chore(i8n,learn): processed translations

curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/averages-pythagorean-means.md

@@ -1,32 +1,52 @@
 ---
 id: 594d966a1467eb84194f0086
-title: 平均值 - 毕达哥拉斯指的是
+title: Averages/Pythagorean means
 challengeType: 5
-videoUrl: ''
+forumTopicId: 302227
 dashedName: averagespythagorean-means
 ---
 
 # --description--
 
-<p class='rosetta__paragraph'>计算整数<big>1</big>到<big>10</big> （包括）的所有三个<a class='rosetta__link--wiki' href='https://en.wikipedia.org/wiki/Pythagorean means' title='wp：毕达哥拉斯意味着'>毕达哥拉斯方法</a> 。 </p><p class='rosetta__paragraph'>为这组正整数显示<big>$ A（x_1，\ ldots，x_n）\ geq G（x_1，\ ldots，x_n）\ geq H（x_1，\ ldots，x_n）$</big> 。 </p>这三种方法中最常见的<a class='rosetta__link--rosetta' href='http://rosettacode.org/wiki/Averages/Arithmetic mean' title='平均值/算术平均值'>算术平均值</a>是列表的总和除以其长度： <big>$ A（x_1，\\ ldots，x_n）= \\ frac {x_1 + \\ cdots + x_n} {n} $</big>  <a class='rosetta__link--wiki' href='https://en.wikipedia.org/wiki/Geometric mean' title='wp：几何平均值'>几何mean</a>是列表产品的$ n $ th根： <big>$ G（x_1，\\ ldots，x_n）= \\ sqrt \[n] {x_1 \\ cdots x_n} $</big>  <a class='rosetta__link--wiki' href='https://en.wikipedia.org/wiki/Harmonic mean' title='wp：谐波意味着'>调和平均值</a>是$ n $除以总和列表中每个项目的倒数： <big>$ H（x_1，\\ ldots，x_n）= \\ frac {n} {\\ frac {1} {x_1} + \\ cdots + \\ frac {1} {x_n}} $</big>  <p class='rosetta__paragraph'>假设输入是包含所有数字的有序数组。 </p><p class='rosetta__paragraph'>要获得答案，请按以下格式输出对象： </p><pre class='rosetta__pre'> {
-  值：{
-    算术：5.5，
-    几何：4.528728688116765，
-    谐波：3.414171521474055
-  }，
-  测试：'是A> = G> = H？是'
+Compute all three of the [Pythagorean means](https://en.wikipedia.org/wiki/Pythagorean means "wp: Pythagorean means") of the set of integers $1$ through $10$ (inclusive).
+
+Show that $A(x_1,\\ldots,x_n) \\geq G(x_1,\\ldots,x_n) \\geq H(x_1,\\ldots,x_n)$ for this set of positive integers.
+
+<ul>
+  <li>The most common of the three means, the <a class='rosetta__link--rosetta' href='https://rosettacode.org/wiki/Averages/Arithmetic mean' title='Averages/Arithmetic mean' target='_blank'>arithmetic mean</a>, is the sum of the list divided by its length:<br>
+  <big>$ A(x_1, \ldots, x_n) = \frac{x_1 + \cdots + x_n}{n}$</big></li>
+  <li>The <a class='rosetta__link--wiki' href='https://en.wikipedia.org/wiki/Geometric mean' title='wp: Geometric mean' target='_blank'>geometric mean</a> is the $n$th root of the product of the list:<br>
+  <big>$ G(x_1, \ldots, x_n) = \sqrt[n]{x_1 \cdots x_n} $</big></li>
+  <li>The <a class='rosetta__link--wiki' href='https://en.wikipedia.org/wiki/Harmonic mean' title='wp: Harmonic mean' target='_blank'>harmonic mean</a> is $n$ divided by the sum of the reciprocal of each item in the list:<br>
+  <big>$ H(x_1, \ldots, x_n) = \frac{n}{\frac{1}{x_1} + \cdots + \frac{1}{x_n}} $</big></li>
+</ul>
+
+# --instructions--
+
+When writing your function, assume the input is an ordered array of all inclusive numbers.
+
+For the answer, please output an object in the following format:
+
+```js
+{
+  values: {
+    Arithmetic: 5.5,
+    Geometric: 4.528728688116765,
+    Harmonic: 3.414171521474055
+  },
+  test: 'is A >= G >= H ? yes'
 }
-</pre>
+```
 
 # --hints--
 
-`pythagoreanMeans`是一种功能。
+`pythagoreanMeans` should be a function.
 
 ```js
 assert(typeof pythagoreanMeans === 'function');
 ```
 
-`pythagoreanMeans([1, 2, ..., 10])`应该等于上面相同的输出。
+`pythagoreanMeans([1, 2, ..., 10])` should equal the same output above.
 
 ```js
 assert.deepEqual(pythagoreanMeans(range1), answer1);