...The Help, a recent 2011 drama film is the adaptation of Kathryn Stockett’s debut novel - The Help. The story expresses the story of black maids working for white homes in Jackson, Mississippi around the late 1960s and early 1970s. In this age before the American Civil Rights Movement, black women commonly held low wage employment such as cleaning, daycare, or maid service. A main protagonist, Eugenia “Skeeter” Phelan is a young recently college educated white woman looking to advance her journalistic abilities. She gladly accepts a local newspaper housekeeping advice column called “Miss Myrna” to progress her journalistic experience toward greater positions. As Skeeter knows little about housekeeping, she turns to her local friend’s maid, Aibileen another main protagonist critical to the true narrative. Skeeter is characterized as very progressive for her time period, she can see the injustice and segregation in the country, and around her town that creates a disgusting, racist, and saddening society. She views her new friend, Mrs. Aibileen and her friendly associates as a miraculous caring median to attempt a change in society for the better. Throughout our diverse readings we have dealt little with race relations in the community context of justice. The community itself however, has been a major fundamental principle of a successful and democratic system of justice. Platos – The Republic attempted to characterize and debate justice between scholarly individuals such...
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...创意策略CREATIVE BRIEF |CONFIDENTIAL INTERNAL CIRCULATION ONLY | | |工作号/Job Number | | |客户名/Client 辉瑞制药有限公司 | | |产品名/Product万艾可 | | |工作描述/Job Description 06年主题创意 | | BACKGROUND/背景 (有关的历史背景及现在所面对的问题) 万艾可作为治疗ED的第一品牌，通过一年的传播，使很多患有ED且知道万艾可的消费者得以正视性生活质量，并且进入药店进行购买。但是从长期发展来看，仍有很多问题需要解决： 1. 2005年的宣传内容过于松散，需要一条核心的品牌主线和可延续性强的年度主题。 2. 有很大一批消费者意识到自己需要采取行动来改善性生活质量，正在积极寻找方案，或者已经采取了一些行动（但这些行动抢占了万艾可的销售机会）。2006年，我们将着重影响这样一批人，使之转换到万艾可品牌上来。 3. 店员的教育：客户对2005年的成绩非常满意。我们需要将教育的目标从“可以开口询问和介绍”升级至“懂得顾客心理，积极主动介绍”，使终端的影响发挥到最大。 4. 在“万艾可”、“伟哥”、“蓝色菱形小药片”三者的视觉关系和使用规则上有更系统的规范，长期以往，建立起独特的视觉识别系统和三者的关联度。 5. 创意在2005年创造出“高雅性文化”的概念，但和消费者的日常生活关联不大，对better erection, better sex, better life这一重要主张呈现不够直接，无法真正打动消费者，唤起行动。 06年市场传播主题: 万艾可，帮助有进取心的男人,释放能量，感受男人的成就。(文案要求修饰，红色部分是关键词) MARKETING OBJECTIVE/市场营销目标 (请精确的描述。请勿只写“提高销售量”。应写明具体提升品牌知名度X%或提升市场份额X%.) 保持万艾可第一品牌的地位 强化万艾可的品牌形象 在药店之外探索新的营销方式 提升在零售药店的销售 ADVERTISING OBJECTIVE/广告目标 (请用一句话简单陈述。它必须是-相关联的、可理解的、可测量的、可信的及可达成的) 保持万艾可第一品牌的地位； 传达“万艾可帮助带来健康男性魅力”的信息； 寻找有效渠道进行传播 TARGET...
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...1 1111 11111111 111 11111 11111 1111111 111111 1111111 111111 11111111 1111 11111111 111 11111 11111 1111111 111111 1111111 111111 1111111 1 1111 11111111 111 11111 11111 1111111 111111 1111111 111111 1111111 1 1111 11111111 111 11111 11111 1111111 111111 1111111 111111 1111111 1 1111 11111111 111 11111 11111 1111111 111111 1111111 111111 1111111 1 1111 11111111 111 11111 11111 1111111 111111 1111111 111111 1111111 1 1111 11111111 111 11111 11111 1111111 111111 1111111 111111 1111111 1 1111 11111111 111 11111 11111 1111111 111111 1111111 111111 1111111 1 1111 11111111 111 11111 11111 1111111 111111 1111111 111111 1111111 1 1111 11111111 111 11111 11111 1111111 111111 1111111 111111 1111111 1 1111 11111111 111 11111 11111 1111111 111111 1111111 111111 1111111 1 1111 11111111 111 11111 11111 1111111 111111 1111111 111111 1111111 1 1111 11111111 111 11111 11111 1111111 111111 1111111 111111 1111111 1 1111 11111111 111 11111 11111 1111111 111111 1111111 111111 1111111 1 1111 11111111 111 11111 11111 1111111 111111 1111111 111111 1111111 1 1111 11111111 111 11111 11111 1111111 111111 1111111 111111 1111111 1 1111 11111111 111 11111 11111 1111111 111111 1111111 111111 1111111 1 1111 11111111 111 11111 11111 1111111 111111 1111111 111111 1111111 1 1111 11111111 111 11111 11111 1111111 111111 1111111 111111 1111111 1 1111 11111111 111 11111 11111 1111111 111111 1111111 111111 1111111 1 1111 11111111 111 11111 11111 1111111 111111 1111111 111111 1111111 1 1111 11111111...
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...will lose 50%. Your goal is to subnet them with as little subnet as possible but still meeting the requirement. In other word, maximize the number of hosts that is available for each subnet. 26. 134.84.0.0/16 subnetted to 97 subnets and provide information for subnets #1, #8, #65, and #97 134.84.0.0/16 10000110 01010100 00000000 00000000 Network address 11111111 11111111 00000000 00000000 Subnet Mask The formula to find out how many bits to borrow 2 to the power off 7 = 128 need to borrow 7 bits. 7 bits = 0000001 New 10000110 01010100 00000000 00000000 Network Address 134.84.0.0/16 11111111 11111111 11111110 00000000 Subnet Mask 255.255.254.0 Subnet Mask #1 Network Address 10000110 01010100 00000010 00000000 Network Address 134.84.2.0 11111111 11111111 11111110 00000000 Subnet Mask 255.255.254.0 10000110 01010100 00000011 11111111 Broadcast Address 134.84.3.255 Available IP Addresses are 134.84.2.1 until 134.84.3.254 #8 Network Address Convert 8 to Binary =1000 10000110 01010100 00010000 00000000 Network address 134.84.16.0 11111111...
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...1.Find the CY and AC flags for each of the following: (a)MOV A, #3FHCY=0 No carry out from D7 ADD A, #45HAC=1 carry from D3 to D4 00111111 +01000101 10000100 (b)MOV A, #99HCY=0 No carry out from D7 ADD A, #58HAC=1 carry from D3 to D4 10011001 +01011000 11110001 (c)MOV A, #0FFHCY=1 SETB CAC=0 ADDC A, #00 SETB C-> CY=1 11111111 +CY 1 100000000 +00000000 100000000 (d)MOV A, #OFFHCY=1 ADD A, #1AC=1 11111111 + 1 100000000 (e)MOV A, #OFEHCY=1 SETB CAC=1 ADDC A, #01 11111110 + 1 11111111 + 1 100000000 (f)CLR CCY=1 MOV A, #0FFHAC=1 ADDC A, #01 ADDC A, #0 11111111 +00000001 100000000 +00000000 100000000 2. Write a program to add all the digits of your ID number and save the result in R3. The result must be in BCD. 3. Write a program to add the following numbers and save the result in R2, R3. The data is stored in on-chip ROM. ORG 250H MYDATA: DB 53, 94, 56, 92, 74, 65, 43, 23, 83 4. Modify Problem 3 to make the result in BCD. 6. State the steps that the SUBB instruction will go through for each of the following: 1. Take the 2’s complement of the subtrahend 2. Add it to the minuend 3. Invert the carry (a) 23H-12H 23H -> 00100011 00100011 12H -> 00010010 – 2’s complement+11101110 100010001 Step 3: CY=1 -> CF=0 Positive number 11H (b) 43H-53H 43H -> 01000011 01000011 53H -> 01010011 – 2’s complement+10101101 11110000 Step 3: CY=0 -> CF=1 Negative number -F0H (c) 99-99 99 -> 10011001 10011001 99 ->...
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...Week 1 Homework Chapter 0 1-Convert the following decimal numbers to binary: 12 = 00001100 123 = 01111011 63 = 00111111 128 = 10000000 1000 = 00000011 11101000 4-Convert the following hex numbers to binary and decimal: 2B9H = 001010111001, 697 F44H = 00001111 01000100, 3908 912H = 00001001 00010010, 2322 2BH = 00101011, 43 FFFFH = 11111111 11111111, 65535 6-Find the 2’s compliment of the following binary numbers: 1001010 = 110110 111001 = 111 10000010 = 1111110 111110001 = 1111 7-Add the following hex values: 2CH + 3FH = 6BH F34H + 5D6H = 150AH 20000H + 12FFH = 212FFH FFFFH + 2222H = 12221H 21-Answer the following: How many nibbles are 16 bits? 4 How many bytes are 32 bits? 4 If a word is defined in 16 bits, how many words is a 64-bit data item? 4 What is the exact value (in decimal) of 1 meg? 1,000,000 How many kilobytes is 1 meg? 1,000 What is the exact value (in decimal) of 1 gigabyte? 1,000,000,000 How many kilobytes is 1 gigabyte? 1,048,576 How many megs is 1 gigabyte? 1,024 If a given computer has a total of 8 megabytes of memory, how many bytes (in decimal) is this? 8388608 How many kilobytes is this? 8,192 Chapter 1 1-True or false. A general purpose microprocessor has on-chip ROM? False 2-True or false. A microprocessor has on-chip ROM? True 3-True or false. A microprocessor has on-chip I/O ports? True 4-True or false. A microprocessor has a fixed amount of RAM on the chip? True Chapter 2 1-In the 8051, looping action with...
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...Subnetting Made Simple IP Subnetting without Tables, Tools, or Tribulations Larry Newcomer The Pennsylvania State University York Campus Abstract Every networking professional should have a thorough understanding of TCP/IP subnetting. Subnetting can improve network performance by splitting up collision and broadcast domains. Subnets can reflect organizational structure and help support security policies. WAN links typically join different subnets. Subnets can define administrative units and hence support the structuring and delegation of administrative tasks. Unfortunately, mastering subnetting can pose difficulties for both professionals and students because of the binary mathematics that underlies the technology. While it is imperative to present subnetting concepts in terms of the underlying binary representation, most texts also present subnetting procedures in binary terms. Such an approach can make it difficult for students to learn how to actually carry out subnetting without tables or other reference materials, even when they understand the basic concepts. This paper presents a simple, alternative method for understanding and implementing subnetting without software, calculators, tables, or other aids. The only knowledge of binary arithmetic required is familiarity with the powers of 2 from 0 to 8 (2x for x = 0, 1, …, 8). With a little decimal arithmetic thrown in, the whole process is simple enough to be carried out mentally. This paper assumes the reader...
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...Subnetting Made Simple IP Subnetting without Tables, Tools, or Tribulations Larry Newcomer The Pennsylvania State University York Campus Abstract Every networking professional should have a thorough understanding of TCP/IP subnetting. Subnetting can improve network performance by splitting up collision and broadcast domains. Subnets can reflect organizational structure and help support security policies. WAN links typically join different subnets. Subnets can define administrative units and hence support the structuring and delegation of administrative tasks. Unfortunately, mastering subnetting can pose difficulties for both professionals and students because of the binary mathematics that underlies the technology. While it is imperative to present subnetting concepts in terms of the underlying binary representation, most texts also present subnetting procedures in binary terms. Such an approach can make it difficult for students to learn how to actually carry out subnetting without tables or other reference materials, even when they understand the basic concepts. This paper presents a simple, alternative method for understanding and implementing subnetting without software, calculators, tables, or other aids. The only knowledge of binary arithmetic required is familiarity with the powers of 2 from 0 to 8 (2x for x = 0, 1, …, 8). With a little decimal arithmetic thrown in, the whole process is simple enough to be carried out mentally. This paper assumes the reader...
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...Monday 1 Tuesday 2 Wednesday 3 Thursday 4 Friday 5 Saturday 6 Answer: Monday Sequential Inputs of numbers with 8 1 x 8 + 1 = 9 12 x 8 + 2 = 98 123 x 8 + 3 = 987 1234 x 8 + 4 = 9876 12345 x 8 + 5 = 98765 123456 x 8 + 6 = 987654 1234567 x 8 + 7 = 9876543 12345678 x 8 + 8 = 98765432 123456789 x 8 + 9 = 987654321 Sequential 1's with 9 1 x 9 + 2 = 11 12 x 9 + 3 = 111 123 x 9 + 4 = 1111 1234 x 9 + 5 = 11111 12345 x 9 + 6 = 111111 123456 x 9 + 7 = 1111111 1234567 x 9 + 8 = 11111111 12345678 x 9 + 9 = 111111111 123456789 x 9 + 10 = 1111111111 Sequential 8's with 9 9 x 9 + 7 = 88 98 x 9 + 6 = 888 987 x 9 + 5 = 8888 9876 x 9 + 4 = 88888 98765 x 9 + 3 = 888888 987654 x 9 + 2 = 8888888 9876543 x 9 + 1 = 88888888 98765432 x 9 + 0 = 888888888 Numeric Palindrome with 1's 1 x 1 = 1 11 x 11 = 121 111 x 111 = 12321 1111 x 1111 = 1234321 11111 x 11111 = 123454321 111111 x 111111 = 12345654321 1111111 x 1111111 = 1234567654321 11111111 x 11111111 = 123456787654321 111111111 x 111111111 = 12345678987654321...
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...subnets #1, #4, #5, and #46 Step 1: Convert the network address to binary 215.251.145.0/24 is a class C address with a subnet mask of 24 bits of ones First Octet 128.64.32.16.8.4.2.1, 128=1, 64=1, 32=0, 16=1, 8=0, 4=1, 2=1, 1=1 215 = 11010111. Second Octet = 128.64.32.16.8.4.2.1, 128=1, 64=1, 32=1, 16=1, 8=1, 4=0, 2=1, 1=1 251 = 11111011 Third Octet = 128.64.32.16.8.4.2.1, 128=1, 64=0, 32=0, 16=1, 8=0, 4=0, 2=0, 1=1 145 = 10010001 Fourth Octet = 0 11010111.11111011.10010001.00000000 Network address Step 2: Convert Subnet mask into binary First Octet 128.64.32.16.8.4.2.1, 128=1, 64=1, 32=1, 16=1, 8=1, 4=1, 2=1, 1=1 255 = 11111111 Second Octet 128.64.32.16.8.4.2.1, 128=1, 64=1, 32=1, 16=1, 8=1, 4=1, 2=1, 1=1 255 = 11111111 Third Octet 128.64.32.16.8.4.2.1, 128=1, 64=1, 32=1, 16=1, 8=1, 4=1, 2=1, 1=1 255 = 11111111 Forth Octet = 0 11111111.11111111.11111111.00000000 Subnet mask Step 3: Calculate the subnets required 46 subnets required per instructions: Use the formula 2 to the power of X to find out how many bits to borrow from the subnet mask. 2^5= 32 and is insufficient to cover the subnets required so, 2^6 = 64 x is 6 we need 6 bits from last octet of the subnet mask to cover our subnets. I need 46 subnets but I have total 64 available so that leaves me with; 64-46 = 18 extra subnets to use at a later date. Original Network address in binary: 11010111.11111011.10010001.00000000 Replace subnet bits with the subnet number 1, in binary is...
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...The Netmask When setting up each node with its IP address, the Netmask ask must also be specified. This mask is used to specify which part of the address is the network number part, and which is the host part. This is accomplished by a logical bitwise-AND between the Netmask ask and the IP address. The result specifies the network number. For Class C, the Netmask ask will always be 255.255.255.0; for Class B, the Netmask ask will always be 255.255.0.0; and so on. When A sent a packet to E in the last example, A knew that E wasn’t on its network segment by comparing A’s network number 200.1.2 to the value resulting from the bitwise-AND between the Netmask ask 255.255.255.0 and the IP address of E, 200.1.3.2, which is 200.1.3. The Netmask ask becomes very important, and more complicated, when “classless” addressing is used. Hierarchical Sub-Allocation of Class C Addresses To make more efficient use of Class C addresses in the Internet community, these addresses are subnetted hierarchically from the service provider to the organization. They are allocated bi™ ask-oriented subsets of the provider’s address space [4, 5]. These are classless addresses. Consider the following example of a small organization consisting of two Ethernet segments connecting to an Internet service provider using a WAN router that emulates an additional network segment, such as WANPIPE®;. The service provider has been allocated several different Class C addresses to be used for its clients. This particular...
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..._____________________________ ALWAYS ALBUM: MORE THAN LIFE ---------------------------------------------- Hillsongs United Tabbed By: Vhadz mORA www.vhadz101@yahoo.com Tuning: Standard Tuning INTRO: Bm-G-D-A (2x) Bm G D A e|--2----3------10-10---------*12--*9--*5---| B|---3----3---10------10---*12--------------| G|------------------------------------------| (2x) D|------------------------------------------| A|------------------------------------------| E|------------------------------------------| Verse: Bm Did You rise the sun for me? G D A Or paint a million stars that I might know Your majesty? Bm Is Your voice upon the wind? G D Is everything I've known marked with my Maker's A fingerprints? Pre-Chorus: G Bm ...
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...Lab 1.1 1.1.1 10⁰ 10⁰ 10¹ 10¹ 10² 10² 10³ 10³ 1 1 10 10 100 100 1000 1000 2931 2931 30 30 1 1 2000 2000 900 900 2 2 1 1 3 3 9 9 X X X X 1.1.2 2² 2¹ 2º 4 2 1 1 1 X X X 0 0 1 1 5 5 4 4 0 0 1 1 1.1.3 2² 2¹ 2º 4 2 1 0 0 X X X 1 1 1 1 3 3 0 0 2 2 1 1 1.1.4 24 2³ 2² 2¹ 2º 16 8 4 2 1 1 1 0 0 0 0 1 1 0 0 X X X X X 0 0 2 2 16 16 18 18 0 0 0 0 1.1.5 27 26 25 24 2³ 2² 2¹ 2º 128 64 32 16 8 4 2 1 1 1 1 1 1 1 0 0 0 0 0 0 1 1 0 0 X X X X X X X X 226 226 128 128 64 64 32 32 0 0 0 0 0 0 2 2 0 0 1.1.6 28 28 156 156 28 28 28 28 12 12 4 4 0 0 0 0 ˅ ˅ ˅ ˅ ˅ ˅ ˅ ˅ 27 26 25 24 2³ 2² 2¹ 2º 128 64 32 16 8 4 2 1 10011100 10011100 1 1 0 0 0 0 1 1 1 1 1 1 0 0 0 0 X X X X ...
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...1. Use what you’ve learned about the binary numbering in this chapter to convert the following decimal numbers to binary? * 11= 00001011 * 65= 01000001 * 100= 01100100 * 255= 11111111 2. Use what you’ve learned about the binary numbering system in this chapter to convert the following binary numbers to decimal? * 00001101=13 * 00001000=8 * 00101011=43 3. Look at the ASCII chart in Appendix A on CD that accompanies this book and determine the codes for each letter of your name? * 67 97 114 108 111 115 78 97 106 101 114 97 * C a r l o s N a j e r a 4. Use the web to research the history of the BASIC, C++, Java, and Python programming languages, and answer the following questions? * Who was the creator of each language? * When was each of these languages created? * Was there a specific motivation behind the creation of these languages? If so, what was it? * Basic was created by John George Kemeny. It was created on 1964. * C++ was created by Bjarne Stroustrup. It was created on 1979. * Java was created by Dr. James A. Gosling. It was created on 1995. * Python was created by Guido van Rossum. It was created on 1991. * The motivation behind the development of computer programming languages such as C ++, Basic, Java, and Python is that once written, a function may be used several times without rewriting it over and over. Programming languages can be used to...
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...One of Pitterno’s demands on joining SWU had been a new stadium. With attendance increasing, SWU administrators began to face the issue head-on. After 6 months of study, much political arm wrestling, and some serious financial analysis, Dr. Joel Wisner, president of Southwestern University, had reached a decision to expand the capacity at its on-campus stadium. * This integrated study runs throughout the text. Other issues facing Southwestern’s football expansion include (B) forecasting game attendance (Chapter 4); (C) quality of facilities (Chapter 6); (D) break-even analysis for food services (Supplement 7 Web site); (E) location of the new stadium (Chapter 8 Web site); (F) inventory planning of football programs (Chapter 12 Web site); and (G) scheduling of campus security officers/staff for game days (Chapter 13). Adding thousands of seats, including dozens of luxury skyboxes, would not please everyone. The influential Pitterno had argued the need for a first-class stadium, one with built-in dormitory rooms for his players and a palatial office appropriate for the coach of a future NCAA champion team. But the decision was made, and everyone, including the coach, would learn to live with it. TABLE 3.6 Southwestern University Project Time Estimates (days) ________________________________________ Activity Description Predecessor(s) Optimistic Most Likely Pessimistic Crash Cost/Day A Bonding, insurance, tax structuring — 20 30 40 $1,500 B Foundation, concrete footings...
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