P(all in some semicircle)=∑i=1NP(Ei)=N⋅12N−1=N2N−1P(\text{all in some semicircle}) = \sum_{i=1}^{N} P(E_i) = N \cdot \frac{1}{2^{N-1}} = \frac{N}{2^{N-1}}P(all in some semicircle)=∑i=1NP(Ei)=N⋅2N−11=2N−1N
The only catch was this had a planned six-month mission, extending the astronauts stay until now.
。下载安装 谷歌浏览器 开启极速安全的 上网之旅。对此有专业解读
Server throughput before and after the fix. x-axis shows the batch size on a logarithmic scale; y-axis shows the response rate.
У берегов популярного среди россиян курорта появились опасные медузы08:45