Da 1155: Fill & Download for Free

The Army uses them in KIA/MIA instances. Each soldier should carry an DA-1155 and DA-1156 which have been filled out as much as possible, and then completed when there is an incident that requires them.

Because “classical” composers in China were not held in especially high esteem as a profession, and there were very few from before 400 years ago that were prominent, a lot of the early music was anonymous, and Chinese didn't write most music down, and composers and even if they did, the Chinese music notation they used was descriptive, but not precise like European notation, which was already modern by the 400 years ago since it's the same system almost everyone uses now. Chinese notation served as primarily as a reminder, and it required you to already know what the music sounds like, which is why modern interpretations can all sound very different, especially when it comes to rhythm.Still, many thousands of pieces from before 400 years ago have survived.As well, there are many old texts in China that simply remain to be studied by Chinese scholars, for whatever reason.This used to be the case as well in Europe in the 1800s to early 1900s, when much older music was just hidden away and never heard during all that time, and even now, old music is sometimes still rediscovered in Europe, such as when the complete Viola de Gamba Fantasias by Telemann were found in 2015 in an old palace in Germany, as well as rediscovered music by other composers. The only reason anyone even knew after sll that timd that those had actually existed was because they were published in the 1730s and there was a record of this, and not a single copy made it to this day except for one hifden away for almost 300 years, so just imagine how bad the chances are fiyr music that was never published that we will never know about, and that we have no trace of.Notwithstanding, the majority of music from the 1200s, 1300d, 1400s, 1500s, 1600s, and even 1700s from both places is gone forever because it was very rarely published and after all that time even then most of it gets lost or destroyed, especially during the late 19th and early 20th century due to WWI, WWII, other wars and disasters, and the Cultural Revolution.Here are some of the early chinese composers:Bo Ya (Yu Boya) Warring States Period (5th Century BC – 221 BC)Cai Yong (132-193) Han Dynasty – Cai Shi Wunong (Five Melodies)Cai Wenji (Cai Yan) (177-250) Han Dynasty – Da HuijiaPu’An (1115-1169) – Song Dynasty – Pu’An ZhouXi Kang (Ji Kang) (223-262) Six Dynasties PeriodJiang Kui (1155-1221) Song DynastyGuo Chuwang – Song Dynasty

Thanks Henrik and Michael for your answers. After posting this question I continued to work on it with my friend using the following notation and relative probabilities:[math]a = \dfrac{P_a}{P_a [/math][math][/math][math]+ P_b [/math][math][/math][math]+ P_c [/math][math][/math][math]+ P_d}[/math][math]b = \dfrac{P_b}{P_a [/math][math][/math][math]+ P_b [/math][math][/math][math]+ P_c [/math][math][/math][math]+ P_d}[/math][math]c = \dfrac{P_c}{P_a [/math][math][/math][math]+ P_b [/math][math][/math][math]+ P_c [/math][math][/math][math]+ P_d}[/math][math]d = \dfrac{P_d}{P_a [/math][math][/math][math]+ P_b [/math][math][/math][math]+ P_c [/math][math][/math][math]+ P_d}[/math]where [math]a [/math][math][/math][math]+ b [/math][math][/math][math]+ c [/math][math][/math][math]+ d = 100\%[/math]We broke the problem down into smaller pieces as follows:What is the probability of not rolling a before d as:[math]d(b+c) [/math][math][/math][math]+ d(b+c)^2 [/math][math][/math][math]+ d(b+c)^3 [/math][math][/math][math]+ \cdots [/math][math][/math][math]+ d(b+c)^\infty[/math]which simplifies to:[math]d[ (b+c) [/math][math][/math][math]+ (b+c)^2 [/math][math][/math][math]+ (b+c)^3 [/math][math][/math][math]+ \cdots [/math][math][/math][math]+ (b+c)^\infty [/math][math][/math][math]][/math]We defined V as the infinite series[math]V = (b+c) [/math][math][/math][math]+ (b+c)^2 [/math][math][/math][math]+ (b+c)^3 [/math][math][/math][math]+ \cdots +(b+c)^\infty[/math]multiplying both sides by (b+c) results in:[math]V(b+c) = (b+c)^2 [/math][math][/math][math]+ (b+c)^3 [/math][math][/math][math]+ (b+c)^4 [/math][math][/math][math]+ \cdots [/math][math][/math][math]+ (b+c)^\infty[/math]subtracting V(b+c) from V results in:[math]V - V(b+c) = (b+c)[/math][math]V(1-b-c) = (b+c)[/math][math]V = \dfrac{b+c}{1-b-c}[/math]substituting this in place of the infinite series resulted in:[math]d( \dfrac{b+c}{1-b-c} [/math][math][/math][math])[/math]By symmetry, not rolling b before d is:[math]d( \dfrac{a+c}{1-a-c} [/math][math][/math][math])[/math]and, not rolling c before d is:[math]d( \dfrac{a+b}{1-a-b} [/math][math][/math][math])[/math]At this point we had the following formula:[math]P = 1 - [d [/math][math][/math][math]+ d(\dfrac{b+c}{1-b-c}) [/math][math][/math][math]+ d(\dfrac{a+c}{1-a-c}) [/math][math][/math][math]+ d(\dfrac{a+b}{1-a-b}) [/math][math][/math][math]][/math]however we were double counting all of the instances where a or b or c occur individually and removed them as follows:the times that a is double counted are:[math]da[/math][math] [/math][math][/math][math]+ da^2 [/math][math][/math][math]+ da^3 [/math][math][/math][math]+ \cdots [/math][math][/math][math]+ da^\infty[/math]using the above method to remove the infinite series results in:[math]d(\dfrac{a}{1-a})[/math]By symmetry b is double counted:[math]d(\dfrac{b}{1-b})[/math]and, c is double counted:[math]d(\dfrac{c}{1-c})[/math]this resulted in the following formula:[math]P = 1 - [d [/math][math][/math][math]+ d(\dfrac{b+c}{1-b-c}) [/math][math][/math][math]+ d(\dfrac{a+c}{1-a-c}) [/math][math][/math][math]+ d(\dfrac{a+b}{1-a-b}] - d(\dfrac{a}{1-a}) - d(\dfrac{b}{1-b}) - d(\dfrac{c}{1-c}) [/math][math][/math][math])[/math]which we simplified through these steps:[math]P = 1 - d(1 [/math][math][/math][math]+ \dfrac{b+c}{1-b-c} [/math][math][/math][math]+ \dfrac{a+c}{1-a-c} [/math][math][/math][math]+ \dfrac{a+b}{1-a-b} - \dfrac{a}{1-a} - \dfrac{b}{1-b} - \dfrac{c}{1-c} [/math][math][/math][math])[/math][math]P = 1 - d(\dfrac{1-b-c}{1-b-c} [/math][math][/math][math]+ \dfrac{b+c}{1-b-c} [/math][math][/math][math]+ \dfrac{a+c}{1-a-c} [/math][math][/math][math]+ \dfrac{a+b}{1-a-b} - \dfrac{a}{1-a} - \dfrac{b}{1-b} - \dfrac{c}{1-c} [/math][math][/math][math])[/math][math]P = 1 - d(\dfrac{1}{1-b-c} [/math][math][/math][math]+ \dfrac{a+c}{1-a-c} [/math][math][/math][math]+ \dfrac{a+b}{1-a-b} - \dfrac{a}{1-a} - \dfrac{b}{1-b} - \dfrac{c}{1-c} [/math][math][/math][math])[/math]adding 1 and subtracting 1 inside the brackets[math]P = 1 - d(\dfrac{1}{1-b-c} [/math][math][/math][math]+ \dfrac{a+c}{1-a-c} [/math][math][/math][math]+ \dfrac{a+b}{1-a-b} - \dfrac{a}{1-a} - \dfrac{b}{1-b} - \dfrac{c}{1-c} +1 -1 +1 -1 +1 -1)[/math][math]P = 1 - d(\dfrac{1}{1-b-c} [/math][math][/math][math]+ \dfrac{1}{1-a-c} [/math][math][/math][math]+ \dfrac{1}{1-a-b} - \dfrac{1}{1-a} - \dfrac{1}{1-b} - \dfrac{1}{1-c} [/math][math][/math][math]+ 1)[/math]further simplification provides:[math]P = 1 - d(\dfrac{1}{a+d} [/math][math][/math][math]+ \dfrac{1}{b+d} [/math][math][/math][math]+ \dfrac{1}{c+d} [/math][math][/math][math]+ \dfrac{1}{a-1} [/math][math][/math][math]+ \dfrac{1}{b-1} [/math][math][/math][math]+ \dfrac{1}{c-1} [/math][math][/math][math]+ 1)[/math]plugging in the original probabilities[math]P_a = 3/36[/math] , [math]P_b = 2/36[/math] , [math]P_c = 1/36[/math] , [math]P_d = 6/36[/math][math]a = 3/12 = 1/4[/math] , [math]b = 2/12 = 1/6[/math] , [math]c = 1/12[/math] , [math]d = 6/12 = 1/2[/math][math]P = 1 - \frac{1}{2}(\frac{4}{3} [/math][math][/math][math]+ \frac{3}{2} [/math][math][/math][math]+ \frac{12}{7} - \frac{4}{3} - \frac{6}{5} - \frac{12}{11} [/math][math][/math][math]+ 1)[/math][math]P = 1 - \frac{1}{2}(\frac{1155}{770} [/math][math][/math][math]+ \frac{1320}{770} - \frac{924}{770} - \frac{840}{770} [/math][math][/math][math]+ \frac{770}{770})[/math][math]P = 1 - \frac{1}{2}*\frac{1481}{770}[/math][math]P = \frac{1540}{1540} - \frac{1481}{1540}[/math][math]P = \frac{59}{1540} \approx 0.038312[/math]Thanks again to Michael and Henrik for supplying answers, I'm glad we all have the same result.