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Furkat [3]
2 years ago
14

The IT department has rejected the password you submitted for approval to join the network. Your suggestion was "bobbie." Why di

d IT reject your choice?
1 Someone else is already using "bobbie."
2 The word does not meet your company's policy for choice of passwords.
3 It will be confusing because it could be written down as "Bobby" by mistake.
4 Because everyone knows you have a dog called Bobbie, so the password is too predictable.
Computers and Technology
2 answers:
vodomira [7]2 years ago
8 0

4. Although all of them could be true, this one is most acurate :)

Nana76 [90]2 years ago
4 0

My choice would be option 2.

1: Multiple people with the same password is technically not a problem, as long as they don't know it of eachother.

2: Every company has password complexity rules, such as minimum length requirements and the use of special characters, so it makes sense that your choice does not meet those requirements.

3: Doesn't make sense. Passwords are generally case sensitive so it is up to you to get the casing right.

4: Passwords are confidential, so validity should be checked by an algorithm, not a person. No other human with knowledge about your pets  should be viewing your password.

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Write only in C, not C++.
yawa3891 [41]

Answer:

#include <iostream>

using namespace std;

int main() {

  int k;

double d;

string s;

cin >> k >> d >> s;

cout << s << " " << d << " " << k << "\n" << k << " " << d << " " << s; }

                                                                   

Explanation:

k is int type variable that stores integer values.

d is double type variable that stores real number.

s is string type variable that stores word.

cin statement is used to take input from user. cin takes an integer, a real number and a word from user. The user first enters an integer value, then a real number and then a small word as input.

cout statement is used to display the output on the screen. cout displays the value of k, d and s which entered by user.

First the values of k, d and s are displayed in reverse order. This means the word is displayed first, then the real number and then the integer separated again by EXACTLY one space from each other. " " used to represent a single space.

Then next line \n is used to produce a new line.

So in the next line values of k, d and s are displayed in original order (the integer , the real, and the word), separated again by EXACTLY one space from each other.

The program along with the output is attached.

7 0
3 years ago
__ means having a current knowledge and understanding of computers, mobile devices, the web, and related technologies.
Brrunno [24]

Answer:

Digital literacy

Explanation:

Digital Literacy means having a current knowledge and understanding of computers, mobile devices, the web, and related technologies.

Brainliest plz

8 0
3 years ago
Which of the following is a list of input devices?
charle [14.2K]
The Correct answer would be C. <span>microphone,mouse,scanner

microphone you talk into to give info (data)

a mouse you use to click on objects  to give input 

a scanner you input a paper/object 

hope this helps!! :)


</span>
7 0
3 years ago
During a user’s onboarding process, many designers focus on a gradual release of information. This process is called what?
mrs_skeptik [129]

progressive disclosure

3 0
2 years ago
Write a MIPS assembly language program that accomplishes the following tasks:The program will prompt the user to enter an intege
Svetlanka [38]

Answer:

Here's the code below

Explanation:

# All the comment will start with #

# MIPS is assembly language program.

# we store the all the data inside .data

.data:

k: .asciiz "Enter the kth value:\n"

.text

.globl main

li $s0, 0 # $s0 = 0

li $t0, 0 # $t0 = 0

la $n0, k # $n0 is kth value address

syscall # call print_string()

li $v0, 0 # $v0

syscall

move $n0, $v0 # $a0 = user input or "int n"

addi $sp, $sp, -4 # move the stack pointer down

sw $ra, 0 ($sp) # save the return address for main

jal fncheck # call fncheck(n);

fncheck:

beq $n, 0, fnfirst #if n==0 then fnfirst

beq $n, 0, fnfirst #if n==0 then fnfirst

ble $n0, 5, fnrec # if (n <=5) then 5

bge $n0, 5, fnjoke

#first function for n 0 and 1

fnfirst:

li $t0, 0 # initialize $t0 = 0

addi $t0, $t0, 20 # $t0 = 20

move $v0, $t0 # return 20

jr $ra

#function to execute 5*function1(n-2) + n;

fnrec:

sub $sp, $sp, 12 # store 3 registers

sw $ra, 0($sp) # $ra is the first register

sw $n0, 4($sp) # $n0 is the second register

addi $n0, $n0, -2 # $a0 = n - 2

jal function1

sw $v0, 8($sp) # store $v0, the 3rd register to be stored

lw $n0, 4($sp) # retrieve original value of n

lw $t0, 8($sp) # retrieve first function result (function1 (n-2))

mul $t0, $t0, 5 # $t0 = 5 * function1(n-2)

add $v0, $v0, $t0

lw $ra, 0($sp) # retrieve return address

addi $sp, $sp, 12

jr $ra

#function to print joke

fnjoke:

la "Joke section"

4 0
3 years ago
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