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alexandr402 [8]
2 years ago
15

A network administrator is using packet tracer to mock up a network that includes iot devices. What can the administrator do fro

m the physical tab of any iot device?
Computers and Technology
1 answer:
Vikki [24]2 years ago
6 0

Answer:

give some instructions for iot device

Explanation:

on how to go about something that you would be able to control physically

You might be interested in
Which cloud computing element that gives a service provider the ability to dynamically allocate shared physical resources to mul
borishaifa [10]

Answer:

Resource Polling is the correct answer to the given question .

Explanation:

The Resource pooling is used in cloud computing frameworks that describe a situation where companies offer temporary and flexible resources to different clients .In the resource polling the programs can be changed to fit the needs of each client without making any adjustments visible to the client.

The main function of resource pooling that it grants a service provider the opportunity to dynamically allocate common physical resources to different clients based on the utilization criteria of each client.

5 0
2 years ago
Write a program, NumberStatistics.java, that reads an unspecified number of integers. The program ends with the input 0. The pro
Lerok [7]

Answer:

Following are the program, which can be below:

import java.util.*; //defining header file

public class NumberStatistics  //defining class NumberStatistics

{

public static void main(String[] arc) //defining main method

{

int total_number=0,max_val=0,min_val=0,neg_val=0,pos_val=0,neg_count=0,pos_count=0,x; //defining variables

Scanner ox= new Scanner(System.in); //creating Scanner class Object

System.out.println("Enter positive and negative value with sign and when complete inter 0:"); //print message  

x= ox.nextInt(); //input value from user

while(x!=0) //defining loop to check input value is 0

{

total_number++; // increment value

if(total_number == 1)  //defining conditional statement to check minimum and maximum value

{

max_val = x; //assign value in max  

min_val = x;//assign value in min

}

else  //defining else block

{

if(max_val < x) //check maximum value

{

max_val = x; //assign maximum value

}

if(min_val > x) //check minimum value

{

min_val = x; //assign minimum value

}

}

if(x > 0) //input values  

{

pos_count++; //count positive value

pos_val+=x; // add positive values

}

else  

{

neg_count++;//count negative value

neg_val+=x; // add negative values

}

x = ox.nextInt(); //input values

}

// defining conditionl statement

if(total_number != 0)  // defining if block to print all value

{

System.out.println("positive value average= "+(pos_val/(double)pos_count));

System.out.println("negative values Average= "+(neg_val/(double)neg_count));

System.out.println("all values average= "+(pos_val+neg_val)/(double)total_number);

System.out.println("maximum value = "+max_val+"\n"+" minimum value = "+min_val);

}

else  

{

System.out.println("no positive values found");

System.out.println("no negative values found");

}

}

}

Output:

Enter positive and negative value with sign and when complete inter 0:

12

8

04

22

-22

-5

9

0

positive value average= 11.0

negative values Average= -13.5

all values average= 4.0

maximum value = 22

minimum value = -22

Explanation:

The Description of the above java program can be described as follows:

  • In the above code first we import the package for user input, then defines the variable "total_number,max_val,min_val,neg_val,pos_val,neg_count, pos_count,x" to store the value prints its value.
  • In the next step, a scanner class object is created for user input and defines a while loop, that checks input value isn't equal to 0, inside the loop a conditional statement defines that cont and calculates all positive, negative, and all values addition.
  • At the another conditional statement is defines, that prints all value average, and also prints maximum, minimum value.
6 0
3 years ago
Create a macro named mReadInt that reads a 16- or 32-bit signed integer from standard input and returns the value in an argument
timofeeve [1]

Answer:

;Macro mReadInt definition, which take two parameters

;one is the variable to save the number and other is the length

;of the number to read (2 for 16 bit and 4 for 32 bit) .

%macro mReadInt 2

mov eax,%2

cmp eax, "4"

je read2

cmp eax, "2"

je read1

read1:

mReadInt16 %1

cmp eax, "2"

je exitm

read2:

mReadInt32 %1

exitm:

xor eax, eax

%endmacro

;macro to read the 16 bit number, parameter is number variable

%macro mReadInt16 1

mov eax, 3

mov ebx, 2

mov ecx, %1

mov edx, 5

int 80h

%endmacro

;macro to read the 32 bit number, parameter is number variable

%macro mReadInt32 1

mov eax, 3

mov ebx, 2

mov ecx, %1

mov edx, 5

int 80h

%endmacro

;program to test the macro.

;data section, defining the user messages and lenths

section .data

userMsg db 'Please enter the 32 bit number: '

lenUserMsg equ $-userMsg

userMsg1 db 'Please enter the 16 bit number: '

lenUserMsg1 equ $-userMsg1

dispMsg db 'You have entered: '

lenDispMsg equ $-dispMsg

;.bss section to declare variables

section .bss

;num to read 32 bit number and num1 to rad 16-bit number

num resb 5

num1 resb 3

;.text section

section .text

;program start instruction

global _start

_start:

;Displaying the message to enter 32bit number

mov eax, 4

mov ebx, 1

mov ecx, userMsg

mov edx, lenUserMsg

int 80h

;calling the micro to read the number

mReadInt num, 4

;Printing the display message

mov eax, 4

mov ebx, 1

mov ecx, dispMsg

mov edx, lenDispMsg

int 80h

;Printing the 32-bit number

mov eax, 4

mov ebx, 1

mov ecx, num

mov edx, 4

int 80h

;displaying message to enter the 16 bit number

mov eax, 4

mov ebx, 1

mov ecx, userMsg1

mov edx, lenUserMsg1

int 80h

;macro call to read 16 bit number and to assign that number to num1

;mReadInt num1,2

;calling the display mesage function

mov eax, 4

mov ebx, 1

mov ecx, dispMsg

mov edx, lenDispMsg

int 80h

;Displaying the 16-bit number

mov eax, 4

mov ebx, 1

mov ecx, num1

mov edx, 2

int 80h

;exit from the loop

mov eax, 1

mov ebx, 0

int 80h

Explanation:

For an assembly code/language that has the conditions given in the question, the program that tests the macro, passing it operands of various sizes is given below;

;Macro mReadInt definition, which take two parameters

;one is the variable to save the number and other is the length

;of the number to read (2 for 16 bit and 4 for 32 bit) .

%macro mReadInt 2

mov eax,%2

cmp eax, "4"

je read2

cmp eax, "2"

je read1

read1:

mReadInt16 %1

cmp eax, "2"

je exitm

read2:

mReadInt32 %1

exitm:

xor eax, eax

%endmacro

;macro to read the 16 bit number, parameter is number variable

%macro mReadInt16 1

mov eax, 3

mov ebx, 2

mov ecx, %1

mov edx, 5

int 80h

%endmacro

;macro to read the 32 bit number, parameter is number variable

%macro mReadInt32 1

mov eax, 3

mov ebx, 2

mov ecx, %1

mov edx, 5

int 80h

%endmacro

;program to test the macro.

;data section, defining the user messages and lenths

section .data

userMsg db 'Please enter the 32 bit number: '

lenUserMsg equ $-userMsg

userMsg1 db 'Please enter the 16 bit number: '

lenUserMsg1 equ $-userMsg1

dispMsg db 'You have entered: '

lenDispMsg equ $-dispMsg

;.bss section to declare variables

section .bss

;num to read 32 bit number and num1 to rad 16-bit number

num resb 5

num1 resb 3

;.text section

section .text

;program start instruction

global _start

_start:

;Displaying the message to enter 32bit number

mov eax, 4

mov ebx, 1

mov ecx, userMsg

mov edx, lenUserMsg

int 80h

;calling the micro to read the number

mReadInt num, 4

;Printing the display message

mov eax, 4

mov ebx, 1

mov ecx, dispMsg

mov edx, lenDispMsg

int 80h

;Printing the 32-bit number

mov eax, 4

mov ebx, 1

mov ecx, num

mov edx, 4

int 80h

;displaying message to enter the 16 bit number

mov eax, 4

mov ebx, 1

mov ecx, userMsg1

mov edx, lenUserMsg1

int 80h

;macro call to read 16 bit number and to assign that number to num1

;mReadInt num1,2

;calling the display mesage function

mov eax, 4

mov ebx, 1

mov ecx, dispMsg

mov edx, lenDispMsg

int 80h

;Displaying the 16-bit number

mov eax, 4

mov ebx, 1

mov ecx, num1

mov edx, 2

int 80h

;exit from the loop

mov eax, 1

mov ebx, 0

int 80h

7 0
3 years ago
What are the three fundamental principals of mnemonics??
olasank [31]
The three fundamental principles underlying the use of mnemonics are imagination, association and location
4 0
2 years ago
Read 2 more answers
Bob flys a drone which has a 20 megapixel camera attached, what is the definition of "megapixel in this context? Why does it mat
Verizon [17]

Answer:

megapixel refers to the unit of resolution i.e. one million

Explanation:

Interestingly the higher the pixels does not mean higher quality of image, it's more about the camera and it's sensor.

I am a photographer and a licensed drone pilot and have researched the subject to help with camera choice, both DSLR and drone.

5 0
2 years ago
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