1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
gogolik [260]
3 years ago
6

Moore's Law postulates that:This task contains the radio buttons and checkboxes for options. The shortcut keys to perform this t

ask are A to H and alt 1 to alt 9. A a paradigm shift will take place in the technical sciences every two years. B the number of transistors per square inch on integrated circuits will double every two years. C the rate at which new ideas spread through cultures depends on distribution channels. D the members of a given population will usually adopt or accept a new product or innovation.
Computers and Technology
1 answer:
Brums [2.3K]3 years ago
5 0

Answer:

B. the number of transistors per square inch on integrated circuits will double every two years.

Explanation:

The law has been staying true to its postulation as it has been used in so many fields and for every measure of the capabilities of digital electronic devices, including computer processing speed and memory capacity.

It has led to a paradigm shift , technology adoption life cycle and diffusion innovation theory.

You might be interested in
Which shortcut brings up the Print screen?
sashaice [31]

The answer is alt + prt sc

5 0
3 years ago
Read 2 more answers
Which formula should Gemima use to show the
Shalnov [3]
The answer is
Sum(above)
3 0
3 years ago
Read 2 more answers
Which of the following statements are true concerning abstraction and refinement? Select 3 options:
Akimi4 [234]

Answer:

The answer to this question is given below in the explanation section

Explanation:

First,  we need to know what is abstraction and refinement.

Abstraction and refinement are used in application/system development in the field of computing. Abstraction is used to abstract the details from the model. It means that displaying only essential information about the the model and hiding other details. While refinement is the idea that is used in software development by moving through each level of abstraction and incrementally refining each level of abstraction, providing more detail at each increment. Refinement is the inverse of abstraction.  

So the correct options to this question are:

  • Models may be abstracted and refined depending on the goals of the project.
  • Abstraction is a technique used in computational thinking.
  • Refinement is the inverse of abstraction.

5 0
3 years ago
(a) How many locations of memory can you address with 12-bit memory address? (b) How many bits are required to address a 2-Mega-
I am Lyosha [343]

Answer:

Follows are the solution to this question:

Explanation:

In point a:

Let,

The address of 1-bit  memory  to add in 2 location:

\to \frac{0}{1}  =2^1  \ (\frac{m}{m}  \ location)

The address of 2-bit  memory to add in 4 location:

\to \frac{\frac{00}{01}}{\frac{10}{11}}  =2^2  \ (\frac{m}{m}  \ location)

similarly,

Complete 'n'-bit memory address' location number is = 2^n.Here, 12-bit memory address, i.e. n = 12, hence the numeral. of the addressable locations of the memory:

= 2^n \\\\ = 2^{12} \\\\ = 4096

In point b:

\to Let \  Mega= 10^6

              =10^3\times 10^3\\\\= 2^{10} \times 2^{10}

So,

\to 2 \ Mega =2 \times 2^{20}

                 = 2^1 \times 2^{20}\\\\= 2^{21}

The memory position for '2^n' could be 'n' m bits'  

It can use 2^{21} bits to address the memory location of 21.  

That is to say, the 2-mega-location memory needs '21' bits.  

Memory Length = 21 bit Address

In point c:

i^{th} element array addresses are given by:

\to address [i] = B+w \times (i-LB)

_{where}, \\\\B = \text {Base  address}\\w= \text{size of the element}\\L B = \text{lower array bound}

\to B=\$ 52\\\to w= 4 byte\\ \to L B= 0\\\to address  = 10

\to address  [10] = \$ 52 + 4 \times (10-0)\\

                       =   \$ 52   + 40 \ bytes\\

1 term is 4 bytes in 'MIPS,' that is:

= \$ 52  + 10 \ words\\\\ = \$ 512

In point d:

\to  base \ address = \$ t 5

When MIPS is 1 word which equals to 32 bit :

In Unicode, its value is = 2 byte

In ASCII code its value is = 1 byte

both sizes are  < 4 byte

Calculating address:

\to address  [5] = \$ t5 + 4 \times (5-0)\\

                     = \$ t5 + 4 \times 5\\ \\ = \$ t5 + 20 \\\\= \$ t5 + 20  \ bytes  \\\\= \$ t5 + 5 \ words  \\\\= \$ t 10  \ words  \\\\

3 0
4 years ago
Hey, I am in need of urgent help! My teacher is REALLY bad at his job (no exaggeration) and if I dont get a 100 on this lab, I w
anygoal [31]

Answer:

public class CircularList

{

  private ListNode head; // front of the LinkedList

  private ListNode tail; // last node of the LinkedList

  private int size; // size of the LinkedList

   

  // constructs a new CircularList

  public CircularList()

  {

    head = tail = null;

    size = 0;

  }

   

  // returns the size of the array

  public int size()

  {

     return size;

  }

   

  // returns whether the list is empty

  public boolean isEmpty()

  {

     return (size == 0);

  }

   

  // returns the value of the first node

  public Integer first()

  {

     if (head != null) {

       return head.getValue();

     }

     return -1;

  }

   

  // returns the value of the last node

  public Integer last()

  {

     if (tail != null) {

       return tail.getValue();

     }

     return -1;

  }

  // adds a node to the front of the list

  public void addFirst(Integer value)

  {

    head = new ListNode(value, head);

    if (tail == null) {

      tail = head;

    }

    size++;

  }

   

  // adds a node to the end of the list

  public void addLast(Integer value)

  {

    ListNode newTail = new ListNode(value, null);

    if (tail != null) {

      tail.setNext(newTail);

      tail = newTail;

    } else {

      head = tail = newTail;

    }

     

    size++;

  }

   

  // adds a node at the position pos  

  public void addAtPos(int pos, Integer value)

  {

     if (pos == 0) { // Add at the start

       addFirst(value);

       return;

     }

     if (pos <= 0 || pos > size) { // Ignore attempts to add beyond the ends

       return;

     }

     if (pos == size) { // Special case, tail has to be adjusted

       addLast(value);

       return;

     }

     // size and pos are guaranteed both non-zero

     ListNode ptr = head; // ptr is the node before the new one

     for(int i=0; i<pos-1; i++) {

       ptr = ptr.getNext();

     }

     ListNode newNode = new ListNode(value, ptr.getNext());

     ptr.setNext(newNode);

     size++;

  }

   

  // removes the first node and returns the value of the removed node or -1 if the list is empty

  public Integer removeFirst()

  {

     Integer retVal = -1;

     if (head != null) {

       retVal = head.getValue();

       head = head.getNext();

       size--;

     }

     if (size == 0) {

       head = tail = null;

     }

     return retVal;

  }

   

  // removes the node at position pos and returns the value of the removed node or -1 if pos is not a valid position

  public Integer removeNode(int pos)

  {

     Integer retVal = -1;

     if (head == null || pos < 0 || pos >= size) {

       return retVal;

     }

     if (pos == 0) {

       return removeFirst();

     }

     ListNode ptr = head; // ptr is the node before the deleted

     for(int i=0; i<pos-1; i++) {

       ptr = ptr.getNext();

     }

     retVal = ptr.getNext().getValue();

     if (pos == size-1) { // Is it the last element?      

       tail = ptr;

       tail.setNext(null);

     } else {

       ptr.setNext(ptr.getNext().getNext());

     }

     

     size--;

     return retVal;

  }  

   

  // finds and returns the position of find, or -1 if not found

  public int findNode(Integer find)

  {

     ListNode ptr = head;

     for(int pos=0; pos<size; pos++) {

       if (ptr.getValue() == find) {

         return pos;

       }

       ptr = ptr.getNext();

     }

     return -1;

  }  

   

  // rotates the list by placing the first element at the end

  public void rotate()

  {

     addLast(removeFirst());

  }

   

  // returns the list of values in the LinkedList

  public String toString()

  {

     String output = "";

     ListNode iter = head;

     while(iter != null) {

       output += String.format("%d ", iter.getValue());

       iter = iter.getNext();

     }

     return output;

  }

         

}

Explanation:

Enjoy. Linked list are always more complex than you expect. It is a good exercise to try once, then start using libraries. Life is too short to debug linked lists!

7 0
3 years ago
Other questions:
  • Describe one reason you would import data into a database.
    9·2 answers
  • What is the main function of the motherboard?
    12·1 answer
  • ......acts as a platform on which application software runs.
    10·1 answer
  • What process does a system use to officially permit access to a file or a program CIS 502 ?
    11·1 answer
  • This is an attack that depends on the fact that duplicate values, also called collisions, appear much faster than you would expe
    11·1 answer
  • What types of issues can you most likely resolve by knowing how to access and use the control panel choose all that apply?
    10·1 answer
  • Can you help me in this question
    13·1 answer
  • The following questions requires manipulation an analysis of String objects. a. Create a class named String_Manipulation with a
    6·1 answer
  • Draw the 2-3 tree that results when you insert the keys E A S Y Q U T I O N in that order into an initially empty tree
    12·1 answer
  • You need to apply security settings to the registry on a Windows server. Which command should you use
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!