Oceanography is described as an interdisciplinary science- Describe wh.docx
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Oceanography is an interdisciplinary science that draws from multiple disciplines including biology, chemistry, geology, and physics to study the ocean. The major disciplines of oceanography are physical, chemical, biological and geological oceanography. Ocean currents are studied by both physical and biological oceanography as they influence water properties and distributions of marine organisms.
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PARTICIPANTS PROFILE ANSWER ABOVE MENTIONED QUESTIONS BY TAKING CARE.docx
PARTICIPANTS PROFILE
ANSWER ABOVE MENTIONED QUESTIONS BY TAKING CAREGIVERS OF ADULTS IN LONG TERM FACILITY AS AUDIENCE
TRAINING FOR [ PRESSURE ULCER ]
PLEASE PROVIDE CITATIONS AND REFERENCES - VERYIMPORTANT
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Part3- Offline traffic monitoring In this part will use a PCAP file to.docx
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:
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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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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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Stay on topic, and ensure thoughts and ideas are logically constructed,
Show respect and inclusion, and
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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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Overton Company has gathered the following information-Compute equival.docx
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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
ook at the mean, median, mode, and standard deviation and describe how these measures of central tendency indicates whether the distribution of women are normal or skewed. How does the mean, median, and mode scores indicate if the distribution is normal or skewed?
Mean 19.93
Median 20
Mode 19
Standard Deviation 4.30
Kurtosis 0.03
Skewness -0.46
Minimum 11
Maximum 27
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JPanel
JButton
ActionListener
Submit screenshots of your program's execution and output. Include all appropriate source code in a zip file.
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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.)
ii. Time and print how long it takes to insert the same random integers into an
initially empty AVL tree. Do not print the tree.
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);
}
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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 == .
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