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erik
25 days ago
6

Which kind of image is indispensable and needs added text to how with it A. a map B. a chart C. a graph or D. a photograph

Computers and Technology
2 answers:
Natasha_Volkova [897]25 days ago
8 0

The correct answer is: a chart

Clarification:

Amiraneli [921]25 days ago
4 0
My initial thought would be a photograph. A graph would be my second guess. I trust this reply resolves your inquiry!
You might be interested in
Which of the following is true of how the Internet has responded to the increasing number of devices now using the network? a) T
Harlamova29_29 [932]

Answer:

A

Explanation:

Every year, internet protocols are adjusted to accommodate the influx of new devices on the network. In the 1990s, traffic primarily utilized a few protocols.  IPv4 managed packet routing, TCP handled those packets to establish connections, SSL (later TLS) secured those connections, DNS resolved hostnames, and HTTP was the main application layer protocol utilized.

For years, there were minimal modifications to the fundamental internet protocols; HTTP saw the addition of some new headers and methods, TLS underwent gradual updates, TCP improved congestion management, and DNS incorporated features like DNSSEC. Over a lengthy period, these protocols remained consistent as seen on the wire — with the exception of IPv6, which is regularly discussed among network operators.

Consequently, network administrators, vendors, and policymakers seeking to understand (and sometimes regulate) the Internet have implemented various practices based on the protocols’ wire ‘footprint’ — whether to troubleshoot issues, enhance service quality, or enforce policies.

Currently, there are considerable changes happening in core internet protocols. Although these updates aim to remain compatible with the wider Internet to ensure adoption, they might disrupt entities that have exploited undocumented features of existing protocols or assumed stability in certain aspects.

8 0
1 month ago
Read 2 more answers
The following is a string of ASCII characters whose bit patterns have been converted into hexadecimal for compactness: 73 F4 E5
Rzqust [894]

Answer:

a) Transforming each character into its binary equivalent:

73:

The digits are 7 and 3

7 in binary is: 111

3 in binary is: 11

Thus

73: 0_111_0011

Likewise

F4:

F stands for 15 and its binary is: 1111

4 in binary: 100

Consequently

F4:  1_111_0100

E5:

E corresponds to 14 and its binary form is: 1110

5 in binary: 101

Therefore

E5:  1_110_0101

76:

7 in binary: 111

6 in binary: 110

Hence

76:  0_111_0110

E5:

E has a binary representation of: 1110

5 in binary: 101

Consequently

E5:  1_110_0101

4A:

4 in binary: 100

A represents 10

A in binary form: 1010

Therefore

4A:  0_100_1010

EF:

E in binary is: 1110

F in binary is: 1111

Thus

EF: 1_110_1111

62:

6 in binary form: 110

2 in binary form: 10

Therefore

62:  0_110_0010

73:

The digits are 7 and 3

7 in binary is: 111

3 in binary is: 11

Thus

73: 0_111_0011

for 0_1110011: the decimal equivalent is: 115 which translates to s

for 1_1110100:  the decimal equivalent is: 116 which translates to t

for 1_1100101:  the decimal equivalent is: 101 which translates to e

for 0_1110110: the decimal equivalent is:  118 which translates to v

for 1_1100101:  the decimal equivalent is: 101 which translates to e

for 0_1001010: the decimal equivalent is:  74 which translates to j

for 1_1101111: the decimal equivalent is: 111 which translates to o

for 0_1100010:  the decimal equivalent is: 98 which translates to b

for 0_1110011: the decimal equivalent is:  115 which translates to s

Thus the decoded sequence is:  stevejobs

b) The parity being utilized is odd.

for 0_1110011:  There are 5 instances of 1s and the parity is 0 indicating it is odd.

for 1_1110100: There are 4 instances of 1s and the parity is 1 indicating it is odd.

Thus, we count the number of 1s and then verify if the parity is odd or even.

Likewise

for 1_1100101:  the parity is odd

for 0_1110110: the parity is odd

for 1_1100101:  the parity is odd

for 0_1001010: the parity is odd

for 1_1101111: the parity is odd

for 0_1100010: the parity is odd

for 0_1110011: the parity is odd

Therefore, the parity being used is odd.

5 0
16 days ago
Test if a number grade is an A (greater than or equal to 90). If so, print "Great!". Sample Run Enter a Number: 98 Sample Output
Harlamova29_29 [932]
To use Python: a = eval(input("Please enter a mark")) if (a < 0): print("Invalid mark") elif (a >= 90 and a <= 100): print("GREAT") Explanation:
7 0
16 days ago
Wendy is an attacker who recently gained access to a vulnerable web server running Microsoft Windows. What command can she use t
ivann1987 [930]

Answer:

If she has successfully connected to the server and obtained access, Wendy should use the shortcut win+r (to run) or alternatively, she can utilize the "Type here to search" option.

Then she would enter "cmd" to access the command prompt.

Finally, she needs to employ the scp protocol for file transfers. Wendy may need to specify a destination for the files to be sent to.

7 0
1 month ago
Complete function PrintPopcornTime(), with int parameter bagOunces, and void return type. If bagOunces is less than 3, print "To
oksian1 [797]

Response:

Function 1:

#include <stdio.h> //import input-output functionality

// this begins the function PrintPopcornTime body, accepting an integer variable //bagOunces as a parameter

void PrintPopcornTime(int bagOunces){

if (bagOunces < 3){ //check if bagOunces equals less than 3

printf("Too small"); //outputs Too small message

printf("\n"); } //prints a new line//The next else if executes when prior conditions are false and bagOunces surpasses 10

else if (bagOunces > 10){

    printf("Too large"); //outputs the message:  Too large     printf("\n");

//prints a new line

}/*else will run when bagOunces is nether under 3 nor over 10 */

else {

/* The subsequent three statements could store bagOunces * 6 in result for printing. However, you can simply use one print statement printf("%d",bagOunces * 6) instead */

    //int result;

    //result = bagOunces * 6;

    //printf("%d",result);

 printf("%d",bagOunces * 6); /calculates the value of bagOunces multiplied by 6

 printf(" seconds"); // "seconds" follows the value of bagOunces * 6

 printf("\n"); }}

//outputs a new line

int main(){

//initiating main function body

int userOunces; //defines integer variable userOunces

scanf("%d", &userOunces); //captures userOunces input

PrintPopcornTime(userOunces);

//calls PrintPopcornTime function with userOunces argument

return 0; }

Function 2:

#include <stdio.h>

//header for using input-output functions// this starts the function PrintShampooInstructions body with integer variable numCycles as

a parametervoid PrintShampooInstructions(int numCycles){

if(numCycles < 1){

//checks if numCycles is below 1 or not

printf("Too few.");

//outputs Too few if the condition is metprintf("\n"); }

//prints a new line

//else if executes when the condition fails, checking if numCycles exceeds 4

else if(numCycles > 4){

//outputs Too many if the above condition is accepted

printf("Too many.");

printf("\n"); } //prints a new line//else will handle when previous if and else if evaluations return false

else{

//outputs "N: Lather and rinse." numCycles times, with N presenting the cycle //number, finished with Done

for(int N = 1; N <= numCycles; N++){

printf("%d",N);

printf(": Lather and rinse. \n");}

printf("Done.");

printf("\n");} }

int main()

//start of the main function body

{    int userCycles;

//defines integer variable userCycles

   scanf("%d", &userCycles); //receives input for userCycles

   PrintShampooInstructions(userCycles); //executes the PrintShampooInstructions function with userCycles argument

   return 0;}

I will clarify the for loop present in PrintShampooInstructions() function. The loop initializes variable N to 1 and evaluates if N remains lesser than or equal to numCycles.

Suppose numCycles = 2. The condition holds true since N < numCycles confirming 1 < 2, thus entering the loop's body. Following statements appear:

printf("%d",N); outputs the value of N followed by printf(": Lather and rinse. \n"); which will print Lather and rinse along with a newline \n.

Initially, this line emerges on the screen as:

1: Lather and rinse.After which, N is enhanced by 1 making it 2 (i.e. N = 2).[TAG_191] During the second iteration:

The loop assesses if N remains below or equal to numCycles. We know numCycles = 2. As the condition is satisfied, the flow of control commences into the loop’s body.

printf("Done."); articulates Done after the preceding two lines are printed.Now, the subsequent check leads to if N remains lower than or equal to numCycles. Noticing that N now reads 3 or greater than 2 thus terminating the loop. The result is:

printf("Done.");

displays Done on the screen.

The collective output is visualized as:1: Lather and rinse.2: Lather and rinse.

Done.

The accompanying programs with their outputs are attached.

6 0
13 days ago
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