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2 years ago

Why is it important for element IDs to have meaningful names?

Computers and Technology

1 answer:

allochka39001 [22]2 years ago

4 0

Answer:

Explanation:

In programming, it is very important that element ID's have meaningful names because it allows anyone reading the code to be able to easily identify what that element is and where it is most likely to be found in the code. This also applies to variables in every programming language. The clearer and more unique the name is the easier it becomes to know what that variable or element ID is representing which ultimately makes reading and debugging the code that much easier. This means that the entire process becomes more efficient as well.

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2 years ago

A regional bank implemented an automated solution to streamline their operations for receiving and processing checks/cheques. Th

Sphinxa [80]

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Explanation:

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1 year ago

Problem 2 - K-Best Values - 30 points Find the k-best (i.e. largest) values in a set of data. Assume you are given a sequence of

masya89 [10]

Answer:

See explaination

Explanation:

/**KBestCounter.java**/

import java.util.ArrayList;

import java.util.List;

import java.util.PriorityQueue;

public class KBestCounter<T extends Comparable<? super T>>

{

PriorityQueue heap;

int k;

public KBestCounter(int k)

{

heap = new PriorityQueue < Integer > ();

this.k=k;

}

//Inserts an element into heap.

//also takes O(log k) worst time to insert an element

//into a heap of size k.

public void count(T x)

{

//Always the heap has not more than k elements

if(heap.size()<k)

{

heap.add(x);

}

//if already has k elements, then compare the new element

//with the minimum element. if the new element is greater than the

//Minimum element, remove the minimum element and insert the new element

//otherwise, don't insert the new element.

else if ( (x.compareTo((T) heap.peek()) > 0 ) && heap.size()==k)

{

heap.remove();

heap.add(x);

}

//Returns a list of the k largest elements( in descending order)

public List kbest()

{

List al = new ArrayList();

int heapSize=heap.size();

//runs O(k)

for(int i=0;i<heapSize;i++)

{

al.add(0,heap.poll());

}

//Restoring the the priority queue.

//runs in O(k log k) time

for(int j=0;j<al.size();j++) //repeats k times

{

heap.add(al.get(j)); //takes O(log k) in worst case

}

return al;

}

public class TestKBest

{

public static void main(String[] args)

{

int k = 5;

KBestCounter<Integer> counter = new KBestCounter<>(k);

System.out.println("Inserting 1,2,3.");

for(int i = 1; i<=3; i++)

counter.count(i);

System.out.println("5-best should be [3,2,1]: "+counter.kbest());

counter.count(2);

System.out.println("Inserting another 2.");

System.out.println("5-best should be [3,2,2,1]: "+counter.kbest());

System.out.println("Inserting 4..99.");

for (int i = 4; i < 100; i++)

counter.count(i);

System.out.println("5-best should be [99,98,97,96,95]: " + counter.kbest());

System.out.println("Inserting 100, 20, 101.");

counter.count(100);

counter.count(20);

counter.count(101);

System.out.println("5-best should be [101,100,99,98,97]: " + counter.kbest());

}

5 0

2 years ago

Create a single list that contains the following collection of data in the order provided:

ivann1987 [24]

Answer:

A code was created to single list that contains the following collection of data in the order provided

Explanation:

Solution

#list that stores employee numbers

employeee_numbers=[1121,1122,1127,1132,1137,1138,1152,1157]

#list that stores employee names

employee_name=["Jackie Grainger","Jignesh Thrakkar","Dion Green","Jacob Gerber","Sarah Sanderson","Brandon Heck","David Toma","Charles King"]

#list that stores wages per hour employee

hourly_rate=[22.22,25.25,28.75,24.32,23.45,25.84,22.65,23.75]

#list that stores salary employee

company_raises=[]

max_value=0

#list used to store total_hourly_rate

total_hourly_rate=[]

underpaid_salaries=[]

#loop used to musltiply values in hourly_wages with 1.3

#and store in new list called total_hourly_rate

#len() is used to get length of list

for i in range(0,len(hourly_rate)):

#append value to the list total_hourly_rate

total_hourly_rate.append(hourly_rate[i]*1.3)

for i in range(0,len(total_hourly_rate)):

#finds the maximum value

if(total_hourly_rate[i]>max_value):

max_value=total_hourly_rate[i] #stores the maximum value

if(max_value>37.30):

print("\nSomeone's salary may be a budget concern.\n")

for i in range(0,len(total_hourly_rate)):

#check anyone's total_hourly_rate is between 28.15 and 30.65

if(total_hourly_rate[i]>=28.15 and total_hourly_rate[i]<=28.15):

underpaid_salaries.append(total_hourly_rate[i])

for i in range(0,len(hourly_rate)):

#check anyone's hourly_rate is between 22 and 24

if(hourly_rate[i]>22 and hourly_rate[i]<24):

#adding 5%

hourly_rate[i]+=((hourly_rate[i]/100)*5)

elif(hourly_rate[i]>24 and hourly_rate[i]<26):

#adding 4%

hourly_rate[i]+=((hourly_rate[i]/100)*4)

elif(hourly_rate[i]>26 and hourly_rate[i]<28):

#adding 3%

hourly_rate[i]+=((hourly_rate[i]/100)*3)

else:

#adding 2% for all other salaries

hourly_rate[i]+=((hourly_rate[i]/100)*2)

#adding all raised value to company_raises

company_raises.extend(hourly_rate)

#The comparison in single line

for i in range(0,len(hourly_rate)):

For this exercise i used ternary

operator hourly_rate[i]= (hourly_rate[i]+((hourly_rate[i]/100)*5)) if (hourly_rate[i]>22 and hourly_rate[i]<24) else (hourly_rate[i]+((hourly_rate[i]/100)*4)) if (hourly_rate[i]>24 and hourly_rate[i]<26) else hourly_rate[i]+((hourly_rate[i]/100)*2)

8 0

2 years ago