作者查詢 / ice80712
作者 ice80712 在 PTT [ Grad-ProbAsk ] 看板的留言(推文), 共73則
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1F→:bernoulli02/28 23:41
6F→:u(2-t)在t=0~2是1 t>2是0 然後再拉氏轉換...02/28 15:31
1F推:利用級數的基本定義下去做 應該是可以直接寫出答案02/25 13:58
1F→:A贏的機率為1/302/21 08:41
1F推:reduction of order 已知一解求另一解 直接背02/20 11:01
7F→:方法是對的 可以看一下公式的推導過程 以後就用公式02/20 12:50
3F推:逆運算本來就有可能會跑出齊性解 方法用對就只會有特解02/16 20:01
1F→:如果是5*6.5的話應該是有多算到02/07 22:54
3F→:假設有兩個人的目的地是在同一樓的話 只能算一次02/07 23:06
4F→:停在2~11樓的機率應該是一樣的 我是用這方法算02/07 23:06
5F→:算出來的答案是26.XX 也是選E...02/07 23:07
8F→:題目是要電梯停下來的期望值 不是人02/07 23:13
10F→:如果大家都在4樓下 那Y=20 但電梯只能算402/07 23:15
14F→:我想錯了吧 不過加起來確實是不在ABCD02/07 23:26
15F→:如果是問電梯停下次數的期望值的話 我剛才算是4.50461...02/07 23:38
16F→:慘...這樣就有答案可以選了XD02/07 23:38
19F→:可以整除喔 先算出電梯停在每一樓的機率02/07 23:46
21F→:更正 我算的是4.0951 腦筋已經錯亂了...02/07 23:52
23F→:令Y=X2+X3+...+X11 10個iid 求E[Y]02/07 23:55
24F→:我一開始看題目也是這麼覺得阿XD02/07 23:56
26F→:就想成電梯會停幾次 最少1次 最多5次 平均大概就4次02/07 23:58
32F→:平均次數應該是4次附近 電梯最多只會停5次02/08 00:25
33F→:我也覺得我想法怪怪的 但我用2人 3人驗證結果都對02/08 00:25
35F→:S是電梯停的次數期望值02/08 09:58
61F→:這題厲害的高中生應該也做得出來 不過我考試時沒寫對02/08 11:02
65F→:最快的方法應該就停在某一樓的機率為1-(0.9)^502/08 11:04
67F→:再乘10 應該就是答案了吧02/08 11:05
71F→:我是考K卷02/08 11:09
74F→:不 但是如果確定有些選項不能選的話 我會猜02/08 11:11
77F→:猜e吧 如果覺得期望值比0大就猜下去02/08 11:13
79F→:我應該也是選a 有點忘了...02/08 11:17
2F→:哪一題?02/04 23:42
4F→:transfer function算出來 再用基本定義去求3dB02/05 01:09
12F→:題目有說 exactly 只能精確解 時間常數法不能用吧02/05 17:48
3F→:迴旋積不好 碰到更高次的很難算02/01 20:04
3F→:這些選項史密斯都找的到01/29 23:07