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115年 - 115 公共工程品管人員(土建班)題庫 751-800#130668(50題)

115年 - 115 公共工程品管人員(土建班)題庫 851-900#130666(50題)

115年 - 115 公共工程品管人員(土建班)題庫 801-850#130665(50題)

115年 - 115 公共工程品管人員(土建班)題庫 701-750#130663(50題)

115年 - 115 公共工程品管人員(土建班)題庫 651-700#130662(50題)

115年 - 115 公共工程品管人員(土建班)題庫 601-650#130661(50題)

115年 - 115 公共工程品管人員(土建班)題庫 551-600#130659(50題)

115年 - 115 公共工程品管人員(土建班)題庫 501-550#130657(50題)

115年 - 115 公共工程品管人員(土建班)題庫 451-500#130655(50題)

115年 - 115 公共工程品管人員(土建班)題庫 401-450#130645(50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）。

國中一年級數學

講師：Joyce

簡介：第一章 整數運算與科學記號 第二章 因數分解與分數運算 第三章 一元一次方程式 第四章 簡單圖形與幾...

【矮袋鼠共筆】物理治療技術學｜熱療學 筆記/題庫

講師：矮袋鼠(114-2已上榜！）

簡介：考科包括電療學、熱療學、操作治療學與輔具學 『本課程考試筆記僅包含熱療學』，限時詳解卡解鎖筆記！

計算機概論(含網路概論）

講師：Terry Tung

簡介：計算機概論考試範圍涵蓋數位邏輯、計算機組織、資料表示法、程式語言、作業系統、資料庫、資料結構、電腦網...

Dream And Reality

課程：【WriteMol英文】青春作家的英文寫作班：國中篇
章節：章節

描述：The passage explores the relationship between dreams and reality, highlighting the similarities and...

Father Has Quit Smoking

課程：【WriteMol英文】青春作家的英文寫作班：國中篇
章節：章節

描述：The passage describes how the author's father quit smoking and the positive impact it has had on his...

Generation Gap

課程：【WriteMol英文】青春作家的英文寫作班：國中篇
章節：章節

描述：The passage discusses the generation gap, which is the differences in attitudes, beliefs, and values...

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.