# Tag Info

7

\begin{align} \int_0^1\int_x^1\sin\left(t^2\right)\,\mathrm{d}t\,\mathrm{d}x &=\int_0^1\int_0^t\sin\left(t^2\right)\,\mathrm{d}x\,\mathrm{d}t\\ &=\int_0^1\sin\left(t^2\right)\,t\,\mathrm{d}t\\ &=\frac12\int_0^1\sin\left(t^2\right)\,\mathrm{d}t^2\\ &=\frac12\left[-\cos\left(t^2\right)\right]_0^1\\[3pt] &=\frac{1-\cos(1)}2 \end{align}

7

You can use $\displaystyle f_\text{avg} = \frac{1}{1-0} \int_0^1 \int_x^1 \sin(t^2) \, dt\,dx = \int_0^1 \int_0^t \sin(t^2) \, dx\,dt$

6

Let $x$ represent the average of the $71\%$ subset. Then the following equation holds: $$(0.29) \cdot 52 + (0.71) \cdot x = 55$$ Does that make sense, and can you take it from here?

4

Let $x$ denote your average and let $y$ denote the average of the rest. The overall average is $0.29x+0.71y=55$. You know that $x=52$. Just solve for $y$.

3

In your example, you have $u_1, u_2, u_3$, $v_1, v_2$, and you have correctly showed that $$\text{mean}(u_1,u_2,u_3) + \text{mean}(v_1,v_2)$$ is not necessarily equal to $$\text{mean}(u_1 + u_2 + u_3, v_1 + v_2),$$ so in that sense you are exactly correct. However, this is not what the statement was intended to express. What is intended is that if you ...

2

Suppose you have observations on income from work $w_i$ and income from benefits $b_i$ for a population of $n$ people The text is simply saying that $$\frac{1}{n}\sum_{i=1}^nw_i+\frac{1}{n}\sum_{i=1}^nb_i=\frac{1}{n}\sum_{i=1}^n(w_i+b_i)$$ where the left hand side is the sum of the mean work income and mean benefit income, while the right side is the ...

2

Hint: let there be $g$ girls. What is the total number of children? What is the total weight? What is the total weight of boys? What is the total weight of girls? What is $g$?

2

The mean of a multiset of $n$ numbers, $\{a_1,a_2,a_3,\dots,a_n\}$ is defined as $$\mu = \frac{a_1+a_2+\dots+a_n}{n}$$ This is regardless of whether or not some of the elements in the set are zero, repeated, negative, or otherwise uncommon for whatever reason. Examples: The average of $\{1,3,5\}$ is $\frac{1+3+5}{3}=\frac{9}{3}=3$ The average of $\{-5,1,... 2 In one dimension, the minimum distance is$0$and the maximum is$1$. The average is less than$\frac 12$because it is hard to get large distances. One point needs to be near one end of the interval and the other needs to be near the other. If we call the points$x,y$, the average distance is $$\int_0^1 \int_0^1 |x-y| \;dx \;dy=2\int_0^1 \int _y^1 x-y \; ... 2 Your transfer of the solution from 3D to 2D is wrong. What was a plane is now a line; what was a panel is now a strip. There are two strips of width \mathrm dt, with lengths 10-t and 5-t, so the probability distribution function is$$ f_T(t)=\frac{15-2t}{50} $$and the expected waiting time is$$ E(T)=\int_0^5tF_T(t)\,\mathrm dt=\frac{25}{12}\;. ... 1 Your contribution to the total sum is$0.29\times52$, so what you need is$(55-0.29\times52)/(1-0.29)\approx 56.23$1 Firstly note that "random" is meaningless without specifying a distribution. I'll assume that you mean "uniformly random". It seems the average distance between each coordinate component (x,y,z) shouold be around 0.5. Why should this be? If you choose two uniformly random points$p,q$from$[0,1]$, if$p = 0$then indeed the average distance between ... 1 The average distance in a unit cube is given by the multiple integral $$I=\int_{[0,1]^6}\sqrt{(x-X)^2+(y-Y)^2+(z-Z)^2}\,d\mu$$ whose numerical value is around$\color{red}{0.6617}$. You simulations are correct. We may also notice that the average value of$(x-X)^2$is$\frac{1}{6}$, hence the inequality$\$ I \leq \sqrt{\frac{1}{6}+\frac{1}{6}+\frac{1}{6}}=...

1

I guess my idea is similar to Christian Blatter above. I did a R simulation with the assumptions that buses arrive promptly every 10 and 15 minutes, respectively. However, the interval time among buses is fixed with uniformly distributed starting times. I simulate buses between [0,10000] minutes and the guy arriving at the bus stop at time t in [50,9950] and ...

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