◦ Cover letter ◦ Personal statement/ statement of purpose (PS) ◦ Proposal/ research plan

复旦大学：《学术英语（管理科学）Academic English for Business》学术英语综合素养教学课件_Unit 2 Marketing Text B The 5Ps in action

Reading Assignments Lectures #5-6& PS#2: Chapter 3 of Oppenheim and Willsky(O& w) Lectures #3-4 &z PS#3: Chapters 3&4 of Oppenheim and Willsky(O&w) Exercise for home study (not to be turned in, although we will provide solutions)

Eco514 Game Theory Lecture 8.5: More on Auctions; PS#1 Marciano Siniscalchi October 14, 1999 Introduction These notes essentially tie up a few loose ends in Lecture 8; in particular, I exhibit examples of inefficiencies in first-and second-price auctions I would also like to briefly comment on Questions 1 and 2 in Problem Set 2

delta=beta*1.554*1e-9/(2*pi*3e-7); disp2=23: ps 2/km for ml=1: fix(z/h) DI=exp(( j*(i*w1)+i*0. 5*disp2*(i*w1).2)*h/2) ul=fft(u).*D1 u2=ifft(ul) D2=exp( deltas*(i*w1)+i*0.5*dsp2*(*w1).2)*h/2) vl=fft(v).米D2 v2=ifft(v1)

REMINDER: Quiz #2 will be held from 7: 30-9: 30 p. m. Thursday, November 13 in Walker Memorial. The quiz will cover materials in Chapters 4-7(through Section 7. 4)of O&w, Lectures and Recitations through October 29, Problem Sets #4-6, and that part of Problem Set 7 involving problems from Chapter 7 Reading Assignments

一、键盘 1.基本外设. 控制器集成在芯片组中.主板提供接口 2.键盘的分布.排列.108键 （p51） 需求是第一位的!快捷键位... 3.串行接口：AT或PS/2或USB 4.无线键盘. （集成鼠标功能）

学习目的与要求：掌握塑料的分类、性能及应用；掌握PE、PMMA、PVC、PS、ABS树脂、酚醛树脂、环氧树脂、聚氨酯树脂等生产工艺技术、结构、性能及应用。 §8-1 塑料的分类与应用 §8-2 聚乙烯树脂及塑料 §8-3 聚甲基丙烯酸甲酯 §8-4 聚氯乙烯树脂及塑料 §8-5 聚苯乙烯、ABS树脂及塑料 §8-6 酚醛树脂及塑料 §8-7 环氧树脂及塑料 §8-8 聚氨酯及塑料

一、名词解释 市场营销 目标市场 营销渠道 市场调查 二、简答题 1、产品的整体概念是什么？有哪些特点？ 2、简述市场营销的“4PS”组合。 3、试分析产品生命周期四个阶段的特点及企业应采取的基本策略

The lowpass filter H(u) has a cutoff frequency wc=205T rad/ sec. Thus, c(t)is r(t) where all terms with frequency above we are removed by the lowpass filter. The terms which are kept have kwol 205T rad /sec k|< 10.25, so the output, ac(t),is r(t)= To obtain n, we sample c(t) every T=5 10-3 seconds with an impulse train The sampling frequency is 400T=2 x maximum frequency in c(t). Therefore