Discusses at least five emerging threats to information security..pdf

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Discusses at least five emerging threats to information security. Provides guidelines for creating strong passwords. Solution Information security means protecting information (data) and information systems from unauthorized access, use, disclosure, disruption, modification, or destruction. Information Security management is a process of defining the security controls in order to protect the information assets. 1.Any potential danger to information or systems. 2.A threat is a possibility that someone (person, s/w) would identify and exploit the vulnerability. 3.The entity that takes advantage of vulnerability is referred to as a threat agent. E.g.: A threat agent could be an intruder accessing the network through a port on the firewall. 4.Exploiting information via remote access software 5.Sharing sensitive files on P2P networks. 6.Careless use of wireless networks Use a unique password for each of your important accounts. Use a mix of letters, numbers, and symbols in your password. Don\'t use personal information or common words as a password..

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can someone proofread this for me? As a child, I used to watch T.V. all the time, and found myself fascinated by not only the stories but the movement of the characters. I would find myself watching different things the characters would do and wonder how they were able to make my favorite characters do all the crazy things. That curiosity continues to this day, and is able to bring me so much joy. To be able to bring things to life that I create is part of the reason that I love animation. The thought that I could actually go to school and learn something that I love did not occur to me right away. Right out of high school, I joined the Marine Corps, and spent much of my time in Okinawa, Japan. While my job did not pertain to anything with computers, I was surrounded by the creativity of the Japanese people. I was given a chance to go to Tokyo and explore the city. This only served to inspire me even more, and to decide that it was something that animation was something I wanted to do. Getting out of the Marines and getting back into school has been a bit of a rough start. Trying to relearn math and science has been a bit of a struggle, the only thing so far that I have exceeded greatly in, is the animation class here at the University of Iowa. I feel like while those classes are important, I don’t feel that some of my scores in classes such as chemistry are a good reflection of my dedication to my studies. My current animation teacher, Peter Chanthanakone, has been someone I could look up to, and has helped guide me so far. He helped encourage me to pursue this dream I have of becoming an animator. I greatly value getting an education because without it, I will never really be given a chance to work in a place that growing up brought me so much happiness. Solution Trying to relearn math and science has been a bit of a struggle,;the only thing so far that I have exceeded greatly in, is the animation class here at the University of Iowa My current animation teacher, Peter Chanthanakone, has been someone I could look up to, and has helped guide guided me so far. He helped encourage me to pursue this dream I have of becoming an animator. I greatly value getting an this education because without it..

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Describe the arrangement of Regulations for lifting machines and accessories in The Mineral Act 50 of 1991? Solution The purpose of this act is to regulate the prospecting for and the optimal expoitation , processing and utiisation of mineras ; to reguate the orderly utilization and the rehabiliation of the land surface during and after prospecting and mining operations. CHAPTER I INTRODUCTION CHAPTER II ADMINISTRATION CHAPTER III AUTHORIZATION TO PROSPECT AND TO MINE CHAPTER IV OPTIMAL EXPLOITATION AND UTILIZATION OF MINERALS CHAPTER V SAFETY AND HEALTH CHAPTER VI REHABILITATION OF SURFACE The rehabilitation of the surface of land in any prospecting or mining shall be carried out by the holder of the prospecting permit or mining authorization concerned in accordance with the environmental management programme or/and as an integral part of the prospecting or mining operations concerned. The holder of the prospecting permit or mining authorization concerned must submit an environmental management programme in respect of the surface of land concerned in any prospecting or mining operations or such intended operations. In order to limit any damage to or the disturbance of the surface, vegetation, environment or water sources to the minimum, the Director: Mineral Development may issue directives and determine conditions in relation to the use of the surface of land comprising the subject of any prospecting permit or mining authorization CHAPTER VII TRANSITIONAL PROVISIONS CHAPTER VIII GENERAL AND MISCELLANEOUS PROVISIONS.

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A nurse palpated enlarged lymph nodes. Describe signs and symptoms that help to distinguish cancerous lymph nodes from infected lymph nodes. Solution Cancerous lymph nodes VS infected lymph nodes: If the selling in the lymph node is temporary it is due to infection but if it lasts it and is sustained that biopsy and Contrast-enhanced ultrasonography can tell if the swelling i.e. enlargement is due to infection or cancer. The contrast-enhanced ultrasonography suggests metastatic lesions as defective shadows will be seen enlarged cancerous lymph nodes. In case of inflammation- induced enlargement contrast imaging shows; comparatively uniform staining but the staining is non-uniform in chronic-phase lymph nodes..

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Solve given LP problem using simplex method and find maximum value of objective function. Write all steps. max : = 4.71 + 6x2 s.t. -x1 + 2 11 12s 27 21 52 90 i 1.2 Solution the LPP problem according to Simplex method is given as follows: Let, S1, S2, and S3 are slack variables three constraints. The standard form of given LPP is given as follows: Max Z = 4X1 + 5X2 + 0S1 + 0S2 + 0S3 Subject To: -X1 + X2 + S1 + 0S2 + 0S3 = 11 X1 + X2 + 0S1 + S2 + 0S3 = 27 2X1 + 5X2 + 0S1 + 0S2 + S3 = 90 X1, X2, S1, S2, S3 >= 0 2) To form an initial basic solution, consider the structural variables as zero, in other words X1 = X2 = 0. Thus initial basic variables remains are S1, S2, and S3. Z = 0, S1 = 11, S2 = 27, and S2 = 90 3) The Initial Tableau is constructed as follows: Cj 4 5 0 0 0 B (Quantity) Cb Xb x1 x2 S1 S2 S3 0 S1 -1 1 1 0 0 11 0 S2 1 1 0 1 0 27 0 S3 2 5 0 0 1 90 Iteration 1: Cj 4 5 0 0 0 B (Quantity) RR Cb Xb x1 x2 S1 S2 S3 0 S1 -1 1 1 0 0 11 11/1 = 11 0 S2 1 1 0 1 0 27 27/1 = 27 0 S3 2 5 0 0 1 90 90/5 = 18 Zj 0 0 0 0 0 0 j = Cj – Zj 4 5 0 0 0 0 zj=cBj x xj j= (cj - zj) As the j is positive the optimal solution is not obtained. X2 is having highest j, thus incoming variable is X2. It realizes highest profit for next iteration. Pivot Column = X2, pivot elements are 1, 1, and 5. Determine replacement ratio by dividing B value by respective pivot column elements. The replacement ratio for basis variable S1 is lowest positive, this variable is outgoing variable. S1 is completely utilized if 11 units of X2 are to be produced. Pivot row = S1 Key element (KE) = 1 The elements of next variables are obtained as follows: Incoming Variable value for X1 is obtained as: Other basis variable is obtained as: 1st row/KE Old element – (Key column element x I.V. row element) Cj 4 5 0 0 0 B (Quantity) RR Cb Xb x1 x2 S1 S2 S3 5 X2 -1 1 1 0 0 11 11/-1 = -11 0 S2 1-(1*-1) =2 1-(1*1) = 0 0-(1*1) = - 1 1-(1*0) = 1 0-(1*0) = 0 27-(1*11) 16 16/2 = 8 0 S3 2-(5*-1) = 7 5-(5*1) = 0 0-(5*1) = -5 0-(5*0) = 0 1-(5*0) = 1 90-(5*11) = 35 35/7 = 5 Zj -5 5 5 0 0 55 j = Cj – Zj 9 0 -5 0 0 As the j is positive the optimal solution is not obtained. X1 is having highest j, thus incoming variable is X1. It realizes highest profit for next iteration. Pivot Column = X1, pivot elements are -1, 2, and 7. The replacement ratio for basis variable S3 is lowest positive, this variable is outgoing variable. S3 is completely utilized if 5 units of X1 are to be produced. Pivot row = S3 Key element (KE) = 7 Cj 4 5 0 0 0 B (Quantity) RR Cb Xb x1 x2 S1 S2 S3 5 X2 -1-(-1*1) = 0 1-(-1*0) = 1 1-(-1*-5/7) = 2/7 0 0-(-1*1/7) = 1/7 11-(-1*5) = 16 16/(2/7) = 56 0 S2 2-(2*1) = 0 0-(2*0) = 0 - 1-(2*-5/7) = 3/7 1 0-(2*1/7) = -2/7 16-(2*5) = 6 16/(3/7) = 37.33 4 X1 7/7 = 1 0/7 = 0 -5/7 0 1/7 35/7 = 5 5/(-5/7) = -7 Zj 4 5 -10/7 0 9/7 (5*16+4*5) = 100 j = Cj – Zj 0 0 10/7 0 -9/7 As the j is positive the optimal solution is not obtained. S1 is having highest j, thus incoming variable is S1. It realizes highest profi.

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