Question 1. Who was the first to use an abacus

A file with a com extension is most likely to be a(n) ___ file.

olasank [31]

I think that the answer is: A executable

4 0

3 years ago

1. Implement a method factorial(int n), that takes a number (int) and returns its factorial. Factorial of an integer is the prod

marysya [2.9K]

Answer:

import java.io.*;

import java.util.*;

public class Main

{

public static int factorial(int n){

if (n == 0)

return 1;

else

return(n * factorial(n-1)); //recursively calling factorial function

}

public static void fibbonaci(int n)

{

if(n>0)

{

c=a+b; //next term in series

a=b;

b=c;

System.out.print(" "+c);

fibbonaci(n-1); //recursively calling fibbonaci function

}

static int a=0,b=1,c;

public static void main(String[] args) {

int list_num,fact;

List<Integer> num=Arrays.asList(1,3,5,7,9); //arraylist with 1,3,5,7,9

for(Integer i:num) //for-each loop

{

a=0;

b=1;

list_num=i.intValue();

System.out.println("Fibbonaci series for "+list_num);

System.out.print(a+" "+b);

fibbonaci(list_num-2); //calling fibbonaci for values in list

fact=factorial(list_num); //calling factorial for values in list

System.out.println();

System.out.println("Factorial for "+list_num+" is "+fact);

System.out.println();

}

6 0

3 years ago

1. How many bits would you need to address a 2M × 32 memory if:

Dominik [7]

Answer:

a) 23 b) 21
a) 43 b) 42
a) 0 b) 0

Explanation:

1) How many bits is needed to address a 2M * 32 memory

2M = 2^1*2^20, while item =32 bit long word

hence ; L = 2^21 ; w = 32

a) when the memory is byte addressable

w = 8; L = ( 2M * 32 ) / 8 = 2M * 4

hence number of bits = log2(2M * 4)= log2 ( 2 * 2^20 * 2^2 ) = 23 bits

b) when the memory is word addressable

W = 32 ; L = ( 2M * 32 )/ 32 = 2M

hence the number of bits = log2 ( 2M ) = Log2 (2 * 2^20 ) = 21 bits

2) How many bits are required to address a 4M × 16 main memory

4M = 4^1*4^20 while item = 16 bit long word

hence L ( length ) = 4^21 ; w = 16

a) when the memory is byte addressable

w = 8 ; L = ( 4M * 16 ) / 8 = 4M * 2

hence number of bits = log 2 ( 4M * 2 ) = log 2 ( 4^1*4^20*2^1 ) ≈ 43 bits

b) when the memory is word addressable

w = 16 ; L = ( 4M * 16 ) / 16 = 4M

hence number of bits = log 2 ( 4M ) = log2 ( 4^1*4^20 ) ≈ 42 bits

3) How many bits are required to address a 1M * 8 main memory

1M = 1^1 * 1^20 , item = 8

L = 1^21 ; w = 8

a) when the memory is byte addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

hence number of bits = log 2 ( 1M ) = log2 ( 1^1 * 1^20 ) = 0 bit

b) when memory is word addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

number of bits = 0

5 0

3 years ago

My duties included reviewing

ycow [4]

Answer:

Please check the explanation.

Explanation:

I carried my logical and administrative skills to stand in implementing the succeeding tasks allocated to me:

swotting acquiring trends
investigating client data
concocting yearly inclination reports.

The bullets and numbering and proper indentation make the reading stuff more easily understandable. And that is what we have done above.

4 0

3 years ago

Suppose there are two ISPs providing WiFi service in a café. Each ISP operates its own AP and has its own IP address block. If b

Alinara [238K]

Answer:

As a design rule, access points within range of each other should be set to channel frequencies with minimal signal overlap. Users will find that roaming doesn’t work well, and performance will degrade because of interference between access points.

Explanation:

8 0

3 years ago