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國營事業◆1.工作研究 2.精實生產

精實生產

國營事業◆1.專案管理 2.英文翻譯及寫作

115年 - 115-1 專技高考_社會工作師：社會工作研究方法#137528(42題)

115年 - 115-1 專技高考_社會工作師：社會政策與社會立法#137527(42題)

115年 - 115-1 專技高考_社會工作師：社會工作直接服務#137526(42題)

115年 - 115-1 專技高考_社會工作師：社會工作#137525(42題)

115年 - 115 大學入學美術術科考試試題：美術鑑賞#137367(50題)

115年 - 115-1 專技高考_藥師(一)：藥學(一)(包括藥理學與藥物化學)（重複）#137365(80題)

115年 - 115-1 專技高考_物理治療師：物理治療技術學（包括電療學、熱療學、操作治療學與輔具學）（重複）#137362(80題)

115年 - 115-1 專技高考_牙醫師(二)：牙醫學(四)（包括口腔顎面外科學、牙科放射線學等科目及其相關臨床實例與醫學倫理）#137358(80題)

115年 - 115-1 專技高考_牙醫師(一)：牙醫學(一)（包括口腔解剖學、牙體形態學、口腔組織與胚胎學、生物化學等科目及其臨床相關知識）#137357(80題)

115年 - 115-1 專技高考_藥師(一)：藥學(二)(包括藥物分析與生藥學(含中藥學))#137356(80題)

25. 配電箱內之任何過電流保護裝置，如所裝接負載在正常狀態下須連續滿載三小時以上者，該負載電流不可超過其額定值之(A)70%(B)80%(C)90%(D)125%。

24. 低壓3Φ4W線路中不宜單獨裝開關或斷路器之導線為(A)R相線(B)S相線(C)T相線(D)被接地導線。

23. NFB (No Fuse Breaker)係表示(A)油斷路器(B)無熔線開關(C)燈用分電盤(D)隔離開關。

4. 請根據上述結果求得 F 值（可參考下表）。

3. 請計算組內離均差平方和（SSW, Within Sum of Squares）。

2. 請計算總離均差平方和（SST, Total Sum of Squares）。

國民教育法條

講師：【站僕】摩檸Morning.

簡介：國民教育法條

教育基本法

講師：【站僕】摩檸Morning.

簡介：教育基本法

師資培育法

講師：【站僕】摩檸Morning.

簡介：師資培育法

應收票據與貼現

描述：應收票據介紹 貼現方式

國一數學

描述：MATH

管理學

描述：管理

15. (A) acknowledged by (B) allocated to (C) confined to (D) distinguished from

1. 根據刑事訴訟法規定，告訴乃論之罪，應在知悉犯人之時起，多久內提出告訴？ (A)3個月 (B)4個月 (C)5個月 (D)6個月

65. 下列何者不是現象本位學習所具有的特色？(A)引導學生針對真實的問題進行探究(B)課程很在乎 what ,why 和 how(C)現象本位學習以真實世界中的現象為起點(D)學生會進行各種調查、探究、彙整、訪談等等，試圖對所研究的現象有進一步的了解，甚至解決問題

13.適應體育的 STEP 原則有利於幫助教師構思動作難易度的調整，並有利於規劃差異化教學。請問下列變化何者錯誤? (A)Space:調整投籃距離的遠近 (B)Task:調整投籃的動作 (如: 單手、雙手) (C)Equipment:調整活動進行的速度、對抗強度 (D)People:調整比賽場上的對抗人數

43. During the mitotic phase, the enzyme separase is responsible for cleaving the cohesin proteins that hold sister chromatids together, a step crucial for the initiation of anaphase. If a cell possesseda dominant mutation causing separase to be constitutively active and begin cleaving all cohesins prematurely during prometaphase (before the M checkpoint criteria were met), which outcome is the most likely and immediate consequence? (A) The resulting daughter cells would skip the G1 phase and immediately enter the S phase due to premature activation of maturation-promoting factor (MPF). (B) The cell would successfully complete mitosis, but the spindle poles would fail to move apart due to inactive non-kinetochore microtubules. (C) The M checkpoint would fall, and the liberated chromatids would segregate randomly, resulting in genetically unequal daughter cells. (D) The nuclear envelope would reform immediately, triggering early telophase and preventing the remaining spindle microtubules from attaching to kinetochores. (E) Cytokinesis would initiate during prophase, leading to the formation of multiple, small nuclei within a single parent cell.

44. In an E. coli cell undergoing rapid replication, a newly identified chemical agent completely and specifically inhibits the function of DNA ligase. Assuming all other replication proteins (Helicase, Primase, DNA pol I, and DNA pol III) are fully functional, what structural consequence would be immediately observable following the completion of the first round of DNA synthesis? (A) Both parental DNA strands would remain permanently associated, halting the replication process before the replication fork could open fully. (B) The leading strand would fail to elongate beyond the initial RNA primer because DNA pol III requires DNA ligase to begin continuous synthesis. (C) The lagging strand would consist of multiple Okazaki fragments that are fully synthesized DNA segments but lack covalent bonds between them. NTHU115 (D) The ends of the circular bacterial chromosome would shorten significantly because the replication machinery cannot replace the terminal RNA primers. (E) The concentration of thymine dimers would increase dramatically due to the inability of the DNA replication complex to proofread mismatched bases.