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kirza4 [7]
3 years ago
10

Clusters always occur in multiples of what? A.)10 B.)5 C.)2 D.)100

Computers and Technology
2 answers:
kotykmax [81]3 years ago
8 0
I would say ten would be the answer
djverab [1.8K]3 years ago
4 0
2 ..........................
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Data mining must usestatistics to analyze data.<br> True<br> False
Rasek [7]

Answer: True

Explanation: In data mining, they use statistics component for the analyzing of the large amount of data and helps to deal with it. Statistics provide the techniques for the analyzing, evaluating and dealing with the data that is known as data mining.Data statistics provide the help to organize large data in a proper form and then analyze it. Thus, the statement given is true that data mining must use statistics to analyse data.

4 0
3 years ago
In this exercise, use the following variables : i,lo, hi, and result. Assume that lo and hi each are associated with an int and
Assoli18 [71]

Answer:

result = 0

i = lo

while i <= hi:

       result = result + i

       i += 1

Explanation:

Initialize the <em>result</em> as 0 to hold the summation value.

Since we are asked not to change the value of <em>lo</em> and <em>hi</em>, our loop control variable is <em>i</em> and initially it starts from <em>lo.</em>

Since we are asked to add the number from <em>lo</em> to <em>hi, </em>while loop condition checks it.

While the condition satisfies (during each iteration), the value of <em>i</em> is added to the result and the value of <em>i</em> is incremented by one.

7 0
3 years ago
Write a Bare Bones program that takes as input two variables X and Y. (Again, assume these values are set before your program be
Sauron [17]

Answer:

import java.util.Scanner;

public class num8 {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       System.out.println("Enter first number");

       int X = in.nextInt();

       System.out.println("Enter second number");

       int Y = in.nextInt();

       int Z;

       if(X <= Y){

           Z = 0;

       }

       else if(X >= Y){

           Z = 1;

       }

   }

}

Explanation:

  • The program is implemented in Java
  • In order to set the values for X and Y, The Scanner class is used to receive and store the values in the variables (Although the questions says you should assume these values are set)
  • The if conditional statement is used to assign values (either 0 or 1) to variable Z as required by the question.
5 0
3 years ago
Read 2 more answers
Write a (one) program for your microcontroller so that it:
AleksandrR [38]

Answer:

see explaination

Explanation:

1.) Each time the main program execution starts at address 0000 - Reset Vector. The address 0004 is “reserved” for the “interrupt service routine” (ISR).

ORG 0x000 ; processor reset vector

goto main ; go to beginning of main program

ORG 0x004 ; interrupt vector location

movwf w_temp ; save off current W register contents

movf STATUS,w ; move status register into W register

movwf status_temp ; save off contents of STATUS register

.

.

RETFIE

main

3.) Push button switch is connected to the first bit of PORT B (RB0) which is configured as an input pin. When the switch is pressed this pin RB0 will be grounded. The LED is connected to the first bit of PORT B (RA0)

4.) Push button switch is connected to the first bit of PORT B (RB0) which is configured as an input pin. When the switch is pressed this pin RB0 will be grounded. The LED is connected to the first bit of PORT B (RA0)

5.) Push button switch is connected to the first bit of PORT B (RB0) which is configured as an input pin. When the switch is pressed this pin RB0 will be grounded. The LED is connected to the first bit of PORT B (RA0)

#include <stdio.h>

#include <stdlib.h>

#define _XTAL_FREQ 800000//Declare internal OSC Freq as 8MHz

#include <xc.h>

#include<pic.h>

CONFIG(FOSC_INTOSCIO & WDTE_OFF & PWRTE_OFF & MCLRE_ON & BOREN_OFF & LVP_OFF & CPD_OFF & WRT_OFF & CCPMX_RB0 & CP_OFF);

CONFIG(FCMEN_ON & IESO_ON);

unsigned int count=0;

main()

{

TRISA=0;//Sets all ports on A to be outputs

TRISB=1;//Sets all ports on B to be inputs

for(;;){

if(PORTBbits.RB0==1){//When the button is pressed the LED is off

PORTAbits.RA1 =0;

count=count+1;

}

else {

PORTAbits.RA1=1;

count = count +1;

}

if (count > 6){//if count =6 i.e 6 times button presses the RED LED turns on

PORTAbits.RA0=1;

}

else{

PORTAbits.RA0=0;

}

}

}

#include <stdio.h>

#include <stdlib.h>

#define _XTAL_FREQ 800000//Declare internal OSC Freq as 8MHz

#include <xc.h>

#include<pic.h>

CONFIG(FOSC_INTOSCIO & WDTE_OFF & PWRTE_OFF & MCLRE_ON & BOREN_OFF & LVP_OFF & CPD_OFF & WRT_OFF & CCPMX_RB0 & CP_OFF);

CONFIG(FCMEN_ON & IESO_ON);

unsigned int count=0;

main()

{

TRISA=0;//Sets all ports on A to be outputs

TRISB=1;//Sets all ports on B to be inputs

for(;;){

if(PORTBbits.RB0==1){//When the button is pressed the LED is off

PORTAbits.RA1 =0;

count=count+1;

}

else {

PORTAbits.RA1=1;

count = count +1;

}

if (count > 8){//if count =8 i.e 8 times button presses the RED LED turns on

PORTAbits.RA0=1;

}

else{

PORTAbits.RA0=0;

}

}

}

#include <stdio.h>

#include <stdlib.h>

#define _XTAL_FREQ 800000//Declare internal OSC Freq as 8MHz

#include <xc.h>

#include<pic.h>

CONFIG(FOSC_INTOSCIO & WDTE_OFF & PWRTE_OFF & MCLRE_ON & BOREN_OFF & LVP_OFF & CPD_OFF & WRT_OFF & CCPMX_RB0 & CP_OFF);

CONFIG(FCMEN_ON & IESO_ON);

unsigned int count=0;

main()

{

TRISA=0;//Sets all ports on A to be outputs

TRISB=1;//Sets all ports on B to be inputs

for(;;){

if(PORTBbits.RB0==1){//When the button is pressed the LED is off

PORTAbits.RA1 =0;

count=count+1;

}

else {

PORTAbits.RA1=1;

count = count +1;

}

if (count > 10){//if count =10 i.e 10 times button presses the RED LED turns on

PORTAbits.RA0=1;

}

else{

PORTAbits.RA0=0;

}

}

}

6 0
3 years ago
Part 1 of 4 parts for this set of problems: Given an 4777 byte IP datagram (including IP header and IP data, no options) which i
LekaFEV [45]

Answer:

Fragment 1: size (1332), offset value (0), flag (1)

Fragment 2: size (1332), offset value (164), flag (1)

Fragment 3: size (1332), offset value (328), flag (1)

Fragment 4: size (781), offset value (492), flag (1)

Explanation:

The maximum = 1333 B

the datagram contains a header of 20 bytes and a payload of 8 bits( that is 1 byte)

The data allowed = 1333 - 20 - 1 = 1312 B

The segment also has a header of 20 bytes

the data size = 4777 -20 = 4757 B

Therefore, the total datagram = 4757 / 1312 = 4

6 0
2 years ago
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