1. Create a directory structure IFT383FinalExamActivitiesActivity1.pdf

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1. Create a directory structure IFT383FinalExam/Activities/Activity1 in your home directory 2. Create a file using the cat command named classRoster 3. Add three records to your file with the following fields, separated by a comma Student ID First Name Last Name Grade Program of Study ASURITE ID (username) 4. Display the contents of the file classRoster Move the file classRoster to the directory Activity1 5. Go to the Activity1 directory Display the directory you are in 6. Add read, write and execute permissions for user and group to Activity1.dir 7. Create a new file called header using cat command whose contents are the field names, separated by a tab space Student ID First Name Last Name Grade Program of Study ASURITE ID (username) 8. Display the header file Copy the contents of header into classRoster, as the first line Replace the comma with a tab space using appropriate filters (such as cut, paste, tr, head, etc.) 9. Save this file as finalRoster.txt Display the contents of the file finalRoster.txt Convert the finalRoster to a UNIX file and save as rosterUNIX, 10. perform a byte-by-byte comparison of the two files, and display the differences. Solution 1) cd ~ mkdir -p IFT383FinalExam/Activities/Activity1 2) cat > classRoster 3) 1,bill,gates,cse,bgates 2,larry,page,cse,lpage 3,elon,musk,ece,emusk 4) cat classRoster 1,bill,gates,cse,bgates 2,larry,page,cse,lpage 3,elon,musk,ece,emusk mv classRoster IFT383FinalExam/Activities/Activity1/ 5) cd IFT383FinalExam/Activities/Activity1 pwd /home/vkona/IFT383FinalExam/Activities/Activity1 6) chmod -R 771 ../Activity1 ls -l ../Activity1/ total 4 -rwxrwx--x 1 vkona vkona 69 Mar 1 16:35 classRoster 7) cat >header 1 bill gates cse bgates 2 larry page cse lpage 3 elon musk ece emusk cat header 1 bill gates cse bgates 2 larry page cse lpage 3 elon musk ece emusk 8) cat header >> classRostercat header >> classRoster cat classRoster 1,bill,gates,cse,bgates 2,larry,page,cse,lpage 3,elon,musk,ece,emusk 1 bill gates cse bgates 2 larry page cse lpage 3 elon musk ece emusk.

Education

/home/vkona/IFT383FinalExam/Activities/Activity1
6) chmod -R 771 ../Activity1
ls -l ../Activity1/
total 4
-rwxrwx--x 1 vkona vkona 69 Mar 1 16:35 classRoster
7) cat >header
1 bill gates cse bgates
2 larry page cse lpage
3 elon musk ece emusk
cat header
1 bill gates cse bgates
2 larry page cse lpage
3 elon musk ece emusk
8) cat header >> classRostercat header >> classRoster
cat classRoster
1,bill,gates,cse,bgates
2,larry,page,cse,lpage
3,elon,musk,ece,emusk
1 bill gates cse bgates
2 larry page cse lpage
3 elon musk ece emusk

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All answers must be in your own wordsProvide a good, understandabl.pdf

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C programming tweak needed for a specific program. This is the complete program up and running. except for one modification that my teacher is requiring but it is confusing me. The program needs to allocate disjoint memory and use that. right now im allocating a continous block of memory on the heap but I need it to do disjoing memory instead. // Purpose: Program will allow a reseller to save their sales and calculate their profits. // Also able to view highest profitable items and remove any sales they no longer want in their sales inventory. #define _CRT_SECURE_NO_WARNINGS #include #include #include #define PAUSE system(\"pause\") #define CLS system(\"cls\"); //#define FLUSH fflush(stdin); //////////////// SRUCTS BELOW ////////////////////////// typedef struct { char itemName[40]; float purchasePrice; float soldPrice; float shippingPrice; float ebayFee; float payPalFee; float profit; }SALES; ////////// FUNCTION PROTOTYPE BELOW ///////////// void addSale(SALES *Inventory, int *InventoryCounter); void displayMenu(int *InventoryCounter); void displayProfit(int *InventoryCounter, SALES *Inventory); void displaySales(int *InventoryCounter, SALES *Inventory); int getChoice(int *InventoryCounter); void searchSale(SALES* Inventory, int InventoryCounter); void sortProfits(SALES* Inventory, int InventoryCounter); void removeSale(SALES* Inventory, int *InventoryCounter); main() { int SIZE; int userChoice; int InventoryCounter = 0; //printf(\"*********************\ \"); //printf(\"* Welcome Reseller! *\ \"); //printf(\"*********************\ \"); printf(\"Enter the size of todays inventory:\"); scanf(\"%i\", &SIZE); SALES *Inventory; Inventory = calloc(SIZE, sizeof(SALES));//Using dynamic memory to allocate space on the heap do { userChoice = getChoice(&InventoryCounter); switch (userChoice) { case 1:// Add a Sale if (InventoryCounter == SIZE) // statement checks so we dont exceed the size of the inventory user wants to enter. { printf(\"You Have Reached Max Inventory \ \"); PAUSE; } else { addSale(&(Inventory[InventoryCounter]), &InventoryCounter); } break; case 2: // Display all sales records displaySales(&InventoryCounter, Inventory); PAUSE; break; case 3: // Display all profits sortProfits(Inventory, InventoryCounter); displayProfit(&InventoryCounter, Inventory); PAUSE; break; case 4: // Display Inventory searchSale(Inventory, InventoryCounter); break; case 5:// remove a sale removeSale(Inventory, &InventoryCounter); break; case 6: // Quit break; default: printf(\"Please choose a valid option\ \"); PAUSE; break; }// end of switch } while (userChoice != 6); free(Inventory); } // End of main //////////////////////////////////////////// FUNCTIONS BELOW ////////////////////////////////////////////////// void addSale(SALES *Inventory, int *InventoryCounter)// Function that adds sale and calculates the fees in the background. { float payPalFee = 0; // variable to store the paypal fee from 1 sale float ebayFee = 0; // variable to store ebay fee for 1 sale.

C programming tweak needed for a specific program.This is the comp.pdf

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Add to BST.java a method height() that computes the height of the tree. Develop two implementations: a recursive method (which takes linear time and space proportional to the height), and method like size() that adds a field to each node in the tree (and takes linear space and constant time per query) Solution Answer: /*BST.java a method height() that computes the height of the tree. Developing two implementations: a recursive method, and method like size() that adds a field to each node in the tree **/ public class SplayBST, Value> { private Node root; // root of the BST // BST helper node data type private class Node { private Key key; // key private Value value; // associated data private Node left, right; // left and right subtrees public Node(Key key, Value value) { this.key = key; this.value = value; } } public boolean contains(Key key) { return get(key) != null; } // return value associated with the given key // if no such value, return null public Value get(Key key) { root = splay(root, key); int cmp = key.compareTo(root.key); if (cmp == 0) return root.value; else return null; } /* * Splay tree insertion. **/ public void put(Key key, Value value) { // splay key to root if (root == null) { root = new Node(key, value); return; } root = splay(root, key); int cmp = key.compareTo(root.key); // Insert new node at root if (cmp < 0) { Node n = new Node(key, value); n.left = root.left; n.right = root; root.left = null; root = n; } // Insert new node at root else if (cmp > 0) { Node n = new Node(key, value); n.right = root.right; n.left = root; root.right = null; root = n; } // It was a duplicate key. Simply replace the value else { root.value = value; } } /* * Splay tree deletion. **/ /* *This splays the key, then does a slightly modified Hibbard deletion on the root (if it is the node to be deleted, if it is not, the key was not in the tree). *The modification is that rather than swapping the root (call it node A) with its successor, it\'s successor (call it Node B) is moved to the root position *by splaying for the deletion key in A\'s right subtree. Finally, A\'s right child is made the new root\'s right child. */ public void remove(Key key) { if (root == null) return; // empty tree root = splay(root, key); int cmp = key.compareTo(root.key); if (cmp == 0) { if (root.left == null) { root = root.right; } else { Node x = root.right; root = root.left; splay(root, key); root.right = x; } } // else: it wasn\'t in the tree to remove } /* * Splay tree function. **/ // splay key in the tree rooted at Node h. If a node with that key exists, // it is splayed to the root of the tree. If it does not, the last node // along the search path for the key is splayed to the root. private Node splay(Node h, Key key) { if (h == null) return null; int cmp1 = key.compareTo(h.key); if (cmp1 < 0) { // key not in tree, so we\'re done if (h.left == null) { return h; } int cmp2 = key.compareTo(h.left.key); if (cmp2 < 0) { h.left.left = splay(h.left.left, key); h = rot.

Add to BST.java a method height() that computes the height of the tr.pdf

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Write an awareness objective for a newly formed adolescent chemical dependency program whose target market consists of judeges and social workers who refer to the facility. How would this objective change for the trial level of the hierarchy of effects? Solution Answer: The objectives are: 1. To create primary level awareness to the people who uses this service frequently and also those who refer our facility, 2. To find out the ways to help people in different situations and making an sustainable ecosystem. This objective (twin objectives) would become altered in case of implementation of them in real situation..

Write an awareness objective for a newly formed adolescent chemical .pdf

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which of these prokaryotes are most likely to be found in the immedi.pdf

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2. Why is only one end point observed for citric acid even though it .pdf

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10. Benefits and costs of International Trade : Search for a newspaper article from any English newspaper that discusses the advantages or disadvantages of international trade. Copy and paste the weblink into this Forum and discuss one or two main points of the article. Please give detailed answer with examples and elaborations. Solution The web is a awesome invention. Since of its introduction into schooling and its availability to just about any individual, the world will mainly never be the identical. Persons of all a while can go surfing 24/7 looking for something they want. The elderly woman who\'s unable to drive can log onto the web and shop for groceries, fill a prescription or even chat together with her grandchildren via web cam. Collectors now not have got to scour flea markets and yard revenue on weekends to find coveted gadgets so as to add to their collections all they need to do is go browsing to Ebay and publish a bid. Something can be purchased on line now. Garb, jewellery, furniture...Even take-out. As a instructor i\'ve discovered the web to be extremely valuable in planning classes. The web has a plethora of websites created in particular for lecturers to share suggestions, change information, and share records (thanks enotes!). Probably young people stand to gain the most from making use of the web. Additionally to being in a position to express themselves and show creativity on websites like MySpace and Twitter, they can also use the web for school tasks and homework. Still, for every pro, there is a con. I feel that scholars regularly count too much in the internet. They do not look to have an understanding of that simply considering the fact that some thing is on Google, that does not make it a riskless supply. I are not able to depend the number of occasions a scholar has became in an essay that includes unsuitable data from web sites which were created with the aid of any person who is not an authority on the topic. Chat rooms have emerge as the brand new playground for sexual predators. The web gives them the anonymity they have got to entice unsuspecting youngsters into inappropriate conversations and in some circumstances meeting up with them. There are laws that preserve predators far from parks and schools, however none retaining them off the web. With out suitable supervision, kids can accidentally go online to pornographic web pages. Even adults can fall prey to identity theft and stolen passwords if they don\'t seem to be careful about the matters they are downloading onto their laptop. Advantages and drawbacks of web use amongst kids and adults can be considered from quite a lot of angles - objectively, or subjectively. If a individual is in favor of science, then he/she would find extra advantages than negative aspects for its use. However, if a man or woman believes in guide work and that technology could one day be used to replace human work, then the he/she may just view web use to be more damaging than .

10. Benefits and costs of International Trade Search for a newspap.pdf

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Why does the pattern in a shift register shift only one bit position each time it is clocked? please no hand written answers Solution Beacuse if we move all bits at a time there will be no buffers to store the bits temporarily and then shift the data to next position for each bit. Hence , first Least significant bit(LSB) is moved out the the others follow for each step upto the MSB most significant bit is moved by one bit position..

Why does the pattern in a shift register shift only one bit position.pdf

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Use the Internet to identify three network firewalls, and create a table that compares their features. Note if they are rule-based or settings-based, perform stateless or stateful packet filtering, what additional features they include (IDS, content filtering, and so on), their costs, and so on. Solution Below is the table : Firewall Capability Packet Filters Application-layer Gateways Stateful Inspection Communication Information Partial Partial Yes Communication-derived State No Partial Yes Application-derived State No Yes Yes Information Manipulation Partial Yes Yes If you take packet filters, for example, historically they are implemented on routers, are filters on user defined content, such as IP addresses. packet filters examine a packet at the network layer and are application independent, which allows them to deliver good performance and scaleability. However, they are the least secure type of firewall, especially when filtering services such as FTP. As with application gateways, he security is improved by examining all application layers, bringing context information into the decision process. However, they do this by breaking the client/server model. Every client/server communication requires two connections: one from the client to the firewall and one from the firewall to the server. In addition, each proxy requires a different application process, or daemon, making scaleability and support for new applications a problem. The Stateful Inspection introduced by Check Point overcomes the limitations of the previous two approaches by providing full application-layer awareness without breaking the client/server model. With Stateful Inspection, the packet is intercepted at the network layer. It extracts state- related information required for the security decision from all application layers and maintains this information in dynamic state tables for evaluating subsequent connection attempts. This provides a solution which is highly secure and offers maximum performance, scaleability, and extensibility. Firewall Capability Packet Filters Application-layer Gateways Stateful Inspection Communication Information Partial Partial Yes Communication-derived State No Partial Yes Application-derived State No Yes Yes Information Manipulation Partial Yes Yes.

Use the Internet to identify three network firewalls, and create a t.pdf

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