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Alekssandra [29.7K]
3 years ago
5

What is the term for the era created by the digital revolution?

Computers and Technology
1 answer:
ollegr [7]3 years ago
8 0
We are in the modern era and the information era
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Access time is:________.
worty [1.4K]

Answer:

B) the time it takes for the required sector to position itself under the read/write head.

Explanation:

In Computer science, Access time is the time it takes for the required sector to position itself under the read/write head. It is usually measured in milliseconds.

It is the speed of the storage device.

5 0
3 years ago
Create a new Die object. (Refer to Die.html for documentation.)Create a loop that will iterate at least 100 times. In the loop b
hjlf

Answer:

Java code is given below

Explanation:

import java.util.Random;

class Die{

private int sides;

public Die(){

sides = 6;

}

public Die(int s){

sides = s;

}

public int Roll(){

Random r = new Random();

return r.nextInt(6)+1;

}

}

class DieRoll{

public static void main(String[] args) {

Die die = new Die();

int arr[] = new int[6];

for(int i=0; i<6; i++)

arr[i] = 0;

for(int i=0; i<100; i++){

int r = die.Roll();

arr[r-1]++;

}

for(int i=0; i<6; i++)

System.out.println((i+1)+" was rolled "+arr[i]+" times.");

}

}

8 0
4 years ago
Charles sends Julia text messages every morning insulting her appearance and threatening to hurt her. He writes unflattering des
aalyn [17]
What is the question?
4 0
3 years ago
Read 2 more answers
Computer Networks - Queues
lyudmila [28]

Answer:

the average arrival rate \lambda in units of packets/second is 15.24 kbps

the average number of packets w waiting to be serviced in the buffer is 762 bits

Explanation:

Given that:

A single channel with a capacity of 64 kbps.

Average packet waiting time T_w in the buffer = 0.05 second

Average number of packets in residence = 1 packet

Average packet length r = 1000 bits

What are the average arrival rate \lambda in units of packets/second and the average number of packets w waiting to be serviced in the buffer?

The Conservation of Time and Messages ;

E(R) = E(W) + ρ

r = w + ρ

Using Little law ;

r = λ × T_r

w =  λ × T_w

r /  λ = w / λ  +  ρ / λ

T_r =T_w + 1 / μ

T_r = T_w +T_s

where ;

ρ = utilisation fraction of time facility

r = mean number of item in the system waiting to be served

w = mean number of packet waiting to be served

λ = mean number of arrival per second

T_r =mean time an item spent in the system

T_w = mean waiting time

μ = traffic intensity

T_s = mean service time for each arrival

the average arrival rate \lambda in units of packets/second; we have the following.

First let's determine the serving time T_s

the serving time T_s  = \dfrac{1000}{64*1000}

= 0.015625

now; the mean time an item spent in the system T_r = T_w +T_s

where;

T_w = 0.05    (i.e the average packet waiting time)

T_s = 0.015625

T_r =  0.05 + 0.015625

T_r =  0.065625

However; the  mean number of arrival per second λ is;

r = λ × T_r

λ = r /  T_r

λ = 1000 / 0.065625

λ = 15238.09524 bps

λ ≅ 15.24 kbps

Thus;  the average arrival rate \lambda in units of packets/second is 15.24 kbps

b) Determine the average number of packets w waiting to be serviced in the buffer.

mean number of packets  w waiting to be served is calculated using the formula

w =  λ × T_w

where;

T_w = 0.05

w = 15238.09524 × 0.05

w = 761.904762

w ≅ 762 bits

Thus; the average number of packets w waiting to be serviced in the buffer is 762 bits

4 0
3 years ago
B. Find Addition of Binary Numbers: 1100112 + 11012
schepotkina [342]
Hello, I assume you mean 110011_2 + 1101_2

To add 110011+ 1101 we can setup the problem like the following:

   110011
+     1101 
-------------
  1000000

When we add 1 + 1 this equals 10 so we just carry the 1 and we keep doing that from right to left. Also 1 + 0 = 1 

7 0
3 years ago
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