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Lady bird [3.3K]
3 years ago
15

Which support function under Tech Mahindra is governing data privacy and protection related requirements

Computers and Technology
1 answer:
Alex Ar [27]3 years ago
7 0

Answer:

Privacy Policy

Explanation:

PRIVACY POLICY is the support function under Tech Mahindra that is governing data privacy and protection-related requirements.

Given that support functions are functions which assist and in a way contribute to the company goal.

Other support functions are human resources, training and development, salaries, IT, auditing, marketing, legal, accounting/credit control, and communications.

The above statement is based on the fact that the Privacy Policy clarifies all the data protection rights, such as the right to object to some of the production processes that TechM may carry out.

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Which of the following scenarios falls into the category of network crimes?
solong [7]

Answer:

B

Explanation:

Hope it helps!

3 0
3 years ago
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Mrs. Patel uses a computer program to balance her checkbook. Which of the following best explains how the
Alexandra [31]

Answer:

c it reduces errors

Explanation:

Instead of Mrs.Patel doing it she has an online program made for checks to do it for her.

6 0
3 years ago
Write a Java test program, named TestGuitar, to create 3different Guitars representing each representing a unique test case and
alina1380 [7]

Answer:

Explanation:

//Guitar.java

import java.awt.Color;

import java.lang.reflect.Field;

import java.util.Random;

public class Guitar

     /**

     * these two fields are used to generate random note and duration

     */

     static char[] validNotes = 'A', 'B', 'C', 'D', 'E', 'F', 'G' ;

     static double[] validDuration = 0.25, 0.5, 1, 2, 4 ;

     /**

     * basic guitar attributes

     */

     private int numStrings;

     private double guitarLength;

     private String guitarManufacturer;

     private Color guitarColor;

     public Guitar()

           /**

           * default constructor

           */

           numStrings = 6;

           guitarLength = 28.2;

           guitarManufacturer = Gibson;

           guitarColor = Color.RED;

     

     public Guitar(int numStrings, double guitarLength,

                 String guitarManufacturer, Color guitarColor)

           /**

           * parameterized constructor

           */

           this.numStrings = numStrings;

           this.guitarLength = guitarLength;

           this.guitarManufacturer = guitarManufacturer;

           this.guitarColor = guitarColor;

     

     /**

     * required getters and setters

     */

     public static char[] getValidNotes()

           return validNotes;

     public static void setValidNotes(char[] validNotes)

           Guitar.validNotes = validNotes;      

     public static double[] getValidDuration()

           return validDuration;

     public

4 0
3 years ago
Computer Networks - Queues
lyudmila [28]

Answer:

the average arrival rate \lambda in units of packets/second is 15.24 kbps

the average number of packets w waiting to be serviced in the buffer is 762 bits

Explanation:

Given that:

A single channel with a capacity of 64 kbps.

Average packet waiting time T_w in the buffer = 0.05 second

Average number of packets in residence = 1 packet

Average packet length r = 1000 bits

What are the average arrival rate \lambda in units of packets/second and the average number of packets w waiting to be serviced in the buffer?

The Conservation of Time and Messages ;

E(R) = E(W) + ρ

r = w + ρ

Using Little law ;

r = λ × T_r

w =  λ × T_w

r /  λ = w / λ  +  ρ / λ

T_r =T_w + 1 / μ

T_r = T_w +T_s

where ;

ρ = utilisation fraction of time facility

r = mean number of item in the system waiting to be served

w = mean number of packet waiting to be served

λ = mean number of arrival per second

T_r =mean time an item spent in the system

T_w = mean waiting time

μ = traffic intensity

T_s = mean service time for each arrival

the average arrival rate \lambda in units of packets/second; we have the following.

First let's determine the serving time T_s

the serving time T_s  = \dfrac{1000}{64*1000}

= 0.015625

now; the mean time an item spent in the system T_r = T_w +T_s

where;

T_w = 0.05    (i.e the average packet waiting time)

T_s = 0.015625

T_r =  0.05 + 0.015625

T_r =  0.065625

However; the  mean number of arrival per second λ is;

r = λ × T_r

λ = r /  T_r

λ = 1000 / 0.065625

λ = 15238.09524 bps

λ ≅ 15.24 kbps

Thus;  the average arrival rate \lambda in units of packets/second is 15.24 kbps

b) Determine the average number of packets w waiting to be serviced in the buffer.

mean number of packets  w waiting to be served is calculated using the formula

w =  λ × T_w

where;

T_w = 0.05

w = 15238.09524 × 0.05

w = 761.904762

w ≅ 762 bits

Thus; the average number of packets w waiting to be serviced in the buffer is 762 bits

4 0
3 years ago
can somebody tell me what straykids 2021 collab is gonna be with because i think its gonna be with blackpink
valkas [14]

Answer:

Red velvet: Red Velvet made its debut in 2014, so it is highly probable that SM Entertainment might announce whether or not members of the group will continue to stay in the company by the end of 2021.

Explanation:

Probably black and white, or gold and white Or Gucci clothes.

7 0
2 years ago
Read 2 more answers
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