Answer:
The algorithm to this question as follows:
Algorithm:
finding_small_element(element,Key xa) //defining method that accepts parameter
{
if (element.value>= xa) //check value
{
//skiping node
return;
}
print(element.value);//print value
if (element.left != NULL) //check left node value
{
finding_small_element(element.left,xa); //using method
}
if (element.right != NULL) //check right node value
{
finding_small_element(element.right,xa); //using method
}
}
Explanation:
In the above pre-order traversing algorithm a method "finding_small_element" is defined, that accepts two parameter, that is "element and ax', in which "element" is node value and ax is its "key".
Inside this if block is used that check element variable value is greater then equal to 0. In the next step, two if block is defined, that check element left and the right value is not equal to null, if both conditions are true, it will check its right and left value. In this algorithm, there is no extra need for traversing items.Final answer:
An efficient algorithm to find keys in a min heap smaller than a given value X is to use DFS, starting at the root and adding each key less than X to a result list, with the time complexity being O(n) in the worst case.
Explanation:
Finding Keys Smaller Than X in a Min Heap
To find all keys in a min heap that are smaller than a provided value X, we can use a depth-first search (DFS) algorithm. Since a min heap is a complete binary tree where the key at the root is less than or equal to the keys in its children, and this property applies recursively to subtrees, we can traverse the min heap efficiently with the following approach:
Start at the root of the min heap.If the current node's key is greater than or equal to X, return, as all keys in the subtree rooted at the current node will also be greater than or equal to X (due to heap property).If the current node's key is smaller than X, add it to the result list.Recursively apply this logic to the left and right children of the current node.The time complexity of this algorithm is O(n), where n is the number of elements in the min heap, because in the worst case, we might have to visit every node. However, thanks to the properties of the min heap, we can often avoid exploring all nodes, making the algorithm more efficient in practice than O(n).
Explain how increasingly standardized data, access to third-party datasets, and current trends in hardware and software are collectively enabling a new age of decision making?
Answer and Explanation:
The information revolution has had profound impacts on decision-making allowing more informed decision as a result of "stone throw" information reach- in our pockets, the desk, the TV, and the vast number of technologies that make this possible.
Standardized data which involves data formatted to bring uniformity and be easily understood and compared by programs and people , access to rich, outsider dataset and less tasking and flexible plus powerful programming have all contributed to empowering another time of information driven decision-making that are less prone to errors and mistakes.
In this exercise, you’ll design a "starter" HealthProfile class for a person. The class attributes should include the person’s first name, last name, gender, date of birth (consisting of separate attributes for the month, day and year of birth), height (in inches) and weight (in pounds). Your class should have a constructor that receives this data. For each attribute, provide setters and getters.The class should include methods that calculate and return the user’s age in years, maximum heart rate and target heart rate range, and body mass index (BMI). Write a Java application that prompts for the person’s information, instantiates an object of class HealthProfile for that person and prints the information from that object—including the person’s first name, last name, gender, date of birth, height and weight—then calculates and prints the person’s age in years, BMI, maximum heart rate and target-heart-rate range. It should also display the BMI values chart.
Answer:
package healthcare;
import java.util.Calendar;
public class HealthProfile {
private String firstName;
private String lastName;
private char gender;
private int day;
private int month;
private int year;
private double height;
private double weight;
public HealthProfile(String firstName, String lastName, char gender, int day, int month, int year, double height,
double weight) {
super();
this.firstName = firstName;
this.lastName = lastName;
this.gender = gender;
this.day = day;
this.month = month;
this.year = year;
this.height = height;
this.weight = weight;
}
public String getFirstName() {
return firstName;
}
public void setFirstName(String firstName) {
this.firstName = firstName;
}
public String getLastName() {
return lastName;
}
public void setLastName(String lastName) {
this.lastName = lastName;
}
public char getGender() {
return gender;
}
public void setGender(char gender) {
this.gender = gender;
}
public int getDay() {
return day;
}
public void setDay(int day) {
this.day = day;
}
public int getMonth() {
return month;
}
public void setMonth(int month) {
this.month = month;
}
public int getYear() {
return year;
}
public void setYear(int year) {
this.year = year;
}
public double getHeight() {
return height;
}
public void setHeight(double height) {
this.height = height;
}
public double getWeight() {
return weight;
}
public void setWeight(double weight) {
this.weight = weight;
}
public int calculateAge() {
Calendar dateOfBirth = Calendar.getInstance();
dateOfBirth.set(year, month, day);
Calendar now = Calendar.getInstance();
return now.get(Calendar.YEAR) - dateOfBirth.get(Calendar.YEAR);
}
public int maximumHeartRate() {
return 220 - calculateAge();
}
public double[] targetHeartRateRange() {
double[] range = new double[2];
// Calculate Stating range(50 % of maximumHeartRate)
range[0] = 0.5 * maximumHeartRate();
// Calculate End range(85 % of maximumHeartRate)
range[1] = 0.85 * maximumHeartRate();
return range;
}
public double calculateBMI() {
return (weight * 703)/(height * height);
}
public String getBMIValue()
{
double bmi=calculateBMI();
if(bmi < 18.5)
{
return "Underweight";
}
else if (bmi>18.5 && bmi<24.9)
{
return "Normal";
}
else if (bmi>25 && bmi<29.9)
{
return "Normal";
}
else if (bmi>=30)
{
return "Obese";
}
return "DafultValue"; //you can give any default value of your choice here if no condition meets the given criteria
}
@Override
public String toString() {
return "HealthProfile [firstName=" + firstName + ", lastName=" + lastName + ", gender=" + gender + ", Date Of Birth="
+ day + "-" + month + "-" + year + ", height=" + height + ", weight=" + weight + "]";
}
}
package healthcare;
import java.util.Scanner;
public class TestHealthCare {
private static Scanner sc;
public static void main(String[] args) {
sc = new Scanner(System.in);
System.out.println("Please enter following details of the Patient");
System.out.println("First Name");
String firstName=sc.nextLine();
System.out.println("Last Name");
String lastName=sc.nextLine();
System.out.println("Gender ...... M or F ?");
char gender=sc.next().charAt(0);
System.out.println("Date of Birth");
System.out.println("Day");
int day=sc.nextInt();
System.out.println("Month");
int month=sc.nextInt();
System.out.println("Year");
int year=sc.nextInt();
System.out.println("Height in inches");
double height =sc.nextDouble();
System.out.println("weight (in pounds)");
double weight =sc.nextDouble();
HealthProfile obj=new HealthProfile(firstName, lastName, gender, day, month, year, height, weight);
int age=obj.calculateAge();
System.out.println("Patient age is "+age + " Years");
int maxHeartRate=obj.maximumHeartRate();
System.out.println("Patient Maximum Heart Rate is "+maxHeartRate + " beats per minute");
//Call targetHeartRateRange
double targetHeartRateRange []=obj.targetHeartRateRange();
System.out.println("Target Heart Range is "+targetHeartRateRange [0] + " - " +targetHeartRateRange [1]);
//Call calculateBMI
double bmi=obj.calculateBMI();
System.out.println("Patient BMI is "+bmi);
//call getBMIValue
System.out.println("Patient BMI Value is "+obj.getBMIValue());
}
}
Explanation:
Inside the calculate the age in years method, create Date of Birth of Object.Create Current Date .Inside the method maximumHeartRate , create a New Object of HealthProfile class and Call its constructor .Compare and Contrast IPv4 and IPv6. Find examples of companies who are currently using IPv6. What are some of the benefits of using IPv6 over IPv4 and vice versa.
Answer/Explanation:
IPv4 address field is of 32-bits while the address field of IPv6 is 128-bits.The IPv4 requires a packet size of 576 bytes with optional fragmentation whereas, IPv6 uses 1280 bytes without fragmentation.IPv6 is preferred over IPv4 because of its large address range. IPv4 uses 32-bit source and destination address and we are running out of addresses. To solve this issue, Internet Engineering Task Force has devised a solution by introducing IPv6 which uses 128-bits address field. It is being said that if we convert the water bodies present on earth into deserts and then we assign each particle of desert with an address even then the addresses of IPv6 won't end.IPv6 provides flexible options and extensions.IPv6 provides multi broadcast while, IPv4 just provides broadcast.Internet has switched from IPv4 to IPv6.
A slow response when opening applications or browsing the Internet, applications that do not work properly, an operating system that does not boot up correctly or does not function normally, and event logs that report numerous, unusual alerts are all:________. A. Signs of computer aging. B. Indications of antivirus software running in the background. C. Symptoms of malware mutation.
Answer:
C. Symptoms of malware mutation
Explanation:
They are signs that could indicate that ones system is infected by virus.
Such signs includes:
1.Slow booting or startup.
2.Low space storage.
3.Blue screen of death.
4.Slow Internet performance.
5.Sending of spam messages.
6.Disabled security.
7.Browsing error.
8.Pop-up messages.
9.Renaming of files.
10.Loss of certain files, music, messages, etc.
Answer:
Option c: Symptoms of malware mutation
Explanation:
The most common symptom of a malware mutation is a slow running computer. The main activity of malware programs is to slow down your operating system and cause problems with your local applications even you are not using the internet. Event log reports unusual alerts and RAM is consumed by virus programs which makes RAM unavailable for your tasks.
Advances in data storage techniques and the rapidly declining costs of data storage threaten ________. A. organizational procedures B. individual privacy C. growth in mobile devices D. network advances E. analytical procedures
Answer:
Option C is the correct option.
Explanation:
In the above scenario, Advancements in storage technologies and fast-declining storage prices are undermining growth in the mobile device because current mobile devices have many useful features of data storage and it can also use in other fields such as cloud storage, also in word processing and many other areas which are useful for the users.
Option A is incorrect because it is not suitable according to the following case. Option B is incorrect because data storage is not related the individual privacy. Options D and E are incorrect because in the above case, it is about the data storage techniques not for the network advances and analytical procedures.
Data governance consists of? A. the processes, methods, and techniques to ensure that data is of high quality, reliable, and unique (not duplicated), so that downstream uses in reports and databases are more trusted and accurate B. the overarching policies and processes to optimize and leverage information while keeping it secure and meeting legal and privacy obligations in alignment with organizationally stated business objectives; C. established frameworks and best practices to gain the most leverage and benefit out of IT investments and support the accomplishment of business objectives D. None of the above
Answer:
A. the processes, methods, and techniques to ensure that data is of high quality, reliable, and unique (not duplicated), so that downstream uses in reports and databases are more trusted and accurate
Explanation:
Data governance consists of the processes, methods, and techniques to ensure that data is of high quality, reliable, and unique (not duplicated), so that downstream uses in reports and databases are more trusted and accurate. Master data management (MDM) tools can assist in this effort.
Once the definitions of these three information-related governance disciplines are clear, their differences become more distinct.