GCI

Schubert, Valenzuela in US women's Amateur final

Sophia Schubert and Albane Valenzuela advanced to the finals of the 117th U.S. Women's Amateur Championship on Saturday and hope to end long droughts in Sunday's 36-hole final at the … Click to Continue »

    
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Ha Na Baek, Yu Rim Lee claim title of women's doubles final
Golf - American Schubert wins U.S. Amateur
Women's amateur golf from today
It was always the women, and above all the young ones, who were the most bigoted adherents of the Party, the swallowers of slogans, the amateur spies and nosers-out of unorthodoxy.
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Spotlight on Schubert The German Romantic composer is the subject of this year's Bard Music Festival.
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Music Review Two Takes on a Schubert Quintet Performances by the Emerson String Quartet and the Danish String Quartet offer perspective on age and artistic insight.
Beatles Pop to the Beat of Schubert
Beatles Pop to the Beat of Schubert Music fans might think that they have heard the Beatles in every conceivable context. The New York Festival of Song found a new one, juxtaposing some of the 1960s pop classics with the work of Franz Schubert.
Schubert, the Yankees and Irresistible Grace
Schubert, the Yankees and Irresistible Grace A lovely dinner, with a trip to Carnegie Hall looming. But then: salvation.
Grant Schubert stars in Australia’s big win
‘Schubert paints feelings with his music’
Love the inclusion of Schubert's symphony...
Horse : Schubert wins main event
  • [30-09] A bit of relaxing Schubert for Saturday night on my @KawaiPianos #Schubert #piano #saturdaynight
  • [27-08] Well done #England you did great in womens ruby world cup final #NewZealand are unbeatable ?
  • [15-10] Brampton U21 Junior Womens Cashspiel (SF):Mackenzie Kiemele 6, Veronica Bernard 5 [FINAL]
  • [15-10] Coldwater Rocktoberfest U18 Womens Spiel (CF):Kelly Middaugh 12, Alex Bertolo 6 [FINAL]
  • [14-10] Red Deer Elks Junior Womens Spiel (1):Zi-Xin Wang 7, Taitan Hagglund 2 [FINAL]
  • [15-10] Brampton U21 Junior Womens Cashspiel (CF):Megan Smith 5, Mackenzie Kiemele 2 [FINAL]
  • [15-10] Coldwater Rocktoberfest U18 Womens Spiel (CF):Bella Croisier 6, Evelyn Robert 5 [FINAL]
  • [14-10] Brampton U21 Junior Womens Cashspiel (6):Quinn Walsh 7, Jestyn Murphy 3 [FINAL]
  • [14-10] Red Deer Elks Junior Womens Spiel (1):Subasthika Thangadurai 6, Rebecca Geary 5 [FINAL]
  • [14-10] Red Deer Elks Junior Womens Spiel (2):Veronica Maschmeyer 4, Ryleigh Bakker 2 [FINAL]
  • [14-10] OVCA Junior Superspiel Womens Qualifier (2):Emma Wallingford 6, Jessica Guibault 0 [FINAL]
  • [16-10] OVCA Junior Superspiel Womens Qualifier (9):Kayla Gray 8, Mackenzie Comeau 1 [FINAL]
  • [15-10] OVCA Junior Superspiel Womens Qualifier (8):Mackenzie Comeau 7, Kaleigh MacKay 5 [FINAL]
  • [15-10] Red Deer Elks Junior Womens Spiel (SF):Lauren Marks 4, Ashton Simard 3 [FINAL]
  • [15-10] Brampton U21 Junior Womens Cashspiel (SF):Mackenzie Kiemele 6, Veronica Bernard 5 [FINAL]
  • [15-10] Coldwater Rocktoberfest U18 Womens Spiel (CF):Kelly Middaugh 12, Alex Bertolo 6 [FINAL]
Schubert, Valenzuela in US women's Amateur final
Sophia Schubert and Albane Valenzuela advanced to the finals of the 117th U.S. Women's Amateur Championship on Saturday and hope to end long droughts in Sunday's 36-hole final at the … Click to Continue »
Valenzuela
Schubert
Amateur
womens
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* Schubert - Der Tod und das Mädchen 13 Pieces for Halloween, no.5. Schubert’s “Death and the Maiden” String Quartet is one of his...
* Piano 2/100Today I spent a looong time on technique, not so much on the Schubert. I got all of the A flat major technique up to tempo for the exam except the arpeggios. I’ll remedy that tomorrow, hopefully before my lesson although I’m staying later at school to tutor for a bit. Thankfully I finish school early on Wednesdays. In Schubert I tried to phrase the solid chord segments in the first section and resolve the broken chord passages to the final chord. Also I tried to get the middle section together but it stretches a bit far for my RH so it gets sore after some time.
* Bez naslova #22 by nihada-1234 featuring womens plus dresses ❤ liked on Polyvore Maggy London womens plus dress / Sandro...
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find the probability of selecting exactly two women and at least two women when a six-person committee is selected from $7$ men and $4$ women? A committee of six members is formed from a group of $7$ men and $4$ women. What is the probability that the committee contains a. exactly two women? b. at least two women? My attempt : given $P(A) = 7/11$ and $P(B) = 4/11$ option a) probability that the commitee contains exactly two women $$= \frac{P(AB)}{P(B)} = \frac{P(A)P(B)}{P(B)} = \frac{\frac{7}{11} \cdot \frac{4}{11}}{\frac{4}{11}} = \frac{7}{11}$$ option b)the probability that the committee contains at least two women = $$\frac{P(AB)}{P(A \cup B)} = \frac{\frac{7}{11} \cdot \frac{4}{11}}{1- \frac{7}{11} \cdot \frac{4}{11}} = \frac{28}{93}$$ If my answer is correct or not, I would be more thankful to those rectifying my mistakes......
How can I get the matrix form for a schubert cell? I am trying to learn how to understand the cell-decomposition for the grassmannian and am following these notes: http://www.math.drexel.edu/~jblasiak/grassmannian.pdf. On page 2 the author considers the cell $e(\lambda)$ in $G(3,7)$ for $\lambda = (3,3,1)$. From the definition of $e(\lambda)$ you can find that \begin{align} \text{dim}(W \cap F_1) = 1 && \text{dim}(W \cap F_2) = 1 && \text{dim}(W \cap F_3) = 2 \\ \text{dim}(W \cap F_5) = 2 && \text{dim}(W \cap F_6) = 3 \end{align} where $F_\bullet$ is a maximal flag in $\mathbb{C}^{n+m}$. The author claims that $e(\lambda)$ is the set of matrices $$ \begin{bmatrix} * & 1 & 0 & 0 & 0 & 0 & 0 \\ * & 0 & 1 & 0 & 0 & 0 & 0 \\ * & 0 & 0 & * & * & 1 & 0 \end{bmatrix} $$ but I am not sure how this argument works. How can I show the dimension conditions imply that every $3$-plane looks like this matrix? What I've been able to figure out is that if we consider a basis $v_1,\ldots, v_7$ defining the standard flag, then we have \begin{align} v_2 \in & W\cap F_2,\ldots F_7 \\ v_3 \in & W \cap F_3, \ldots F_7 \\ v_6 \in & W \cap F_6, F_7 \end{align}
Disabling Schubert's PDF Browser plugin on Safari I use Safari and Firefox. To be able to view PDFs on Firefox on the fly, I use the Schubert's PDF viewer. But, I would like to block it from loading my PDFs on Safari. I like Safari's way of handling ...
Correct logic of permuting 5 men and 5 women to find probability of different highest women rank The problem reads like this: Problem Five men and $5$ women are ranked according to their scores on an examination. Assume that no two scores are alike and all $10!$ possible rankings are equally likely. Let $X$ denote the highest ranking achieved by a woman. (For instance, $X = 1$ if the top-ranked person is female.) Find $P(X = i),i = 1, 2, 3, . . ., 8, 9, 10$. Solution given was: $P(X=1)= \frac{5}{10}= \frac{1}{2}$ because there are 5 women and total of 10 to choose from $P(X=2)=\frac{5}{10}\times \frac{5}{9}=\frac{5}{18}$ because for rank1 thereare 5 men and total of 10 to choose from, for rank 2 (we want awoman) we still have 5 women but only a total of 9 to choosefrom. $P(X=3)=\frac{5}{10}\times \frac{4}{9}\times \frac{5}{8}=\frac{5}{36}$ $P(X=4)=\frac{5}{10}\times \frac{4}{9}\times \frac{3}{8}\times \frac{5}{7}=\frac{10}{168}$ $P(X=5)=\frac{5}{10}\times \frac{4}{9}\times \frac{3}{8}\times \frac{2}{7}\times \frac{5}{6}=\frac{5}{252}$ $P(X=6)=\frac{5}{10}\times \frac{4}{9}\times \frac{3}{8}\times \frac{2}{7}\times \frac{1}{6}\times \frac{5}{5}=\frac{1}{252}$ My solution was $P(X=1)=\frac{5\times 9!}{10!}=\frac{1}{2}$ because there are five women to occupy 1st rank and then there remained 9 which can permute in $9!$ ways. There are total $10!$ ways to permute $10$ people $P(X=2)=\frac{5\times \binom{5}{4}\times 8!}{10!}=\frac{5}{18}$ because there are five women to occupy 2st rank. The 1st rank will be of a man. So we have to select $4$ out of $5$ men which will be ranked after $2$nd rank. These four men and remaining 4 women can be permuted in $8!$ ways. $P(X=3)=\frac{5\times \binom{5}{3}\times 7!}{10!}=\frac{5}{72}$ $P(X=4)=\frac{5\times \binom{5}{2}\times 6!}{10!}=\frac{5}{504}$ $P(X=5)=\frac{5\times \binom{5}{1}\times 5!}{10!}=\frac{5}{6048}$ $P(X=6)=\frac{5\times \binom{5}{0}\times 4!}{10!}=\frac{1}{30240}$ Doubts Where my logic went wrong? When I compared the two approach, I realized that the books solution is permuting ranks higher than the highest ranked girl, while my solution is permuting ranks lower than the highest ranked girl. So I was guessing what makes book solution not permute lower ranks and my solution not permuting higher ranks. Shouldn't we permute on both sides of highest ranked girl?
Bill reading Shakespeare and Maureen singing Schubert satisfy/satisfies me [duplicate] Which verb form is grammatically correct here? My intuition says 'satisfy' but a textbook I'm reading says otherwise (Core Syntax: A Minimalist Approach. If interested, a legal copy is available here)....
In a single word, how can you describe something as being made by women for women? Can we say "for women and by women" in a single word? We're creating a service that is similar to uber but only for women. Thank you very much.
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Who won the 2009 Scottish amateur cup final? Queens Park beat Hurlford Thistle AFC 3-1 in the final at Hampden Park, Glasgow. It was the twelfth time Queens have won the cup since it began in 1909.
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What player has scored in the UEFA Cup final and Champions League final and FA Cup final and League Cup final? Steven Gerrard, except he scored an own goal in the league cup final. actually he scored agaisnt man utd in a 2 : 0 victory in 2002/2003 season the other scorer was m . Owen
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