\frac{1575+2265.8}{32+1+59-1}\approx42.2$


2

First of all, an important thing is that it is impossible to know the answer unless you know the age of the woman that stopped working. The total age is not $48.5 + 39.2$. That is just the sum of two ages. In fact, $48.5$ is the mean age of all male employees, i.e. before the replacement, there are $32$ men with an average age of $48.5$, so if $s$ is the ...


2

We just have to compute: $$ I=\int_{[0,1]^4}\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\,d\mu. \tag{1}$$ Assuming that $X_1$ and $X_2$ are two independent random variables, uniformly distributed over $[0,1]$, the pdf of their difference $\Delta X=X_1-X_2$ is given by: $$ f_{\Delta X}(x) = \left(1-|x|\right)\cdot\mathbb{1}_{[-1,1]}(x)\tag{2}$$ hence: $$\begin{eqnarray*} I ...


2

First of all, we will order the set $A$, i.e. call $b_1$ the smallest element of $A$, $b_2$ the smallest element of $A\backslash\{b_1\}$ etcetera, up to $b_m$ to be the largest element of $A$. Now consider the number $b_1+b_m$. If $b_1+b_m\leq n$, then $b_1+b_m\in A$. But $b_m$ was the largest element of $A$, which is a contradiction since $b_1+b_m>b_m$ ...


1

From my classes last semester, you could do this two ways: 1) Don't average everything, just the last 3~4 points in the data, for example: $avg = \frac {16+15+18}3 = 16,33$ 2) Use a weighted average so the last days have more weight, for example: $avg = \frac{\sum a_i w_i}{\sum w_i} = \frac{0*1 + 0*1 + 1*1 + 0*1 + 1*1 + 3*1 + 7*1 + 16*2 + 15*3 + ...


1

Hint: $$\text{Total average speed}=\frac{\text{total distance travelled}}{\text{total time travelled}}$$ Distance for first part of journey is $3\cdot 55=165$ km/h Distance for second part of journey is $2\cdot 65=130$ km/h


1

Let me write $$ S(x):=\sum_{i=1}^L ( \bar x - x_i) = \sum_{i=1}^L (\sum_{j=1}^K K^{-1}x_j - x_i) = \sum_{i=1}^L (L/K-1)x_i + \sum_{i=L+1}^K K^{-1}x_i. $$ Thus, $S(x)$ is maximal if $x_i$, $i=1\dots L$ are at the lower bound, while $x_i$ are at the uppper bound $i=L+1\dots K$. Hence the maximum is $$ S_{max}=L(L/K-1)a + (K-L)/K b = \frac{K-L}{K}(-La + b). $$


1

The issue that you rised with this question is in the area of robust statistics. In the case of estimating a parameter, it is called robust parameter estimation. There is a good book by Huber. I think this one will help alot. The idea is as follows. When you are estimating a parameter, the regular process first finds the log likelihood ratio of the density ...


1

For $i=1,2$ let $X_i$ be independent random variables Poisson-$\lambda_i$ distributed. You seem to be asking: "Is it true that in this case $\mathbb E\min(X_1,X_2)=\min(\mathbb EX_1,\mathbb EX_2)$?" The answer is: "no". It is evident that $\min(X_1,X_2)\leq X_1$ and $P(\min(X_1,X_2)<X_1)>0$. These observations justify the conclusion that ...


1

Both averages are calculated correctly. They differ because there are two sampling models being employed -- in one approach we sample a class at random and ask what is the average; in the other approach we sample a student at random and ask what is the average class size the student sees. As for which number you would choose to represent the 'average' class ...


1

for example;a=1,b=7,c=5 or a=2,b=14,c=10


1

(I'm going to assume you mean arithmetic average, but the same applies for others too.) This doesn't generally make any sense for several reasons, mainly because you can't look up an infinite amount of values to calculate the average of. We can take the average of the first $n$ values and let $n$ grow, but this means that the order of the values in this ...



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