1) What is ID file? 2) Which is the common backup file? 3) what is desktop6.NDK? 4) What is names.nsf? 5) What is bookmark.nsf? 6) What is the difference between lotus notes 7 and 8.....? 7) What is replica? 8) What is the difference between Server and local? ANS 1) Id file contain the information of user name and password and when his id and certificate will expire. 2) 1) desktop6.NDK 2) bookmark.NSF 3) Names.NSF 4) ID FILE 5) Notes.ini 3) It contain all the information
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FI 512 Week 1 Answer Key Chapter 1 1. [Financing Concepts] The following ventures are at different stages in their life cycles. Identify the likely stage for each venture and describe the type of financing each venture is likely to be seeking and identify potential sources for that financing. A. Phil Young‚ founder of Pedal Pushers‚ has an idea for a pedal replacement for children’s bicycles. The Pedal Pusher will replace existing bicycle pedals with an easy release stirrup to help
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tables are available in 1 inch increments from 24 inches to 96 inches and surface depths starting at 18 inches. Tops can be shaped however you need them- including (but not limited to) designing around computer equipment or with cockpits to improve the user reach envelope. A 1 PICK FRAME - specify 2‚ 3 or 4 leg model straight‚ 90°‚ or 120° corner configuration 3-leg 90° corner D B 2 B SPECIFY DIMENSIONS - length‚ width and depths for all sides - choose foot lengths 30” 3 4 24” 17” DESIGN
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Tasks three and four Deadline: 15/1/13 Recommending a PC A family friend has asked for your advice on selecting a computer for their two children. Ben is a games enthusiast and enjoys playing all the latest computer games. Jerry‚ on the other hand‚ is the creative one and enjoys making videos and will want to edit them on the new computer. He also enjoys digital photography and takes a lot of pictures with his camera and edits them. They would also like to be able to connect the PC to the new
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BUS650:Week 3 Assignment 1 Jocquetta D. Harris BUS 650: managerial Finance (NAJ1240A) Martin Cain October 22‚ 2012 Complete Chapter 12 Closing Case at the end of the chapter and submit answers to your instructor. 1. Most publicly traded corporations are required to submit 10Q (quarterly) and 10K (annual) reports to the SEC detailing their financial operations over the
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15-441: Computer Networks Homework 1 Assigned: Sep 2‚ 2011 Due: Sep 15‚ 2011 1:30 PM in class Lead TAs: Athula Balachandran Wolf Richter 1 1 1 1 Byte KByte Mbps GHz Units 8 bits 210 bytes 106 bits per second 109 Hz 1. [Sec 1.5] Calculate the total time required to transfer a 1500 KByte file in the following cases‚ assuming a RTT of 10ms‚ a packet size of 1500 bytes‚ and an initial 3 RTT of handshaking before the actual data is sent. (a) The bandwidth is 10 Mbps and data packets can
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Mechanism 4 shows the debromination of 2α-bromocholestan-3-one to give 4. This mechanism which is similar to a Favoskii rearrangement‚ is likely to go via this path as other research has shown that there is often a by-product with this reaction. The by-product is cholest-4-en-3-one which requires the intermediate seen in the third molecule in the above mechanism. The intermediate is highly strained and with the presence of base a proton will be extracted which aids the collapse of the three membered
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and your ability to edit or enhance them after importing them to your computer. See a demonstration of the most common photo editing tasks‚ and discover more about the tools you can use to enhance your photos. This activity contains 13 questions. 1. Which of the following is NOT an example of image editing software? a. Paint Shop Pro b. PhotoSuite c. Adobe Photoshop d. Microsoft Expression 2. Which of the following is a free application supplied with Windows 8? a. Window Elements b. Windows Photo
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5% Used dent and ends/dent : 90/3 dent 1) Cover Factor Ccover = C1 + C2 – C1 . C 35 Tex = 16.87 Ne C1 = d1 x n1 C2 = d2 x n2 D1 and D2 = 1/ 28. (Ne)^1/2 (inch) n1 = 30 ends/cm D1 and D2 = 0.0220 cm n2 = 24 picks/cm C1 = 0.0220 x 30 = 0.662 C2 = 0.0220 x 24 = 0.528 Ccover = (0.662 + 0.528) – ( 0.662 x 0.528) Ccover = % 85.2 2) Basis Weight Warp Weight = [30 x 100 x (1 + 0.08)]/ 16.87 = 192‚05 g/m^2 Weft Weight = [24 x 100 x (1 + 0.05)]/ 16.87 = 149. 377 g/m^2
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In order to do so‚ the packet analyzer must “understand” the structure of all messages exchanged by protocols. For example‚ suppose we are interested in displaying the various fields in messages exchanged by the HTTP protocol in Figure 1. The packet analyzer understands the format of Ethernet frames‚ and so can identify the IP datagram within an Ethernet frame. It also understands the IP datagram format‚ so that it can extract the TCP segment within the IP datagram. Finally‚ it understands
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