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2 years ago

Why can a failure in a database environment be more serious than an error in a nondatabase environment?

2 answers:

5 0

Failure in a database environment is more serious then a non data base because if you lose important information you may not get it back and failure in a nondatabase environment the problem may be more easier to solve

Anestetic [448]2 years ago

3 0

Answer:

Failures in database environments can occur due to hardware failures, power failures and software/programming error.

Of these three errors, the non-database environment can survive the hardware failure due to recovery processes and the Software/programming error.

Explanation:

Databases are quite fragile when they hit a hardware failure if you try to recover them (Most of them have raw files that can only be read by certain types of Database Management Softwares) this does not exist in the Non-Database Scenario as most of these records are kept in files.

Software failures can occur when a wrong query is used to retrieve data from a DBMS. This can be frustrating except the user gets the right query.

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A sum amounts to ₹2400 at 15% simple interest per annum after 4 years fond the sum.

Elis [28]

Answer: $1,500

Explanation:

The future value of value using simple interest is:

Future value = Value * ( 1 + rate * time)

2,400 = Value * (1 + 15% * 4)

2,400 = Value * 1.6

Value = 2,400 / 1.6

Value = $1,500

6 0

2 years ago

A major clothing retailer has requested help enhancing their online customer experience. How can Accenture apply Artificial Inte

Harrizon [31]

Answer:

Option D

Explanation:

Artificial intelligence is a technology where the information gathered in the past is processed for the future actions.

Here, based on the preferences of customer in the past and purchase history, AI can suggest the customer new products and services

Hence, option D is correct

3 0

2 years ago

Which type of utp cable is used to connect a pc to a switch port?

rodikova [14]

HDMI Cable, i think that's what it's called.

3 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

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