P(Z-0-86).docx
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La probabilidad de que una variable aleatoria Z sea mayor que 0.86.
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Part3- Offline traffic monitoring In this part will use a PCAP file to.docx
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Participation is a requirement of this course. Students provide one post introducing their key concept from the chapter(s) for the week in 200 words or less, with references to the textbook and two other supporting materials in APA format (e.g., Nelson & Quick, 2018, p. 17).
Provide a thoughtful and meaningful approach to posts and ensure quality contributions:
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Partners Johnny and Daniel each have capital balances of $95-000- Migu.docx
Partners Johnny and Daniel each have capital balances of $95 , 000 . Miguel invests $110 , 000 for a 1/3 interest in the partnership. How much of Miguel's investment is credited to his capital account? $300 , 000 $100 , 000 $95 , 000 $110 , 000 Millie and David have capital accounts totalling $100 , 000 and $150 , 000 respectively in a partnership at the beginning of the year. Regarding the division of net income (losses), the partnership agreement states that each partner earns 10% interest on their capital account balance and gives salary allowances of $25 , 000 to Millie and $10 , 000 to David, with the remaining income (loss) being shared equally. How much of a net income of $80 , 000 will be allocated to Millie? $45 , 000 $40 , 000 $25 , 000 $35 , 000 Elliot makes $10 per hour. He worked 42 hours last week. According to FLSA, how much should his gross pay be? $420 $630 $400 $430
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Parthians- write one to two paragraphs on the significance of this ite.docx
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Explain how this item exemplified some trend or movement or ideology.
Explain how this item ties into one of our class themes.
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package edu-ser222-m03_02- --- - A binary search tree based impleme.docx
package edu.ser222.m03_02;
/**
* A binary search tree based implementation of a symbol table.
*
* Completion time: (your completion time)
*
* @author (your name), Sedgewick, Acuna
* @version (version)
*/
import java.util.Collections;
import java.util.LinkedList;
import java.util.NoSuchElementException;
import java.util.Queue;
public class CompletedBST<Key extends Comparable<Key>, Value> implements BST<Key, Value> {
private Node<Key, Value> root;
@Override
public int size() {
return size(root);
}
private int size(Node x) {
if (x == null)
return 0;
else
return x.N;
}
@Override
public Value get(Key key) {
Node<Key, Value> iter = root;
while(iter != null) {
int cmp = key.compareTo(iter.key);
if (cmp < 0)
iter = iter.left;
else if (cmp > 0)
iter = iter.right;
else
return iter.val;
}
return null;
}
private Value get(Node<Key, Value> x, Key key) {
// Return value associated with key in the subtree rooted at x;
// return null if key not present in subtree rooted at x.
if (x == null) return null;
int cmp = key.compareTo(x.key);
if (cmp < 0) return get(x.left, key);
else if (cmp > 0) return get(x.right, key);
else return x.val;
}
@Override
public void put(Key key, Value val) {
root = put(root, key, val);
}
private Node put(Node<Key, Value> x, Key key, Value val) {
if (x == null)
return new Node(key, val, 1);
int cmp = key.compareTo(x.key);
if (cmp < 0)
x.left = put(x.left, key, val);
else if (cmp > 0)
x.right = put(x.right, key, val);
else
x.val = val;
x.N = size(x.left) + size(x.right) + 1;
return x;
}
@Override
public Key min() {
if(root == null)
throw new NoSuchElementException();
return min(root).key;
}
private Node<Key, Value> min(Node x) {
if (x.left == null)
return x;
return min(x.left);
}
@Override
public Key max() {
if(root == null)
throw new NoSuchElementException();
return max(root).key;
}
private Node<Key, Value> max(Node x) {
if (x.right == null) return x;
return max(x.right);
}
@Override
public Key floor(Key key) {
if(root == null)
throw new NoSuchElementException();
Node<Key, Value> x = floor(root, key);
if (x == null)
return null;
return x.key;
}
private Node<Key, Value> floor(Node<Key, Value> x, Key key) {
if (x == null)
return null;
int cmp = key.compareTo(x.key);
if (cmp == 0) return x;
if (cmp < 0) return floor(x.left, key);
Node<Key, Value> t = floor(x.right, key);
if (t != null) return t;
else return x;
}
@Override
public Key select(int k) {
return select(root, k).key;
}
private Node<Key, Value> select(Node x, int k) {
if (x == null) return null;
int t = size(x.left);
if (t > k) return select(x.left, k);
else if (t < k) return select(x.right, k-t-1);
else return x;
}
@Override
public int rank(Key key) {
return rank(key, root);
}
private int rank(Key key, Node<Key, Value> x) {
// Return number of keys less than x.key in the subtree rooted at x.
if (x == null) return 0;
int cmp = key.compareTo(x.key);
if (cmp < 0) return rank(key, x.left);
else if (cmp > 0) return 1 + size(x.left) + rank(key, x.right);
else return size(x.left);
}
@O.
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Part3: Offline traffic monitoring In this part will use a PCAP file to examine the network traffic (offline traffic monitoring) using three IDS tools - Snort - Suricata - Zeek The PCAP file with the name part3.pcap (attached with the project files) containing a captured network traffic for some company having a web site and 256 addresses in the range (192.168.6.0 - 192.168.6.255). The IP address for the network gateway (edge router) is ( 192.168.6.1 ) . The captured network traffic is for some period of time when an attacker started to perform intrusion against the company network. The suspicious address is 192.168.5.55, which conducted a multi-stage attack starting with reconnaissance. A kill chain representing multi-stage attack is a systematic process to target and engage an attacker to perform the desired attack. The following steps are the typical stages followed by any professional attacker. 1. Reconnaissance - Research, identification and selection of targets, often represented as crawling Internet websites such as conference proceedings and mailing lists for email addresses, social relationships, or information on specific technologies. 2. Weaponization - Coupling a remote access trojan with an exploit into a deliverable payload, typically by means of an automated tool (weaponizer). Increasingly, client application data files such as Adobe Portable Document Format (PDF) or Microsoft Office documents serve as the weaponized deliverable. 3. Delivery - Transmission of the weapon to the targeted environment. The three most prevalent delivery vectors for weaponized payloads by APT actors, as observed by the Lockheed Martin Computer Incident Response Team (LM-CIRT) for the years 2004-2010, are email attachments, websites, and USB removable media. 4. Exploitation - After the weapon is delivered to victim host, exploitation triggers intruders' code. Most often, exploitation targets an application or operating system vulnerability, but it could also more simply exploit the users themselves or leverage an operating system feature that auto-executes code. 5. Installation - Installation of a remote access trojan or backdoor on the victim system allows the adversary to maintain persistence inside the environment. 6. Command and Control (C2) - Typically, compromised hosts must beacon outbound to an Internet controller server to establish a C2 channel. APT malware especially requires manual interaction rather than conduct activity automatically. Once the C 2 channel establishes, intruders have "hands on the keyboard" access inside the target environment. 7. Actions on Objectives - Only now, after progressing through the first six phases, can intruders take actions to achieve their original objectives. Typically, this objective is data exfiltration which involves collecting, encrypting and extracting information from the victim environment; violations of data integrity or availability are potential objectives as well. Alternatively, the intruders may.
Participation is a requirement of this course- Students provide one po.docx
Participation is a requirement of this course. Students provide one post introducing their key concept from the chapter(s) for the week in 200 words or less, with references to the textbook and two other supporting materials in APA format (e.g., Nelson & Quick, 2018, p. 17).
Provide a thoughtful and meaningful approach to posts and ensure quality contributions:
Stay on topic, and ensure thoughts and ideas are logically constructed,
Show respect and inclusion, and
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Partners Johnny and Daniel each have capital balances of $95 , 000 . Miguel invests $110 , 000 for a 1/3 interest in the partnership. How much of Miguel's investment is credited to his capital account? $300 , 000 $100 , 000 $95 , 000 $110 , 000 Millie and David have capital accounts totalling $100 , 000 and $150 , 000 respectively in a partnership at the beginning of the year. Regarding the division of net income (losses), the partnership agreement states that each partner earns 10% interest on their capital account balance and gives salary allowances of $25 , 000 to Millie and $10 , 000 to David, with the remaining income (loss) being shared equally. How much of a net income of $80 , 000 will be allocated to Millie? $45 , 000 $40 , 000 $25 , 000 $35 , 000 Elliot makes $10 per hour. He worked 42 hours last week. According to FLSA, how much should his gross pay be? $420 $630 $400 $430
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Explain how this item is a cause or a consequence of some event.
Explain how this item exemplified some trend or movement or ideology.
Explain how this item ties into one of our class themes.
Explain any other way this item is significant.
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The document contains a single mathematical expression: P ( X = x ), which represents the probability of a random variable X taking on a particular value x. This expression calculates the likelihood that the outcome of a random phenomenon or experiment will be a specific number or result. The document focuses on defining and representing this basic probability concept using mathematical notation.
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Overview You are required to prepare a Data Entry report which provides a full data sheet (using FinSas spread sheets) for a specific non-financial company of your choice using the most recent five years of annual reports / data available. The selected annual reports/data must: 1. Be written in English. 2. Be obtained from any credible source such as Datastream, Bloomberg, Thomson One, the company's website, etc. 3. You can use any type of additional data to support your analysis, taking into consideration its credibility and appropriate referencing. 4. Use accounting rules and standards (IFRS/US GAAP) consistent to the ones used for the competitor company used as your benchmark. Important Notes: 1. You will choose your company via Learn Ultra on a first-come first-served basis and you cannot select the same company as another student. You do not need to register your choice of a competitor company. 2. You must fully check the accessibility of annual reports and data before choosing your company. 3. Calculations for the different ratios and vertical/horizontal analysis are required to be done using the FINSAS software. The output of FINSAS should be attached to your report in the appendices section. 4. You must not depend on the disclosed ratios in the annual reports. You are required to prepare these ratios on your own with the assistance of FINSAS. 5. Graphical presentation of your findings can help in your evaluation. Tables in the body of the work should present key information that you refer to in the text and be clearly set out. 6. The module leader needs to be informed and to approve the company you'll be choosing for your project, using the available Google sheet available on Learn Uitra.
.
package edu-ser222-m03_02- --- - A binary search tree based impleme.docx
package edu.ser222.m03_02;
/**
* A binary search tree based implementation of a symbol table.
*
* Completion time: (your completion time)
*
* @author (your name), Sedgewick, Acuna
* @version (version)
*/
import java.util.Collections;
import java.util.LinkedList;
import java.util.NoSuchElementException;
import java.util.Queue;
public class CompletedBST<Key extends Comparable<Key>, Value> implements BST<Key, Value> {
private Node<Key, Value> root;
@Override
public int size() {
return size(root);
}
private int size(Node x) {
if (x == null)
return 0;
else
return x.N;
}
@Override
public Value get(Key key) {
Node<Key, Value> iter = root;
while(iter != null) {
int cmp = key.compareTo(iter.key);
if (cmp < 0)
iter = iter.left;
else if (cmp > 0)
iter = iter.right;
else
return iter.val;
}
return null;
}
private Value get(Node<Key, Value> x, Key key) {
// Return value associated with key in the subtree rooted at x;
// return null if key not present in subtree rooted at x.
if (x == null) return null;
int cmp = key.compareTo(x.key);
if (cmp < 0) return get(x.left, key);
else if (cmp > 0) return get(x.right, key);
else return x.val;
}
@Override
public void put(Key key, Value val) {
root = put(root, key, val);
}
private Node put(Node<Key, Value> x, Key key, Value val) {
if (x == null)
return new Node(key, val, 1);
int cmp = key.compareTo(x.key);
if (cmp < 0)
x.left = put(x.left, key, val);
else if (cmp > 0)
x.right = put(x.right, key, val);
else
x.val = val;
x.N = size(x.left) + size(x.right) + 1;
return x;
}
@Override
public Key min() {
if(root == null)
throw new NoSuchElementException();
return min(root).key;
}
private Node<Key, Value> min(Node x) {
if (x.left == null)
return x;
return min(x.left);
}
@Override
public Key max() {
if(root == null)
throw new NoSuchElementException();
return max(root).key;
}
private Node<Key, Value> max(Node x) {
if (x.right == null) return x;
return max(x.right);
}
@Override
public Key floor(Key key) {
if(root == null)
throw new NoSuchElementException();
Node<Key, Value> x = floor(root, key);
if (x == null)
return null;
return x.key;
}
private Node<Key, Value> floor(Node<Key, Value> x, Key key) {
if (x == null)
return null;
int cmp = key.compareTo(x.key);
if (cmp == 0) return x;
if (cmp < 0) return floor(x.left, key);
Node<Key, Value> t = floor(x.right, key);
if (t != null) return t;
else return x;
}
@Override
public Key select(int k) {
return select(root, k).key;
}
private Node<Key, Value> select(Node x, int k) {
if (x == null) return null;
int t = size(x.left);
if (t > k) return select(x.left, k);
else if (t < k) return select(x.right, k-t-1);
else return x;
}
@Override
public int rank(Key key) {
return rank(key, root);
}
private int rank(Key key, Node<Key, Value> x) {
// Return number of keys less than x.key in the subtree rooted at x.
if (x == null) return 0;
int cmp = key.compareTo(x.key);
if (cmp < 0) return rank(key, x.left);
else if (cmp > 0) return 1 + size(x.left) + rank(key, x.right);
else return size(x.left);
}
@O.
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Over 1 million Americans make a living working for advertising and pub.docx
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Oscar is suing John for $1000 dollars but the judge sees that their ar.docx
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ook at the mean- median- mode- and standard deviation and describe how.docx
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Median 20
Mode 19
Standard Deviation 4.30
Kurtosis 0.03
Skewness -0.46
Minimum 11
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JButton
ActionListener
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Once you have all the structures working as intended- it is time to co.docx
Once you have all the structures working as intended, it is time to compare the
performances of the 2. Use runBenchmark() to start.
a. First, generate an increasing number (N) of random integers (1000, 5000, 10000,
50000, 75000, 100000, 500000 or as far as your computing power will let you)
i. Time and print how long it takes to insert the random integers into an initially
empty BST. Do not print the tree.
You can get a random list using the java class Random, located in
java.util.Random. To test the speed of the insertion algorithm, you should use
the System.currentTimeMillis() method, which returns a long that contains the
current time (in milliseconds). Call System.currentTimeMillis() before and after
the algorithm runs and subtract the two times. (Instant.now() is an alternative
way of recording the time.)
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iii. If T(N) is the time function, how does the growth of TBST(N) compare with the
growth of TAVL(N)?
Plot a simple graph to of time against N for the two types of BSTs to visualize
your results.
b. Second, generate a list of k random integers. k is also some large value.
i. Time how long it takes to search your various N-node BSTs for all k random
integers. It does not matter whether the search succeeds.
ii. Time how long it takes to search your N-node AVL trees for the same k random
integers.
iii. Compare the growth rates of these two search time-functions with a graph the
same way you did in part a.
public class BinarySearchTree<AnyType extends Comparable<? super AnyType>> {
protected BinaryNode<AnyType> root;
public BinarySearchTree() {
root = null;
}
/**
* Insert into the tree; duplicates are ignored.
*
* @param x the item to insert.
* @param root
* @return
*/
protected BinaryNode<AnyType> insert(AnyType x, BinaryNode<AnyType> root) {
// If the root is null, we've reached an empty leaf node, so we create a new node
// with the value x and return it
if (root == null) {
return new BinaryNode<>(x, null, null);
}
// Compare the value of x to the value stored in the root node
int compareResult = x.compareTo(root.element);
// If x is less than the value stored in the root node, we insert it into the left subtree
if (compareResult < 0) {
root.left = insert(x, root.left);
}
// If x is greater than the value stored in the root node, we insert it into the right subtree
else if (compareResult > 0) {
root.right = insert(x, root.right);
}
// If x is equal to the value stored in the root node, we ignore it since the tree does not allow duplicates
return root;
}
/**
* Counts the number of leaf nodes in this tree.
*
* @param t The root of the tree.
* @return
*/
private int countLeafNodes(BinaryNode<AnyType> root) {
// If the root is null, it means the tree is empty, so we return 0
if (root == null) {
return 0;
}
// If the root has no children, it means it is a leaf node, so we return 1
if (root.left == .
On January 1- 2022- a parent company acquired 100- of the voting commo.docx
On January 1, 2022, a parent company acquired 100% of the voting common stock of a subsidiary company for $1 , 500 , 000 . On the acquisition date, the stockholders' equity of the subsidiary was $1 , 400 , 000 . The fair values of the subsidiary's identifiable net assets were equal to their recorded pre- acquisition carrying values except for the following: 1. Compute the acquisition accounting premium. 2. Assign the acquisition accounting premium to the identifiable net assets. 3. Complete the journal entry to record the acquisition.
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On January 1- 2021- MNO Corp- paid $200-000 for its investment in 10-0.docx
On January 1, 2021, MNO Corp. paid $200,000 for its investment in 10,000 shares of XYZ Inc.
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Dividends declared and paid $60,000
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What amount would MNO Corp report in its year-end 2021 balance sheet for its investment in XYZ?
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On January 1- 2024- Blossom Ltd- had the following shareholders' equit.docx
On January 1, 2024 , Blossom Ltd. had the following shareholders' equity accounts:
Common shares (1,090,000 issued)
$1,635,000
Retained earnings
1,962,000
The company was also authorized to issue an unlimited number of $4 noncumulative preferred shares. As at January 1, 2024 , none had been issued. During 2024, the corporation had the following transactions and events related to its shareholders' equity:
Jan.
2
Issued 109,000 preferred shares for $50 per share.
Mar.
31
Paid quarterly dividend to preferred shareholders.
June
30
Paid quarterly dividend to preferred shareholders.
Aug.
12
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Sept.
30
Paid quarterly dividend to preferred shareholders.
Dec.
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Dec.
31
Loss for the year was $103,000.
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On 1 July 2020- Madrid Ltd leased a processing plant from Reece Ltd- T.docx
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On its December 31-2020 balance sheet- Blossom Company appropriately r.docx
On its December 31,2020 balance sheet, Blossom Company appropriately reported a $10 , 000 debit balance in its Fair Value Adjustment account. There was no change during 2021 in the composition of Blossom's portfolio of debt investments held as available-for-sale debt securities. The following information pertains to that portfolio: The amount of unrealized loss to appear as a component of comprehensive income for the year ending December 31,2021 is $40000 $35000 50. $25000
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On equilibrium- Consult Mod- 15-17 18- What kind(s) of sensations is-a.docx
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public class BinarySearchTree<AnyType extends Comparable<? super AnyType>> {
protected BinaryNode<AnyType> root;
public BinarySearchTree() {
root = null;
}
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*
* @param x the item to insert.
* @param root
* @return
*/
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// If the root is null, we've reached an empty leaf node, so we create a new node
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if (root == null) {
return new BinaryNode<>(x, null, null);
}
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root.left = insert(x, root.left);
}
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else if (compareResult > 0) {
root.right = insert(x, root.right);
}
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12
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