(i) Deques cannot be implemented using arrays. : false
Justification:
We can implement dequeue using circular array. We need to maintain two pointers rear and
front.
Ex :
1. Create an array of any element type like int [] dequeue.
2. Create two indexes front and rear and initialize them to 0.
3. Create an element capacity.(it should be length of an array - 1).
(ii) If we require all add/remove operations to run in O(1) time, we cannot use a Deque
implemented with a singly linked list. False
Justification :
In linklist there will be two pointers head and tail.
Adding an element :
tail -> next = newelement;
newelement -> next = NULL;
tail = newelement;
These all are O(1) operation.
Removing an element :
head = head->next; O(1) operation
(iii) A Deque can be used as a stack as well as a queue. : true
Justification :
Using as stack : You can use dequeue as stack by removing and inserting the element at
one end.
Using as queue : You can use dequeue as queue by inserting the element at front and
removing at rear.
(iv) Assume the Deque contains n elements and n is odd. The middle element is the e lement
equidistant from either end. The middle element cannot be determined in O(1) time in any
implementation : false
Justification :
You can achieve this in O(1) operation with the help of doubly link list. You will have to
maintain a mid pointer to achieve this. You can move this pointer in either direction with previos
and next pointers.
Solution
(i) Deques cannot be implemented using arrays. : false
Justification:
We can implement dequeue using circular array. We need to maintain two pointers rear and
front.
Ex :
1. Create an array of any element type like int [] dequeue.
2. Create two indexes front and rear and initialize them to 0.
3. Create an element capacity.(it should be length of an array - 1).
(ii) If we require all add/remove operations to run in O(1) time, we cannot use a Deque
implemented with a singly linked list. False
Justification :
In linklist there will be two pointers head and tail.
Adding an element :
tail -> next = newelement;
newelement -> next = NULL;
tail = newelement;
These all are O(1) operation.
Removing an element :
head = head->next; O(1) operation
(iii) A Deque can be used as a stack as well as a queue. : true
Justification :
Using as stack : You can use dequeue as stack by removing and inserting the element at
one end.
Using as queue : You can use dequeue as queue by inserting the element at front and
removing at rear.
(iv) Assume the Deque contains n elements and n is odd. The middle element is the e lement
equidistant from either end. The middle element cannot be determined in O(1) time in any
implementation : false
Justification :
You can achieve this in O(1) operation with the help of doubly link list. You will have to
maintain a mid pointer to achieve this. You can move this pointer in either direction with previos
and next pointers..
To complete the task, you need to fill in the missing code. I’ve inc.pdfezycolours78
To complete the task, you need to fill in the missing code. I’ve included code to create an
Iterator. An Iterator is an object that iterates over another object – in this case, a circular linked
list. You can use the .next() method to advance the Iterator to the next item (the first time you
call it, the iterator will travel to node at index 0). Using iterator’s .remove() removes the node the
iterator is currently at. Say that we had a CircularLinkedList that looked like this: A ==> B ==>
C ==> D ==> E ==> Calling .next() three times will advance the iterator to index 2. Calling
.remove() once will remove the node at index 2. A ==> B ==> D ==> E ==> Calling .remove()
once more will remove the node now at index 2. A ==> B ==> E ==> The Iterator methods
handle wrapping around the CircularLinkedList. Be sure to create the iterator using l.iterator()
and after you’ve added all the nodes to the list, where l is your CircularLinkedList
Solution
import java.util.Iterator;
class CircularLinkedList implements Iterable {
// Your variables
// You can include a reference to a tail if you want
Node head;
int size; // BE SURE TO KEEP TRACK OF THE SIZE
// implement this constructor
public CircularLinkedList() {
head=null;
}
// writing helper functions for add and remove, like the book did can help
// but remember, the last element\'s next node will be the head!
// attach a node to the end of the list
// Be sure to handle the adding to an empty list
// always returns true
public boolean add(E e) {
Node newNode=new Node(e);
if(size==0){
head=newNode;
}
else{
Node last=getNode(size-1);
last.next=newNode;
}
newNode.next=head; //last element node is set to head
size++;
return true;
}
// need to handle
// out of bounds
// empty list
// adding to front
// adding to middle
// adding to \"end\"
// REMEMBER TO INCREMENT THE SIZE
public boolean add(int index, E e){
if(index>size) return false;
Node tmp=new Node(e);
if(index==0){
tmp.next=head;
Node last=getNode(size-1);
head=tmp;
last.next=head;
}
else {
Node curr=getNode(index-1);
tmp.next=curr.next;
curr.next=tmp;
}
size++;
return true;
}
// I highly recommend using this helper method
// Return Node found at the specified index
// be sure to handle out of bounds cases
private Node getNode(int index ) {
Node prev=head;
for(int i=0;isize) {
e=null;
}
else if(index==0){
e = head.getElement();
Node last=getNode(size-1);
head=head.next;
last.next=head;
size--;
}
else{
Node prev=getNode(index-1);
Node curr=getNode(index);
e=curr.getElement();
prev.next=curr.next;
size--;
}
return e;
}
// Turns your list into a string
// Useful for debugging
public String toString(){
Node current = head;
StringBuilder result = new StringBuilder();
if(size == 0){
return \"\";
}
if(size == 1) {
return head.getElement().toString();
}
else{
do{
result.append(current.getElement());
result.append(\" ==> \");
current = current.next;
} while(current != head);
}
return result.toString();
}
public Iterator iterator() {
return new ListIterator();
}
/.
Objectives- In this lab- students will practice- 1- Stack with single.pdfAugstore
Objectives: In this lab, students will practice: 1. Stack with single link-list 2. Circular Link-list
Introduction Implement a stack using singly linked list To implement a stack using singly linked
list concept, all the singly linked list operations are performed based on Stack operations
LIFO(last in first out) and with the help of that knowledge we are going to implement a stack
using single linked list. Using singly linked lists, we implement stack by storing the information
in the form of nodes and we need to follow the stack rules and implement using singly linked list
nodes. So we need to follow a simple rule in the implementation of a stack which is last in first
out and all the operations can be performed with the help of a top variable .Let us learn how to
perform Pop, Push, Peek, Display operations in the following article. Page 2 of 5 A stack can be
easily implemented using the linked list. In stack Implementation, a stack contains a top pointer.
which is "head" of the stack where pushing and popping items happens at the head of the list.
First node have null in link field and second node link have first node address in link field and so
on and last node address in "top" pointer. The main advantage of using linked list over an arrays
is that it is possible to implement a stack that can shrink or grow as much as needed. In using
array will put a restriction to the maximum capacity of the array which can lead to stack
overflow. Here each new node will be dynamically allocate. so overflow is not possible. Stack
Operations: 1. push() : Insert a new element into stack i.e just inserting a new element at the
beginning of the linked list. Page 3 of 5 2. pop() : Return top element of the Stack i.e simply
deleting the first element from the linked list. 3. peek(): Return the top element. 4. display():
Print all elements in Stack. Circular linked list Introduction: Circular Linked List is a variation of
Linked list in which the first element points to the last element and the last element points to the
first element. Both Singly Linked List and Doubly Linked List can be made into a circular linked
list. Singly Linked List as Circular In singly linked list, the next pointer of the last node points to
the first node. Doubly Linked List as Circular In doubly linked list, the next pointer of the last
node points to the first node and the previous pointer of the first node points to the last node
making the circular in both directions. Question 1 1. Implement a template-based stack using a
singly linked list. The required member methods are: int size(): returns the count of total element
stored in the stack. bool isEmpty(): returns true if the stack is empty else false. bool top(T\&):
returns, but does not delete, the topmost element from the stack via the parameter passed by
reference. It returns false via a return statement if there is no element in the stack, else it returns
true and assigns the top most element to the parameter passed by ref.
First lets draw out the Lewis structure such that each atom has.pdfaradhana9856
First let\'s draw out the Lewis structure such that each atom has either an octet (C and N), or 2
electrons (H)
The only way to do this is:
H-C=N:
There is a C-N triple bond and a lone pair on the N.
Now, the steric number is defined as the number of bonds to another atom + the number of lone
pairs on the atom.
For H: 1 bond = steric number of 1.
For C: 4 bonds (1 single + 1 triple) = steric number of 4 (carbon almost always has SN = 4)
For N: 3 bonds (1 triple) + 1 lone pair = steric number of 4 (nitrogen also often has SN = 4)
Solution
First let\'s draw out the Lewis structure such that each atom has either an octet (C and N), or 2
electrons (H)
The only way to do this is:
H-C=N:
There is a C-N triple bond and a lone pair on the N.
Now, the steric number is defined as the number of bonds to another atom + the number of lone
pairs on the atom.
For H: 1 bond = steric number of 1.
For C: 4 bonds (1 single + 1 triple) = steric number of 4 (carbon almost always has SN = 4)
For N: 3 bonds (1 triple) + 1 lone pair = steric number of 4 (nitrogen also often has SN = 4).
To complete the task, you need to fill in the missing code. I’ve inc.pdfezycolours78
To complete the task, you need to fill in the missing code. I’ve included code to create an
Iterator. An Iterator is an object that iterates over another object – in this case, a circular linked
list. You can use the .next() method to advance the Iterator to the next item (the first time you
call it, the iterator will travel to node at index 0). Using iterator’s .remove() removes the node the
iterator is currently at. Say that we had a CircularLinkedList that looked like this: A ==> B ==>
C ==> D ==> E ==> Calling .next() three times will advance the iterator to index 2. Calling
.remove() once will remove the node at index 2. A ==> B ==> D ==> E ==> Calling .remove()
once more will remove the node now at index 2. A ==> B ==> E ==> The Iterator methods
handle wrapping around the CircularLinkedList. Be sure to create the iterator using l.iterator()
and after you’ve added all the nodes to the list, where l is your CircularLinkedList
Solution
import java.util.Iterator;
class CircularLinkedList implements Iterable {
// Your variables
// You can include a reference to a tail if you want
Node head;
int size; // BE SURE TO KEEP TRACK OF THE SIZE
// implement this constructor
public CircularLinkedList() {
head=null;
}
// writing helper functions for add and remove, like the book did can help
// but remember, the last element\'s next node will be the head!
// attach a node to the end of the list
// Be sure to handle the adding to an empty list
// always returns true
public boolean add(E e) {
Node newNode=new Node(e);
if(size==0){
head=newNode;
}
else{
Node last=getNode(size-1);
last.next=newNode;
}
newNode.next=head; //last element node is set to head
size++;
return true;
}
// need to handle
// out of bounds
// empty list
// adding to front
// adding to middle
// adding to \"end\"
// REMEMBER TO INCREMENT THE SIZE
public boolean add(int index, E e){
if(index>size) return false;
Node tmp=new Node(e);
if(index==0){
tmp.next=head;
Node last=getNode(size-1);
head=tmp;
last.next=head;
}
else {
Node curr=getNode(index-1);
tmp.next=curr.next;
curr.next=tmp;
}
size++;
return true;
}
// I highly recommend using this helper method
// Return Node found at the specified index
// be sure to handle out of bounds cases
private Node getNode(int index ) {
Node prev=head;
for(int i=0;isize) {
e=null;
}
else if(index==0){
e = head.getElement();
Node last=getNode(size-1);
head=head.next;
last.next=head;
size--;
}
else{
Node prev=getNode(index-1);
Node curr=getNode(index);
e=curr.getElement();
prev.next=curr.next;
size--;
}
return e;
}
// Turns your list into a string
// Useful for debugging
public String toString(){
Node current = head;
StringBuilder result = new StringBuilder();
if(size == 0){
return \"\";
}
if(size == 1) {
return head.getElement().toString();
}
else{
do{
result.append(current.getElement());
result.append(\" ==> \");
current = current.next;
} while(current != head);
}
return result.toString();
}
public Iterator iterator() {
return new ListIterator();
}
/.
Objectives- In this lab- students will practice- 1- Stack with single.pdfAugstore
Objectives: In this lab, students will practice: 1. Stack with single link-list 2. Circular Link-list
Introduction Implement a stack using singly linked list To implement a stack using singly linked
list concept, all the singly linked list operations are performed based on Stack operations
LIFO(last in first out) and with the help of that knowledge we are going to implement a stack
using single linked list. Using singly linked lists, we implement stack by storing the information
in the form of nodes and we need to follow the stack rules and implement using singly linked list
nodes. So we need to follow a simple rule in the implementation of a stack which is last in first
out and all the operations can be performed with the help of a top variable .Let us learn how to
perform Pop, Push, Peek, Display operations in the following article. Page 2 of 5 A stack can be
easily implemented using the linked list. In stack Implementation, a stack contains a top pointer.
which is "head" of the stack where pushing and popping items happens at the head of the list.
First node have null in link field and second node link have first node address in link field and so
on and last node address in "top" pointer. The main advantage of using linked list over an arrays
is that it is possible to implement a stack that can shrink or grow as much as needed. In using
array will put a restriction to the maximum capacity of the array which can lead to stack
overflow. Here each new node will be dynamically allocate. so overflow is not possible. Stack
Operations: 1. push() : Insert a new element into stack i.e just inserting a new element at the
beginning of the linked list. Page 3 of 5 2. pop() : Return top element of the Stack i.e simply
deleting the first element from the linked list. 3. peek(): Return the top element. 4. display():
Print all elements in Stack. Circular linked list Introduction: Circular Linked List is a variation of
Linked list in which the first element points to the last element and the last element points to the
first element. Both Singly Linked List and Doubly Linked List can be made into a circular linked
list. Singly Linked List as Circular In singly linked list, the next pointer of the last node points to
the first node. Doubly Linked List as Circular In doubly linked list, the next pointer of the last
node points to the first node and the previous pointer of the first node points to the last node
making the circular in both directions. Question 1 1. Implement a template-based stack using a
singly linked list. The required member methods are: int size(): returns the count of total element
stored in the stack. bool isEmpty(): returns true if the stack is empty else false. bool top(T\&):
returns, but does not delete, the topmost element from the stack via the parameter passed by
reference. It returns false via a return statement if there is no element in the stack, else it returns
true and assigns the top most element to the parameter passed by ref.
First lets draw out the Lewis structure such that each atom has.pdfaradhana9856
First let\'s draw out the Lewis structure such that each atom has either an octet (C and N), or 2
electrons (H)
The only way to do this is:
H-C=N:
There is a C-N triple bond and a lone pair on the N.
Now, the steric number is defined as the number of bonds to another atom + the number of lone
pairs on the atom.
For H: 1 bond = steric number of 1.
For C: 4 bonds (1 single + 1 triple) = steric number of 4 (carbon almost always has SN = 4)
For N: 3 bonds (1 triple) + 1 lone pair = steric number of 4 (nitrogen also often has SN = 4)
Solution
First let\'s draw out the Lewis structure such that each atom has either an octet (C and N), or 2
electrons (H)
The only way to do this is:
H-C=N:
There is a C-N triple bond and a lone pair on the N.
Now, the steric number is defined as the number of bonds to another atom + the number of lone
pairs on the atom.
For H: 1 bond = steric number of 1.
For C: 4 bonds (1 single + 1 triple) = steric number of 4 (carbon almost always has SN = 4)
For N: 3 bonds (1 triple) + 1 lone pair = steric number of 4 (nitrogen also often has SN = 4).
13. The answer is C. 34Let D be the dominant allele and d be the.pdfaradhana9856
13. The answer is: C. 3/4
Let D be the dominant allele and d be the recessive allele.
Since father and mother are both heterozygous: Dd x DdDdDDD (affected)Dd (affected)dDd
(affected)dd (unaffected)
Solution
13. The answer is: C. 3/4
Let D be the dominant allele and d be the recessive allele.
Since father and mother are both heterozygous: Dd x DdDdDDD (affected)Dd (affected)dDd
(affected)dd (unaffected).
1) The Arrhenius Theory of acids and bases Acids are substances wh.pdfaradhana9856
1) The Arrhenius Theory of acids and bases: Acids are substances which produce hydrogen
ions in solution. Bases are substances which produce hydroxide ions in solution. Neutralisation
happens because hydrogen ions and hydroxide ions react to produce water. HCl(acid) + NaOH
(base) = H2O + NaCl 2) The Bronsted-Lowry Theory of acids and bases An acid is a proton
(hydrogen ion) donor. A base is a proton (hydrogen ion) acceptor. H2O (base) + HCl(acid) =
H3O^+ + Cl^- 3) The Lewis Theory of acids and bases: An acid is an electron pair acceptor. A
base is an electron pair donor. BH3(acid) + NH3(base) = H3B:NH3 (where NH3 donotes one
pair of electrons to the empty orbital of BH3 to form H3BNH3 complex.)
Solution
1) The Arrhenius Theory of acids and bases: Acids are substances which produce hydrogen
ions in solution. Bases are substances which produce hydroxide ions in solution. Neutralisation
happens because hydrogen ions and hydroxide ions react to produce water. HCl(acid) + NaOH
(base) = H2O + NaCl 2) The Bronsted-Lowry Theory of acids and bases An acid is a proton
(hydrogen ion) donor. A base is a proton (hydrogen ion) acceptor. H2O (base) + HCl(acid) =
H3O^+ + Cl^- 3) The Lewis Theory of acids and bases: An acid is an electron pair acceptor. A
base is an electron pair donor. BH3(acid) + NH3(base) = H3B:NH3 (where NH3 donotes one
pair of electrons to the empty orbital of BH3 to form H3BNH3 complex.).
1. Coenzyme A - C. A water soluble acyl group carrier.2. Thiamine .pdfaradhana9856
1. Coenzyme A - C. A water soluble acyl group carrier.
2. Thiamine pyrophosphate -B. A coenzyme frequently used in decarboxylation reactions.
3. Lipoate - A. The disulfide bond at its end enables it to participate in redox reactions as well as
carry acyl groups.
4. Folic acid - D. A one-carbon group donor. Essential to rapidly growing tissues. Recommended
for women during pregnancy.
Solution
1. Coenzyme A - C. A water soluble acyl group carrier.
2. Thiamine pyrophosphate -B. A coenzyme frequently used in decarboxylation reactions.
3. Lipoate - A. The disulfide bond at its end enables it to participate in redox reactions as well as
carry acyl groups.
4. Folic acid - D. A one-carbon group donor. Essential to rapidly growing tissues. Recommended
for women during pregnancy..
Using Clausius-Clapeyron relation Ln(P2P1) = (D.pdfaradhana9856
Using Clausius-Clapeyron relation: Ln(P2/P1) = (DH/R)*(1/T1-1/T2) where DH
reads Delta H, and is the Hvap you are looking for when P1641,P2=1atm, T1=81.2C, and
T2=99C.
Solution
Using Clausius-Clapeyron relation: Ln(P2/P1) = (DH/R)*(1/T1-1/T2) where DH
reads Delta H, and is the Hvap you are looking for when P1641,P2=1atm, T1=81.2C, and
T2=99C..
purines are double ringed nitrogenous bases of DN.pdfaradhana9856
purines are double ringed nitrogenous bases of DNA and RNA including adenine
and guanine. the three others are cytosine and thymine( uracil is present instead of thymine in
RNA) are pyrimidines. So your answer should be (B) Improve Answer
Solution
purines are double ringed nitrogenous bases of DNA and RNA including adenine
and guanine. the three others are cytosine and thymine( uracil is present instead of thymine in
RNA) are pyrimidines. So your answer should be (B) Improve Answer.
no reaction note 1-butanol is a primary alcohol.pdfaradhana9856
no reaction note: 1-butanol is a primary alcohol that cannot react with Lucas
reagent. both tertiary, secondary alcohols can react with this reagent.
Solution
no reaction note: 1-butanol is a primary alcohol that cannot react with Lucas
reagent. both tertiary, secondary alcohols can react with this reagent..
What type of variable is number of people in the roomNominal.pdfaradhana9856
What type of variable is number of people in the room?
Nominal
Ratio
Interval
Ordinal
What type of variable is number of people in the room?
Nominal
Ratio
Interval
Ordinal
Nominal
Ratio
Interval
Ordinal
Nominal
Ratio
Interval
Ordinal
Solution
Ratio. Although the number of people in a room is integral, there is a well-defined 0 and the
difference between numbers of people in a room is meaningful..
What is the oxidation number for oxygen in H2O2; for OF2 Explain..pdfaradhana9856
What is the oxidation number for oxygen in H2O2; for OF2 ? Explain.
Solution
In H2O2
The oxidation number of Oxygen is -1. The oxidation number of Oxygen in compounds is
usually -2, but it is -1 in peroxides.
Another Way
The oxidation number of H is +1. For two H atoms, the sum is +2.
Since H2O2 is a neutral compound (no charge), (The sum of the oxidation numbers of all the
atoms in a neutral compound is 0.)
The total oxidation number of the two O atoms must be -2.
Hence the oxidation number of one oxygen atom must be -1.
For OF2
The electronegativity of F is 4.0 and of O 3.5, so Oxygen is surely (+) in this compound.
Oxidation number of F2 is -1
Let oxidation number of O be n
n+ 2*(-1) = 0
n= 2
oxidation number of Oxyfen is 2..
We all know humans are good at removing the competition and it is di.pdfaradhana9856
We all know humans are good at removing the competition and it is difficult for any other
species to dominate or even evolve to super intelligent like us. With large brains evolved and we
had taught ourselves how to control other animals as they tends to have not so evolved brains
like humans. There can be seen in zoo, circus dancing on the tunes of human’s orders.
Since humans are self-centred, each and everything is designed or developed is based on the
same feeling human is only species which is important, others are just to serve for its pleasure,
food, and etc. we believe we can do anything we want with animals and their habitats.
Due to industrialization and continue growth of big cities, there is war like condition between
animals and cities. And history has teaches us in number of times, the strong one who has more
number wins the war. Same is the case we can see in NYC, we humans has tendency to get
threaten by whole species even an individual attacks one single person, which itself is rare case.
In this regard there tends to forget these lands was initially dominated by animals, later humans
have conquered and started ruling it.
Solution
We all know humans are good at removing the competition and it is difficult for any other
species to dominate or even evolve to super intelligent like us. With large brains evolved and we
had taught ourselves how to control other animals as they tends to have not so evolved brains
like humans. There can be seen in zoo, circus dancing on the tunes of human’s orders.
Since humans are self-centred, each and everything is designed or developed is based on the
same feeling human is only species which is important, others are just to serve for its pleasure,
food, and etc. we believe we can do anything we want with animals and their habitats.
Due to industrialization and continue growth of big cities, there is war like condition between
animals and cities. And history has teaches us in number of times, the strong one who has more
number wins the war. Same is the case we can see in NYC, we humans has tendency to get
threaten by whole species even an individual attacks one single person, which itself is rare case.
In this regard there tends to forget these lands was initially dominated by animals, later humans
have conquered and started ruling it..
Theories devised by keynes and Minsky are similar in concept since M.pdfaradhana9856
Theories devised by keynes and Minsky are similar in concept since Minsky himself was a post-
Keynesian. He supported quite much of Keynes\' theories and rather contributed to the
economics of financial markets by providing theories on why such markets fail and how they
recover. So they are a more advanced extension of Keynes\' theory to explain the behaviour of
financial markets and how they influence the overall economy.
For understanding the similarities between the two theories, we need to look at some of their
main points in brief, without going into detail and then compare them on where they stood
unique shared the same concept. Since nearly very business cycle theory explains through some
common points, their mode can be different, for Keynes\' theory explained business cycle on a
macro economic scale while Minsky\'s was more market related. Some of the main key points
for Keynes\' and Minsky\'s theories are listed down below.
Keynes\' theory of business cycles
Minsky, as I already told, was a Keynesian himself and all his theories have been always been an
extension of Keynesian concepts to explain the economics of today. His most valuable thesis on
the workings of financial markets why they fail becasue of fluctuations is a Kynesian modifed
concept.
Minsky proposed his theory in pure financial terminology. His point was that during propserity
when cash flows are maximum to pay off the debts, the resulting euphoria that develops into
society results in a debt crises that the borrowers are unable to pay. This leads to banks
restructing money supply which in the end contracts the whole economy.
Solution
Theories devised by keynes and Minsky are similar in concept since Minsky himself was a post-
Keynesian. He supported quite much of Keynes\' theories and rather contributed to the
economics of financial markets by providing theories on why such markets fail and how they
recover. So they are a more advanced extension of Keynes\' theory to explain the behaviour of
financial markets and how they influence the overall economy.
For understanding the similarities between the two theories, we need to look at some of their
main points in brief, without going into detail and then compare them on where they stood
unique shared the same concept. Since nearly very business cycle theory explains through some
common points, their mode can be different, for Keynes\' theory explained business cycle on a
macro economic scale while Minsky\'s was more market related. Some of the main key points
for Keynes\' and Minsky\'s theories are listed down below.
Keynes\' theory of business cycles
Minsky, as I already told, was a Keynesian himself and all his theories have been always been an
extension of Keynesian concepts to explain the economics of today. His most valuable thesis on
the workings of financial markets why they fail becasue of fluctuations is a Kynesian modifed
concept.
Minsky proposed his theory in pure financial terminology. His point was that durin.
They might have consumed cyanide.The pit of Apricot contains cyani.pdfaradhana9856
They might have consumed cyanide.
The pit of Apricot contains cyanide.
Cyanide acts by inhibiting the enzyme cytochrome c oxidase of the electron transport chain
which prevents the cell from producing ATP which is important for aerobic respiration of cells.
Hydroxocobalamin, Sodium thiosulfate, Sodium nitrite are the common antedotes used in
cyanide poisoning
Solution
They might have consumed cyanide.
The pit of Apricot contains cyanide.
Cyanide acts by inhibiting the enzyme cytochrome c oxidase of the electron transport chain
which prevents the cell from producing ATP which is important for aerobic respiration of cells.
Hydroxocobalamin, Sodium thiosulfate, Sodium nitrite are the common antedotes used in
cyanide poisoning.
The attacking of the nucleophile formed in the reactionof acetone an.pdfaradhana9856
The attacking of the nucleophile formed in the reactionof acetone and benzaldehyde on most
preferntially on the carbonylcarbon of the benzaldehyde but not on the acetaldehyde. This
isbecause the electrophilic cheractor of the carbonyl carbon of thebenzaldehyde is increases due
to the presence of the electron withdrawing benzene. Hence it readily participate in the
reactioncompare with the acetaldehyde.
Ethanol is polar solvent. When washes a compound withthis solvent remove the impurities has
the polar nature.
Solution
The attacking of the nucleophile formed in the reactionof acetone and benzaldehyde on most
preferntially on the carbonylcarbon of the benzaldehyde but not on the acetaldehyde. This
isbecause the electrophilic cheractor of the carbonyl carbon of thebenzaldehyde is increases due
to the presence of the electron withdrawing benzene. Hence it readily participate in the
reactioncompare with the acetaldehyde.
Ethanol is polar solvent. When washes a compound withthis solvent remove the impurities has
the polar nature..
Terry And Hill can develop a method of profit sharing on below facto.pdfaradhana9856
Terry And Hill can develop a method of profit sharing on below factors
Unequal capital contribution
Compensate by paying interest on capital - Compensate for the uneven contributions towards
capital by paying interest on their capital contributions. Greater the Capital contributed, greater
the interest earned. This would set right the difference in contributions in the form of capital.
Profit equal to the total \"Interest on Capital\" on all the capital of the firm is set aside and is
distributed to the partners in the ratio of their capitals.Rest of the profits can be shared equally if
all other contributions are equal.
Unequal time and efforts -
Compensate by paying Salary to Partner. - Compensate for the uneven contributions of time and
efforts towards the firm by paying salary to partner. The salary paid to Hill would be
compensation for his greater contribution. Profit equal to the total \"Salary to Partners\" to all the
partners who have made a contribution in this respect is set aside and is distributed to them in
proportion to the value of their contributions as agreed upon between partners. Rest of the profits
can be shared equally if all other contributions are equal.
and if terry and hill are not interested in compensating by above methods then profit sharing by
effort in running the partnership is more preferable as its helping in generating income of
partnership.
Solution
Terry And Hill can develop a method of profit sharing on below factors
Unequal capital contribution
Compensate by paying interest on capital - Compensate for the uneven contributions towards
capital by paying interest on their capital contributions. Greater the Capital contributed, greater
the interest earned. This would set right the difference in contributions in the form of capital.
Profit equal to the total \"Interest on Capital\" on all the capital of the firm is set aside and is
distributed to the partners in the ratio of their capitals.Rest of the profits can be shared equally if
all other contributions are equal.
Unequal time and efforts -
Compensate by paying Salary to Partner. - Compensate for the uneven contributions of time and
efforts towards the firm by paying salary to partner. The salary paid to Hill would be
compensation for his greater contribution. Profit equal to the total \"Salary to Partners\" to all the
partners who have made a contribution in this respect is set aside and is distributed to them in
proportion to the value of their contributions as agreed upon between partners. Rest of the profits
can be shared equally if all other contributions are equal.
and if terry and hill are not interested in compensating by above methods then profit sharing by
effort in running the partnership is more preferable as its helping in generating income of
partnership..
The algorithm to reverse a linked list by rearranging the required p.pdfaradhana9856
The algorithm to reverse a linked list by rearranging the required pointers is given as follows:
The pseudo-code to show the working of the algorithm is given as follows:
//Define the function to reverse a list.
node* reverseList(node* head)
{
//Declare and initialize the required pointer variables.
node* prev_node;
node* cur_node;
node* next_node;
prev_node = NULL;
next_node = NULL;
cur_node = head;
//Traverse the whole list using while loop.
while(cur_node != NULL)
{
//Rearrange the pointers to reverse the list.
next_node = cur_node->next;
cur_node->next = prev_node;
prev_node = cur_node;
cur_node = next_node;
}
//Make the recent previous node as head of the list.
head = prev_node;
//Return the head of the list.
return head;
}
Solution
The algorithm to reverse a linked list by rearranging the required pointers is given as follows:
The pseudo-code to show the working of the algorithm is given as follows:
//Define the function to reverse a list.
node* reverseList(node* head)
{
//Declare and initialize the required pointer variables.
node* prev_node;
node* cur_node;
node* next_node;
prev_node = NULL;
next_node = NULL;
cur_node = head;
//Traverse the whole list using while loop.
while(cur_node != NULL)
{
//Rearrange the pointers to reverse the list.
next_node = cur_node->next;
cur_node->next = prev_node;
prev_node = cur_node;
cur_node = next_node;
}
//Make the recent previous node as head of the list.
head = prev_node;
//Return the head of the list.
return head;
}.
Splicing mechanismSmall nuclear ribonucleic acid (snRNA) It is al.pdfaradhana9856
Splicing mechanism
Small nuclear ribonucleic acid (snRNA): It is also known as U-RNA as it contains many uridine
contents in its complex structure. This is composed of 150 nucleotides nearly to produce
spliceosomes and to act along with snRNP. This snRNA has a predominant role in splicing. The
roles of U1snRNA is different to that of U4snRNA and U1snRNA is mainly going to mediate
splicing by binding to 5\'-splice site when snRNP (small nuclear ribonucleoprotein particles) to
remove latriant introns but the U4snRNA is going to
Peptidyl transferase rRNA is different in their function to U1snRNA,U4snRNA in RNP as
ribozymes and these ribozymes are enzymatic RNA molecules & they are going to mediate RNA
splicing. The peptidyl transferase activity is possessed by 5S and 23S rRNA of ribosomes. The
role of U4snRNA is going to form a complex to form as U1/U2/U4/U5 on pre-mRNA as
spliceosome to remove introns. However, U4snRNA is specifically has specific role in 3 prime
end of hnRNA
Solution
Splicing mechanism
Small nuclear ribonucleic acid (snRNA): It is also known as U-RNA as it contains many uridine
contents in its complex structure. This is composed of 150 nucleotides nearly to produce
spliceosomes and to act along with snRNP. This snRNA has a predominant role in splicing. The
roles of U1snRNA is different to that of U4snRNA and U1snRNA is mainly going to mediate
splicing by binding to 5\'-splice site when snRNP (small nuclear ribonucleoprotein particles) to
remove latriant introns but the U4snRNA is going to
Peptidyl transferase rRNA is different in their function to U1snRNA,U4snRNA in RNP as
ribozymes and these ribozymes are enzymatic RNA molecules & they are going to mediate RNA
splicing. The peptidyl transferase activity is possessed by 5S and 23S rRNA of ribosomes. The
role of U4snRNA is going to form a complex to form as U1/U2/U4/U5 on pre-mRNA as
spliceosome to remove introns. However, U4snRNA is specifically has specific role in 3 prime
end of hnRNA.
since both are strong acids we can add up their individual contribut.pdfaradhana9856
since both are strong acids we can add up their individual contributions towards H+
concentration
[H+]=(4.4+5.3)*10-2=(9.7)*10-2
pH=2-log(9.7)=1.013
Solution
since both are strong acids we can add up their individual contributions towards H+
concentration
[H+]=(4.4+5.3)*10-2=(9.7)*10-2
pH=2-log(9.7)=1.013.
public int getPosition(T anObject) { int result = null; result .pdfaradhana9856
public int getPosition(T anObject)
{ int result = null; // result to return
int lastIndex = length - 1;
for(int index=0; index < lastIndex;index++){
if(anObject.equals(list[index]))
{
result = index;
}
}
return result;
} // end getPosition
Solution
public int getPosition(T anObject)
{ int result = null; // result to return
int lastIndex = length - 1;
for(int index=0; index < lastIndex;index++){
if(anObject.equals(list[index]))
{
result = index;
}
}
return result;
} // end getPosition.
PresentationThe capability of enormous information - or the new .pdfaradhana9856
Presentation
The capability of enormous information - or \"the new oil,\" as a few CIOs and industry
specialists have named it - appears as perpetual as it is subtle. Huge information battles are in
their early stages, with endeavors of all stripes making sense of how to utilize new, old,
unstructured and outer information to make a focused procedure.
Despite the fact that the standard procedures for get-together information and investigating its
value are as yet coming to fruition, organizations know they have to get in the diversion. They
are gathering and mining information on clients, workers, market flow, the climate, and so on,
with instruments going from conventional business insight (BI) frameworks to more trial ones,
for example, geospatial and constant versatile following innovations, online networking
investigation and NoSQL databases.
SearchCIO isn\'t remaining on the sidelines, either. Our Essential Guide on enormous
information incorporates a preliminary for beginning with information social affair and
investigation, true contextual investigations from the CIO and business viewpoints, tips on the
best way to beat hindrances experienced by the huge information pioneers, and expectations on
the following huge information boondocks and what it implies for aggressive methodology.
This aide on the development of huge information is a piece of SearchCIO\'s CIO Briefings
arrangement, which is intended to give IT pioneers vital administration and basic leadership
guidance on opportune themes.
The most effective method to Collect Big Data ?
1 year agoby Ayush1 Comment
The most effective method to Collect Big Data ? : Yes we knoe you would have various inquiries
in your psyche like Collection of Big Data, How organizations gather Big Data, how to gather
information for quantitative research so don\'t stress, in the event that you are here to scan for
these inquiries here then you are on the right website page as here we are going to give you a
complete article on Collection of Big Data techniques quickly.
Astounding Facts about Rise of Big Data Collectection
Consistently buyers make around 11.5 million installments by utilizing Paypal
Consistently, Walmart (chain of rebate retail chains) handles more than 1 million client
exchanges
510 remarks, 293000 status and 136000 overhauls are posted on Facebook consistently
Consistently, ~7000 tweets are made on Twitter
Simply picture the measure of information created if the above details are figured for 24 hours?
Whoa! That is huge.
The term \'Enormous Data\' is ordinarily connected with 4V\'s to be specific, Velocity, Volume,
Variety, Veracity. These 4V\'s appropriately speaks to the genuine way of Big Data. Each \"V\"
has a noteworthy part to play in the presence of Big Data. On the off chance that consolidated,
these 4V paints a wonderful clarification of Big Data which can be comprehended as \" Big Data
as an idea alludes to high speed gathering of information in expansive volumes which radia.
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
Antifertility, Toxicity studies as per OECD guidelines
The Roman Empire A Historical Colossus.pdfkaushalkr1407
The Roman Empire, a vast and enduring power, stands as one of history's most remarkable civilizations, leaving an indelible imprint on the world. It emerged from the Roman Republic, transitioning into an imperial powerhouse under the leadership of Augustus Caesar in 27 BCE. This transformation marked the beginning of an era defined by unprecedented territorial expansion, architectural marvels, and profound cultural influence.
The empire's roots lie in the city of Rome, founded, according to legend, by Romulus in 753 BCE. Over centuries, Rome evolved from a small settlement to a formidable republic, characterized by a complex political system with elected officials and checks on power. However, internal strife, class conflicts, and military ambitions paved the way for the end of the Republic. Julius Caesar’s dictatorship and subsequent assassination in 44 BCE created a power vacuum, leading to a civil war. Octavian, later Augustus, emerged victorious, heralding the Roman Empire’s birth.
Under Augustus, the empire experienced the Pax Romana, a 200-year period of relative peace and stability. Augustus reformed the military, established efficient administrative systems, and initiated grand construction projects. The empire's borders expanded, encompassing territories from Britain to Egypt and from Spain to the Euphrates. Roman legions, renowned for their discipline and engineering prowess, secured and maintained these vast territories, building roads, fortifications, and cities that facilitated control and integration.
The Roman Empire’s society was hierarchical, with a rigid class system. At the top were the patricians, wealthy elites who held significant political power. Below them were the plebeians, free citizens with limited political influence, and the vast numbers of slaves who formed the backbone of the economy. The family unit was central, governed by the paterfamilias, the male head who held absolute authority.
Culturally, the Romans were eclectic, absorbing and adapting elements from the civilizations they encountered, particularly the Greeks. Roman art, literature, and philosophy reflected this synthesis, creating a rich cultural tapestry. Latin, the Roman language, became the lingua franca of the Western world, influencing numerous modern languages.
Roman architecture and engineering achievements were monumental. They perfected the arch, vault, and dome, constructing enduring structures like the Colosseum, Pantheon, and aqueducts. These engineering marvels not only showcased Roman ingenuity but also served practical purposes, from public entertainment to water supply.
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdfTechSoup
In this webinar you will learn how your organization can access TechSoup's wide variety of product discount and donation programs. From hardware to software, we'll give you a tour of the tools available to help your nonprofit with productivity, collaboration, financial management, donor tracking, security, and more.
Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
Overview on Edible Vaccine: Pros & Cons with Mechanism
(i) Deques cannot be implemented using arrays. false Justif.pdf
1. (i) Deques cannot be implemented using arrays. : false
Justification:
We can implement dequeue using circular array. We need to maintain two pointers rear and
front.
Ex :
1. Create an array of any element type like int [] dequeue.
2. Create two indexes front and rear and initialize them to 0.
3. Create an element capacity.(it should be length of an array - 1).
(ii) If we require all add/remove operations to run in O(1) time, we cannot use a Deque
implemented with a singly linked list. False
Justification :
In linklist there will be two pointers head and tail.
Adding an element :
tail -> next = newelement;
newelement -> next = NULL;
tail = newelement;
These all are O(1) operation.
Removing an element :
head = head->next; O(1) operation
(iii) A Deque can be used as a stack as well as a queue. : true
Justification :
Using as stack : You can use dequeue as stack by removing and inserting the element at
one end.
Using as queue : You can use dequeue as queue by inserting the element at front and
removing at rear.
(iv) Assume the Deque contains n elements and n is odd. The middle element is the e lement
equidistant from either end. The middle element cannot be determined in O(1) time in any
implementation : false
2. Justification :
You can achieve this in O(1) operation with the help of doubly link list. You will have to
maintain a mid pointer to achieve this. You can move this pointer in either direction with previos
and next pointers.
Solution
(i) Deques cannot be implemented using arrays. : false
Justification:
We can implement dequeue using circular array. We need to maintain two pointers rear and
front.
Ex :
1. Create an array of any element type like int [] dequeue.
2. Create two indexes front and rear and initialize them to 0.
3. Create an element capacity.(it should be length of an array - 1).
(ii) If we require all add/remove operations to run in O(1) time, we cannot use a Deque
implemented with a singly linked list. False
Justification :
In linklist there will be two pointers head and tail.
Adding an element :
tail -> next = newelement;
newelement -> next = NULL;
tail = newelement;
These all are O(1) operation.
Removing an element :
head = head->next; O(1) operation
(iii) A Deque can be used as a stack as well as a queue. : true
Justification :
Using as stack : You can use dequeue as stack by removing and inserting the element at
one end.
Using as queue : You can use dequeue as queue by inserting the element at front and
3. removing at rear.
(iv) Assume the Deque contains n elements and n is odd. The middle element is the e lement
equidistant from either end. The middle element cannot be determined in O(1) time in any
implementation : false
Justification :
You can achieve this in O(1) operation with the help of doubly link list. You will have to
maintain a mid pointer to achieve this. You can move this pointer in either direction with previos
and next pointers.
![(i) Deques cannot be implemented using arrays. : false
Justification:
We can implement dequeue using circular array. We need to maintain two pointers rear and
front.
Ex :
1. Create an array of any element type like int [] dequeue.
2. Create two indexes front and rear and initialize them to 0.
3. Create an element capacity.(it should be length of an array - 1).
(ii) If we require all add/remove operations to run in O(1) time, we cannot use a Deque
implemented with a singly linked list. False
Justification :
In linklist there will be two pointers head and tail.
Adding an element :
tail -> next = newelement;
newelement -> next = NULL;
tail = newelement;
These all are O(1) operation.
Removing an element :
head = head->next; O(1) operation
(iii) A Deque can be used as a stack as well as a queue. : true
Justification :
Using as stack : You can use dequeue as stack by removing and inserting the element at
one end.
Using as queue : You can use dequeue as queue by inserting the element at front and
removing at rear.
(iv) Assume the Deque contains n elements and n is odd. The middle element is the e lement
equidistant from either end. The middle element cannot be determined in O(1) time in any
implementation : false](https://image.slidesharecdn.com/idequescannotbeimplementedusingarrays-230629080809-da70cef4/85/i-Deques-cannot-be-implemented-using-arrays-false-Justif-pdf-1-320.jpg)
![Justification :
You can achieve this in O(1) operation with the help of doubly link list. You will have to
maintain a mid pointer to achieve this. You can move this pointer in either direction with previos
and next pointers.
Solution
(i) Deques cannot be implemented using arrays. : false
Justification:
We can implement dequeue using circular array. We need to maintain two pointers rear and
front.
Ex :
1. Create an array of any element type like int [] dequeue.
2. Create two indexes front and rear and initialize them to 0.
3. Create an element capacity.(it should be length of an array - 1).
(ii) If we require all add/remove operations to run in O(1) time, we cannot use a Deque
implemented with a singly linked list. False
Justification :
In linklist there will be two pointers head and tail.
Adding an element :
tail -> next = newelement;
newelement -> next = NULL;
tail = newelement;
These all are O(1) operation.
Removing an element :
head = head->next; O(1) operation
(iii) A Deque can be used as a stack as well as a queue. : true
Justification :
Using as stack : You can use dequeue as stack by removing and inserting the element at
one end.
Using as queue : You can use dequeue as queue by inserting the element at front and](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)