...MOTHERBOARD#1 1. LGA 11156, All LGA 1156 processors and motherboards made to date are interoperable, making it possible to switch between a Celeron, Pentium, Core i3 or Core i5 with integrated graphics and a Core i5 or Core i7 without graphics. LGA 1156, also known as Socket H or H1, is an Intel desktop CPU socket. LGA stands for land grid array. That's where the CPU goes. 1156 is a socket type. The Intel H55 Chipset is used with the following processors; i7-800, i5, and the i3.Its purpose is to provide the interface for the PCI express lanes on a motherboard. The Gigabyte GA-H55M-UD2H can charge an iPhone, iPad, and an iPod touch even if the PC is on or off MOTHERBOARD#2 The AMD 770 Northbridge chipset is used with the AMD Phenom family processor. Its purpose is to link the hardware to the processor. The AMD SB710 Southbridge chipset is used with embedded AMD Athlon single and dual-core processors, it is also compatible with the quad-core Phenom. The Socket AM3 connector is used with the following processors; Phenom II, Athlon II, Sempron, and the Opteron 138x PROCESSOR CLOCK SPEED CACHE BUS SPEED ARCHITECTURE I7 I7 mobile I5 I5 mobile I3 I3 mobile vpro Vpro mobile Xeon 7000 Xeon 5000 Xeon 3000 Itanium 9000 pentium celeron TASK-3 #2 Intel E5300 is the processor number Pentium dual core means it has 2 cores 2.6 GHz is the clock speed 2M is the L2 cache 800 is the FSB speed The pins are the...
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...Jon Irfan NT1210 Lab_8 8.1.1: The network ID is 110. And the network host is 10.10.1. Binary value of the network ID is 192. And the binary value of the host is 21. 8.1.3: yes it would be. The first octet is 192 which would be in Class C. 8.1.4: The network ID of the IP address is 192.8. The network ID of host address is 8.4. The Binary network ID is 10111110.1000. And the binary host ID is 1000.100. 8.1.5: 255.255.255.0 = /24 11111111.11111111.11111111.00000000. total of 24 1’s. 255.128.0.0 =/9 11111111.00000000.00000000.00000000 = /9 total of 9 1’s. Block size | Starting IP address | Ending IP Address | Subnet Mask (slash notation | 24 bit block | 10.0.0.0 | 10.255.255.255 | 255.0.0.0 = /8 | 20 bit block | 172.16.0.0 | 17.31.255.255 | 255.255.0.0 = /16 | 16 bit block | 192.168.0.0 | 192.168.255.255 | 255.255.255.0 =/24 | 8.1 Review: The IP address is; 10.91.108.165. Subnet Musk is; 255.255.255.0. And the IP address is considered class C.Yes, it is a part of the private address block. 8.2.1: yes, the IP address works. No, the Ping command don’t reach the default gatway. 8.2 Review: using a static IP you can dial into the computer that is having the issue directly. 2. in large scale networking environments. So that you would not have to worry about setting up each individual machine and their IP setting. 8.4 1- AP 2 would be the most reliable. Outside of using AP3 it is in the middle...
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.... a) Virus – A malicious software that attaches itself to or copies itself into, another program for the purpose of causing the computer to follow instructure that were not by the original program developer. b) Worm – A malicious software programs that actively transmit themselves. Generally over networks to infect other computers. c) Spyware – A malicious software that covertly monitors and records pieces of information such as Web surfing activities and all data processed by the browser. d) Malicious code – Software that is designed to infiltrate a target computer and make it do something the attacker has instructed it to do. 2. An update should be done every day 3. Antivirus or something excepted to the program. And they assist to detecting. And by using cd the files are protected for any virus. 4. a) Internet e-mail-messages Shield - set your computer to update Viruses day or weekly and run a complete Virus Scan. b) Web Traffic Shield – Scan or update before using web browsing activities. c) P2P traffic Shield d) Network Traffic Shield 5) Yes they should be tested first before to fill out or put in public environment. 6) It depends on how often you run the scans vs your activity with the internet. Do a quick scan on a daily basis. Once a month I'll do a full scan. The quick scan should be good enough, though.... 7. 8. Yes 9. Home Page Hijacking, Pop up Advertisements, Crashes, firewall and antivirus programs are frequently turned off automatically. 10. what...
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...Lab #4 Operating Systems: Hardening and Security March 1st, 2015 Lab Assessment Questions & Answers 1. What is the one thing that a virus, a worm, spyware, and malicious code have in common? What are the differences among these three threats: a virus, a worm, and spyware? A virus, worm and spyware all have in common that they replicate themselves into the network’s system. Some of the differences between the three are that worms can edit files, viruses need a host file to propagate itself and a spyware can steal very sensitive information within the network. 2. How often should you update your antivirus protection? You should update your antivirus protection daily to always keep your network secured. 3. Why is it a best practice to have and to carry an antivirus boot-up disk or CD? It is best practice to use an antivirus boot-up or CD because it can bypass any files that are meant to protect the malware from being detected during scans. 4. In a corporate environment, should new antivirus definitions be installed as soon as they are available? Yes, in a corporate environment antivirus programs should be installed right away to protect the company’s files. 5. Is the manual quick scan good enough protection for a system? No a manual quick scan is not good enough protection for a system the files also have to be deleted. 6. What best practices for the workstation domain that can mitigate the risks and threats caused by malicious code...
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...Old code import java.util.Scanner; public class lab { /** Main method */ public static void main(String[] args) { // Create a Scanner Scanner input = new Scanner(System.in); // Get number of students System.out.print("Please enter number of students: "); int numberOfStudents = input.nextInt(); int[] scores = new int[numberOfStudents]; // Array scores int best = 0; // The best score char grade; // The grade // Read scores and find the best score for (int i = 0; i < scores.length; i++) { System.out.print("Please enter a score: "); scores[i] = input.nextInt(); if (scores[i] > best) best = scores[i]; } // Declare and initialize output string String output = ""; // Assign and display grades for (int i = 0; i < scores.length; i++) { if (scores[i] >= best - 10) grade = 'A'; else if (scores[i] >= best - 20) grade = 'B'; else if (scores[i] >= best - 30) grade = 'C'; else if (scores[i] >= best - 40) grade = 'D'; else grade = 'F'; output += "Student " + i + " score is " + scores[i] + " and grade is " + grade + "\n"; } // Display the result System.out.println(output); } } Corrected code import java.util.Scanner; public class Grades{ public static void main(String[] args) { // Create a Scanner Scanner input = new Scanner(System.in); // Get number of students System.out.print("Please enter number of students: "); int numberOfStudents = input.nextInt(); int[] scores = new int[numberOfStudents];...
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...740 words. Question 3: “Describe the function of the following Docker commands and briefly explain the purpose of each of the parameters. docker run –d –p 8080:5000 –v $(pwd)/data:/data –name container1 lab4 Docker is a software containerization platform, that runs processes in isolated containers. The docker run command is used to define the containers resources at runtime. When the command is run, the container process is isolated. It is separate from the host itself. The –d command is an option run by Docker to communicate whether the container should be run in “detached” mode or in the “default mode”. The –d command signifies that the container should be run in detached mode, meaning the container will be run in the background. The next option used is –p. This is used to publish all exposed ports to the host interfaces. Docker, in turn, ‘binds’ each of the exposed ports to a host port, addresses of the published ports are 8080:5000....
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...Unit 3 Lab4 Chris Hann Zy Powell 1. Dictionary attacks and rule-based search attacks. 2. Cross-site scripting allows for attackers to bypass client-side security mechanisms normally forced on web content by current browsers. 3. You can do this by disabling scripting when it is not necessary. Do not trust links to other sites on email or message boards, and do not follow links from sites that lead to security-sensitive information directly through its address. 4. The attacker can use redirect vulnerability, where a webpage uses a script to redirect the user somewhere other than the intended site. So then the hacker takes advantage of the script and you are sent to an external page instead of an internal one. 5. It can be denial of access, data loss or corruption, and unnecessary account privileges. 6. Blind SQL injection ask question that can only have a yes or no answer. Yet with normal SQL injection ask questions that will confuse the applications into returning answers in error message and. 7. Because XSS is a hacking technique in which a malicious user enters a short snippet of JavaScript into a textbox so that this script will be saved in the database. Therefore when a user retrieves and displays this later, the browser will execute the script. 8. When a user tries to enter their credentials, the url is explaining that the password is wrong. 9. By removing all unwanted input and accept only expected input, and...
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...tch Lab 4.4.2 VTP Configuration Challenge Topology [pic] Addressing Table |Device |Interface |IP Address |Subnet Mask |Default Gateway | |(Hostname) | | | | | |S2 |VLAN 99 |172.17.99.12 |255.255.255.0 |N/A | |S3 |VLAN 99 |172.17.99.13 |255.255.255.0 |N/A | |PC1 |NIC |172.17.10.1 |255.255.255.0 | | |PC2 |NIC |172.17.20.1 |255.255.255.0 | | |PC3 |NIC |172.17.30.1 |255.255.255.0 | | |PC4 |NIC |172.17.10.2 |255.255.255.0 | | |PC5 |NIC |172.17.20.2 |255.255.255.0 | | |PC6 |NIC |172.17.30.2 |255.255.255.0 | | Port Assignments...
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...http://jozefg.ecs.fullerton.edu/public/CS906/Assignment/ |Ass # |Text Chapter |ASSIGNMENT CPSC 906 FALL 2004 |Due |Max Points | | | |Section 1 | | | | | |Note: Please provide the program assignment documentation according to | | | | | |SyllabusCS901.doc and Project Submittals.doc. | | | |1 |Ch1 |Problems: 26 p.69. Unit conversion. |09-09 |2 | | |Ch2 |Problems: 4 p.153. Stacks separated | | | |2 |Ch2 - 3 |See the exercise assignment description below |09-21 |3 | | | | |Tue | | |3 |Ch2 |Problem: 39 p.156 - provide a Gantt chart for each case and calculate AWT,| 09-30 |3 | | | |ATT and ART. Problems 44. |Thu | | | | |Provide the solution in the PowerPoint slides...
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...This document is exclusive property of Cisco Systems, Inc. Permission is granted to print and copy this document for non-commercial distribution and exclusive use by instructors in the CCNA Exploration: LAN Switching and Wireless course as part of an official Cisco Networking Academy Program. PT Activity 1.2.4: Build a Hierarchical Topology Topology Diagram Learning Objectives • • Add devices to a topology. Connect the devices. Introduction Packet Tracer is integrated throughout this course. You must know how to navigate the Packet Tracer environment to complete this course. Use the tutorials if you need a review of Packet Tracer fundamentals. The tutorials are located in the Packet Tracer Help menu. This activity focuses on building a hierarchical topology, from the core to the distribution and access layers. All contents are Copyright © 1992–2007 Cisco Systems, Inc. All rights reserved. This document is Cisco Public Information. Page 1 of 3 CCNA Exploration LAN Switching and Wireless: LAN Design PT Activity 1.2.4: Build a Hierarchical Topology Task 1: Add Devices to the Topology Step 1. Add the missing distribution layer routers. The routers you need are located in Custom Made Devices. R1 and R3 are 1841 routers. Ctrl-click the 1841 router to add more than one. Press ESC to cancel. R2 is a 2621XM router. Step 2. Add the remaining access layer switches. Following the topology diagram, add nine 2960-24TT switches to complete the rest of the access...
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...3 711 Chapter Tax Accounting TRUE-FALSE QUESTIONSCHAPTER 13 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. A partnership may adopt any tax year without IRS permission. A corporation ling its rst return must annualize its income if the tax period is less than 12 months. A taxable year may be as short as one day and may exceed 366 days. Under no circumstances may a corporation change its scal year without IRS permission. A taxpayer engaged in two or more separate and distinct businesses may use different accounting methods for both businesses. A grocery store may use the cash basis of reporting sales. In general, a CPA on the cash basis method will never have a bad debt deduction. A cash basis taxpayer may deduct prepaid business expenses currently. Both cash and accrual basis taxpayers will be taxed on a dividend when it is actually received. Computing cost of goods soldand being on the accrual basis are independent of each other. If, in the IRSs opinion, the taxpayers books do not clearly reect income, the IRS may revise them so that they do. Taxpayers must generally obtain the permission of the IRS to change accounting methods. A correction of an error in a tax return is usually considered a change in accounting method. The IRS can require a change in accounting methods if the method used by a taxpayer does not clearly reect income. IRS permission is not required for a change from FIFO to LIFO. The installment method cannot be used unless the total selling price...
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