how to estimate the fermi temperature using any value of mSolut.pdf
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how to estimate the fermi temperature using any value of m* Solution All the particles are at the same temperature, regardless of their energy. Temperature is a property of a distribution of energies of particles not a single particle . The Fermi Dirac distribution notice that relation Ef/Kb. The right units to be called temperature. It turns out the system behave more and more differently beyond that temperature. Fermions may be at the same temperature is purely quantum mechanics. Fermi temperature is that it is just another way expressing the Fermi energy. Ef=kbTf. Which is similar to E=nkbT..
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Many primate females form dominance hierarchies becausethey compete .pdf
Many primate females form dominance hierarchies becausethey compete with one another over
access to food.
Solution
In dominance hierchies individuals live in close proximity.it is a type of social structure seen in
primate female in which certain show dominance over others ; and those dominant female
primate claim access to food..
MATLAB CODE Write a function with the header which calculates the p.pdf
MATLAB CODE: Write a function with the header which calculates the partial derivative using
the central difference method of function f with respect to x (ind). The central difference
approximation of a function evaluated at x is: f = f (x + h)-f (x - h)/2epsilon and eps is used to
permute x (ind) such that
Solution
The Code is here for function
------------------------------------------------------
function [df ] = myPartial(f,x,ind,eps )
%MYPARTIAL Summary of this function goes here
% Detailed explanation goes here
if ind==1
h=[eps;0;0];
elseif ind==2
h=[0;eps;0];
elseif ind==3
h=[0;0;eps];
end
df=(f(x+h)-f(x-h))/(2*eps);
end
-----------------------------------------------------------------.
Methods in Behavioral Research 12th EditionIdentify the independen.pdf
Methods in Behavioral Research 12th Edition
Identify the independent and dependent variables in the following descriptions of experiments:
7. Students watched a cartoon either alone or with others and then rated how funny they found
the cartoon to be.
8. A comprehension test was given to students after they had studied textbook material either in
silence or with the television turned on.
9. Some elementary school teachers were told that a child’s parents were college graduates, and
other teachers were told that the child’s parents had not finished high school; they then rated the
child’s academic potential.
10. Workers at a company were assigned to one of two conditions: one group completed a stress
management-training program; another group of workers did not participate in the training. The
number of sick days taken by these workers was examined for the two subsequent months.
Solution
1. Independent variable: Watching cartoon with others
Dependent variable: How funny the cartoon was rated to be
2. Independent variable: Television
Dependent variable: Comprehension test
3. Independent variable: Telling the teachers whether the parents were college graduates or not
finished school
Dependent variable: Rating the child\'s academic potential
4. Independent variable: Completing a stress management-training program
Dependent variable: Number of sick days taken by workers..
look to determine what is the earliest era that fossils have been fo.pdf
look to determine what is the earliest era that fossils have been found here in Georgia. Next, see
if you can find what the oldest fossil that has been found in Georgia.
Solution
The oldest fossils found in Georgia (Trilobites) belong to 500 million years ago. They belong to
the Palaeozoic era (541 to 252.17 million years ago).
The oldest human fossils found in Georgia are dated back to 1.75 million years ago (Cenozoic
era)..
Javathis is my current code i need help to implement public void s.pdf
Java
this is my current code i need help to implement public void sort method (TODO part)
import java.util.List;
import java.util.ArrayList;
import java.util.Collections;
class MergeSort {
public static void sort(List list) {
//TODO call the method private static void sort with parameters(length of input list, 0,0);
}
private static void sort(List list, int startIndex, int endIndex){
int length = endIndex - startIndex;
if(length <= 1) {
return;
}
int middleIndex = startIndex + (length/2);
sort(list,startIndex,middleIndex);
sort(list,middleIndex,endIndex);
int upperStart = startIndex;
int upperEnd = startIndex + (length/2);
int lowerStart = upperEnd;
int lowerEnd = endIndex;
int i = upperStart;
int j = lowerStart;
List tmpList = new ArrayList();
while ( i < upperEnd || j < lowerEnd){
if(i >= upperEnd){
tmpList.add(list.get(j));
j++;
}
else if(j >= lowerEnd) {
tmpList.add(list.get(i));
i++;
}
else{
if(list.get(i) < list.get(j)){
tmpList.add(list.get(i));
i++;
}
else{
tmpList.add(list.get(j));
j++;
}
}
for(int k = startIndex; k < endIndex; k++){
list.set(k,tmpList.get(k - startIndex));
}
}
}
}
Solution
import java.util.List;
import java.util.ArrayList;
import java.util.Collections;
class MergeSort {
public static void sort(List list) {
//call the method private static void sort with parameters(length of input list, 0,0);
sort(list.size(), 0,0);
}
private static void sort(List list, int startIndex, int endIndex){
int length = endIndex - startIndex;
if(length <= 1) {
return;
}
int middleIndex = startIndex + (length/2);
sort(list,startIndex,middleIndex);
sort(list,middleIndex,endIndex);
int upperStart = startIndex;
int upperEnd = startIndex + (length/2);
int lowerStart = upperEnd;
int lowerEnd = endIndex;
int i = upperStart;
int j = lowerStart;
List tmpList = new ArrayList();
while ( i < upperEnd || j < lowerEnd){
if(i >= upperEnd){
tmpList.add(list.get(j));
j++;
}
else if(j >= lowerEnd) {
tmpList.add(list.get(i));
i++;
}
else{
if(list.get(i) < list.get(j)){
tmpList.add(list.get(i));
i++;
}
else{
tmpList.add(list.get(j));
j++;
}
}
for(int k = startIndex; k < endIndex; k++){
list.set(k,tmpList.get(k - startIndex));
}
}
}
}.
java as a beginnerSolutionimport statements import java.io.pdf
java as a beginner
Solution
//import statements
import java.io.*;
import java.util.Scanner;
import java.util.ArrayList;
public class Calculation{
//main method
public static void main(String[] args) {
//try so that we can catch file input exception
try{
//taking input from file called numbers.txt
Scanner scanner = new Scanner(new File(\"numbers.txt\"));
//declaring numbers arraylist and not array since we don\'t know the number of integers
present in the file
ArrayList numbers = new ArrayList();
//reading integers from file and adding it to numbers arraylist
while(scanner.hasNextInt()){
numbers.add(scanner.nextInt());
}
//initialising sum
int sum = 0;
//iterating through the list and adding the number
for (int i=0; i.
Implementing list using linked list Implementing list using linked l.pdf
Implementing list using linked list Implementing list using linked list size java.lang.Integer
size() Determine the size of this list. Returns: the number of elements in this list
Solution
Below is the size() method for LinkedList
public class LinkedList
{ //Start of LinkedList class
private Node first;
public LinkedList()
{
first = null;
}
public Integersize()
{
int s = 0;
Node current = first;
while(current!=null)
{
s++;
current = current.next;
}
return s;
}
}.
In corn, kernels are usually smooth due to the Su allele. The recess.pdf
In corn, kernels are usually smooth due to the Su allele. The recessive su allele gives the kernel a
wrinkled appearance. Likewise, kernel color in corn can either be colorless due to the recessive r
allele or purple due to the dominant R allele. You wish to deduce the genotype of a smooth,
purple kernel for your experiment. What are all the possible genotypes this corn kernel could be
at the smooth and purple loci? What genotype and phenotype of corn should you use in order to
do a testcross? What are the phenotypic ratios expected from this testcross for each of the
possible mystery kernel genotypes?
Solution
Answering the first part of this question we need to see that there are two allele in this Corn, one
for the Kernel smoothness/roughness- Su/ su allele, another is the colour Purple / colorless- R/ r.
So for the corn to be genotypically contain smooth and purple color there should be both the
dominant allele that is Su for smoothness and R for the purple trait, so wr can say that it has to be
SuR type genotype to make smooth and purple appearance. To understand it from the use of
punnet square- Totlly there will be 9 of them which can produce Smooth and Purple appearance.
F2 GENERATION: SuR * sur
SuR Sur suR sur * SuR Sur suR sur
SuSuRR
(Smooth and Purple)
SuSuRr
(Smooth and Purple)
SusuRR
(Smooth and Purple)
SusuRr
(Smooth and Purple)
SuSuRr
(Smooth and Purple)
SuSurr
(Smooth and colorless)
SusurR
(Smooth and Purple)
Susurr
(Smooth and colorless)
suSuRR
(Smooth and Purple)
suSuRr
(Smooth and Purple)
susuRR
(Wrinkled and purple)
susuRr
(Wrinkled and purple)
suSurR
(Smooth and Purple)
suSurr
(Smooth and colorless)
susurR
(Wrinkled and purple)
susurr
(Wrinkled and colourless)
The testcross is shown above in the punnet sqaure, we need to use the dominat allele of SuR
(Smoot and Purple) against recessive allele sur (Wrinkled and colorless).
Phenotypic ratio will be 9:3:3:1.
SuSuRR
(Smooth and Purple)
SuSuRr
(Smooth and Purple)
SusuRR
(Smooth and Purple)
SusuRr
(Smooth and Purple)
SuSuRr
(Smooth and Purple)
SuSurr
(Smooth and colorless)
SusurR
(Smooth and Purple)
Susurr
(Smooth and colorless)
suSuRR
(Smooth and Purple)
suSuRr
(Smooth and Purple)
susuRR
(Wrinkled and purple)
susuRr
(Wrinkled and purple)
suSurR
(Smooth and Purple)
suSurr
(Smooth and colorless)
susurR
(Wrinkled and purple)
susurr
(Wrinkled and colourless).
Recommended
Many primate females form dominance hierarchies becausethey compete .pdf
Many primate females form dominance hierarchies becausethey compete with one another over
access to food.
Solution
In dominance hierchies individuals live in close proximity.it is a type of social structure seen in
primate female in which certain show dominance over others ; and those dominant female
primate claim access to food..
MATLAB CODE Write a function with the header which calculates the p.pdf
MATLAB CODE: Write a function with the header which calculates the partial derivative using
the central difference method of function f with respect to x (ind). The central difference
approximation of a function evaluated at x is: f = f (x + h)-f (x - h)/2epsilon and eps is used to
permute x (ind) such that
Solution
The Code is here for function
------------------------------------------------------
function [df ] = myPartial(f,x,ind,eps )
%MYPARTIAL Summary of this function goes here
% Detailed explanation goes here
if ind==1
h=[eps;0;0];
elseif ind==2
h=[0;eps;0];
elseif ind==3
h=[0;0;eps];
end
df=(f(x+h)-f(x-h))/(2*eps);
end
-----------------------------------------------------------------.
Methods in Behavioral Research 12th EditionIdentify the independen.pdf
Methods in Behavioral Research 12th Edition
Identify the independent and dependent variables in the following descriptions of experiments:
7. Students watched a cartoon either alone or with others and then rated how funny they found
the cartoon to be.
8. A comprehension test was given to students after they had studied textbook material either in
silence or with the television turned on.
9. Some elementary school teachers were told that a child’s parents were college graduates, and
other teachers were told that the child’s parents had not finished high school; they then rated the
child’s academic potential.
10. Workers at a company were assigned to one of two conditions: one group completed a stress
management-training program; another group of workers did not participate in the training. The
number of sick days taken by these workers was examined for the two subsequent months.
Solution
1. Independent variable: Watching cartoon with others
Dependent variable: How funny the cartoon was rated to be
2. Independent variable: Television
Dependent variable: Comprehension test
3. Independent variable: Telling the teachers whether the parents were college graduates or not
finished school
Dependent variable: Rating the child\'s academic potential
4. Independent variable: Completing a stress management-training program
Dependent variable: Number of sick days taken by workers..
look to determine what is the earliest era that fossils have been fo.pdf
look to determine what is the earliest era that fossils have been found here in Georgia. Next, see
if you can find what the oldest fossil that has been found in Georgia.
Solution
The oldest fossils found in Georgia (Trilobites) belong to 500 million years ago. They belong to
the Palaeozoic era (541 to 252.17 million years ago).
The oldest human fossils found in Georgia are dated back to 1.75 million years ago (Cenozoic
era)..
Javathis is my current code i need help to implement public void s.pdf
Java
this is my current code i need help to implement public void sort method (TODO part)
import java.util.List;
import java.util.ArrayList;
import java.util.Collections;
class MergeSort {
public static void sort(List list) {
//TODO call the method private static void sort with parameters(length of input list, 0,0);
}
private static void sort(List list, int startIndex, int endIndex){
int length = endIndex - startIndex;
if(length <= 1) {
return;
}
int middleIndex = startIndex + (length/2);
sort(list,startIndex,middleIndex);
sort(list,middleIndex,endIndex);
int upperStart = startIndex;
int upperEnd = startIndex + (length/2);
int lowerStart = upperEnd;
int lowerEnd = endIndex;
int i = upperStart;
int j = lowerStart;
List tmpList = new ArrayList();
while ( i < upperEnd || j < lowerEnd){
if(i >= upperEnd){
tmpList.add(list.get(j));
j++;
}
else if(j >= lowerEnd) {
tmpList.add(list.get(i));
i++;
}
else{
if(list.get(i) < list.get(j)){
tmpList.add(list.get(i));
i++;
}
else{
tmpList.add(list.get(j));
j++;
}
}
for(int k = startIndex; k < endIndex; k++){
list.set(k,tmpList.get(k - startIndex));
}
}
}
}
Solution
import java.util.List;
import java.util.ArrayList;
import java.util.Collections;
class MergeSort {
public static void sort(List list) {
//call the method private static void sort with parameters(length of input list, 0,0);
sort(list.size(), 0,0);
}
private static void sort(List list, int startIndex, int endIndex){
int length = endIndex - startIndex;
if(length <= 1) {
return;
}
int middleIndex = startIndex + (length/2);
sort(list,startIndex,middleIndex);
sort(list,middleIndex,endIndex);
int upperStart = startIndex;
int upperEnd = startIndex + (length/2);
int lowerStart = upperEnd;
int lowerEnd = endIndex;
int i = upperStart;
int j = lowerStart;
List tmpList = new ArrayList();
while ( i < upperEnd || j < lowerEnd){
if(i >= upperEnd){
tmpList.add(list.get(j));
j++;
}
else if(j >= lowerEnd) {
tmpList.add(list.get(i));
i++;
}
else{
if(list.get(i) < list.get(j)){
tmpList.add(list.get(i));
i++;
}
else{
tmpList.add(list.get(j));
j++;
}
}
for(int k = startIndex; k < endIndex; k++){
list.set(k,tmpList.get(k - startIndex));
}
}
}
}.
java as a beginnerSolutionimport statements import java.io.pdf
java as a beginner
Solution
//import statements
import java.io.*;
import java.util.Scanner;
import java.util.ArrayList;
public class Calculation{
//main method
public static void main(String[] args) {
//try so that we can catch file input exception
try{
//taking input from file called numbers.txt
Scanner scanner = new Scanner(new File(\"numbers.txt\"));
//declaring numbers arraylist and not array since we don\'t know the number of integers
present in the file
ArrayList numbers = new ArrayList();
//reading integers from file and adding it to numbers arraylist
while(scanner.hasNextInt()){
numbers.add(scanner.nextInt());
}
//initialising sum
int sum = 0;
//iterating through the list and adding the number
for (int i=0; i.
Implementing list using linked list Implementing list using linked l.pdf
Implementing list using linked list Implementing list using linked list size java.lang.Integer
size() Determine the size of this list. Returns: the number of elements in this list
Solution
Below is the size() method for LinkedList
public class LinkedList
{ //Start of LinkedList class
private Node first;
public LinkedList()
{
first = null;
}
public Integersize()
{
int s = 0;
Node current = first;
while(current!=null)
{
s++;
current = current.next;
}
return s;
}
}.
In corn, kernels are usually smooth due to the Su allele. The recess.pdf
In corn, kernels are usually smooth due to the Su allele. The recessive su allele gives the kernel a
wrinkled appearance. Likewise, kernel color in corn can either be colorless due to the recessive r
allele or purple due to the dominant R allele. You wish to deduce the genotype of a smooth,
purple kernel for your experiment. What are all the possible genotypes this corn kernel could be
at the smooth and purple loci? What genotype and phenotype of corn should you use in order to
do a testcross? What are the phenotypic ratios expected from this testcross for each of the
possible mystery kernel genotypes?
Solution
Answering the first part of this question we need to see that there are two allele in this Corn, one
for the Kernel smoothness/roughness- Su/ su allele, another is the colour Purple / colorless- R/ r.
So for the corn to be genotypically contain smooth and purple color there should be both the
dominant allele that is Su for smoothness and R for the purple trait, so wr can say that it has to be
SuR type genotype to make smooth and purple appearance. To understand it from the use of
punnet square- Totlly there will be 9 of them which can produce Smooth and Purple appearance.
F2 GENERATION: SuR * sur
SuR Sur suR sur * SuR Sur suR sur
SuSuRR
(Smooth and Purple)
SuSuRr
(Smooth and Purple)
SusuRR
(Smooth and Purple)
SusuRr
(Smooth and Purple)
SuSuRr
(Smooth and Purple)
SuSurr
(Smooth and colorless)
SusurR
(Smooth and Purple)
Susurr
(Smooth and colorless)
suSuRR
(Smooth and Purple)
suSuRr
(Smooth and Purple)
susuRR
(Wrinkled and purple)
susuRr
(Wrinkled and purple)
suSurR
(Smooth and Purple)
suSurr
(Smooth and colorless)
susurR
(Wrinkled and purple)
susurr
(Wrinkled and colourless)
The testcross is shown above in the punnet sqaure, we need to use the dominat allele of SuR
(Smoot and Purple) against recessive allele sur (Wrinkled and colorless).
Phenotypic ratio will be 9:3:3:1.
SuSuRR
(Smooth and Purple)
SuSuRr
(Smooth and Purple)
SusuRR
(Smooth and Purple)
SusuRr
(Smooth and Purple)
SuSuRr
(Smooth and Purple)
SuSurr
(Smooth and colorless)
SusurR
(Smooth and Purple)
Susurr
(Smooth and colorless)
suSuRR
(Smooth and Purple)
suSuRr
(Smooth and Purple)
susuRR
(Wrinkled and purple)
susuRr
(Wrinkled and purple)
suSurR
(Smooth and Purple)
suSurr
(Smooth and colorless)
susurR
(Wrinkled and purple)
susurr
(Wrinkled and colourless).
If a concentration gradient has been established across a membrane, t.pdf
If a concentration gradient has been established across a membrane, the permeable solute will
diffuse across the membrane from the side with the higher concentration to the side with the
lower concentration so slowly that there will be no movement of these molecules, even though
the membrane is permeable to them. from the side with the lower concentration to the side with
the higher concentration. In both directions, but with no \"net\" movement In any particular
direction.
Solution
Answer: (a)
If a concentration gradient has been established across a membrane, the permeable solute will
diffuse across the membrane from the side with the higher concentration to the side with the
lower concentration.
Passive transport is the movement of solutes and molecular substances across cell membranes
without need of energy input. It is the net movement of solutes across the cell membrane along
the concentration gradient and continues until an equilibrium has been reached.
Simple diffusion and osmosis are in someway similar. Simple diffusion is the passive movement
of solute from a high concentration to a lower concentration until the concentration of the solute
is uniform. Osmosis describes the movement of water (not the solute) across a selectively
permeable membrane until there is an equal concentration of water and solute on both sides of
the membrane..
I need help to define each one. thank you. Define Acute enterocoli.pdf
I need help to define each one. thank you. Define: Acute enterocolitis Enterohemorrhagic
Escherichia coli Typhoid fever Bacillary dysentery Travellers\' diarrhea
Solution
Acute enterocolitis: - It is a disease caused by various bacteria, viruses, fungi, and parasites. It is
an inflammation of the colon and small intestine caused by eating or drinking substances that are
contaminated with the disease-causing organism. It may result in fever, nausea, vomiting,
diarrhea, and abdominal pain.
Enterohemorrhagic E.coli: - Enterohemorrhagic Escherichia coli is a subdivision of disease
causing E. coli. which can cause diarrhea or hemorrhagic colitis in humans. It produces a toxin
that is capable of killing intestinal cells by disrupting their protein synthesis and can ultimately
lead to kidney damage, platelet aggregation, anemia, and death.
Typhoid fever: - It is an infection that is spread by contaminated food and drinks. Disease
causing organism is Salmonella typhi. The illness is characterized by a headache, nausea, fever,
and constipation or sometimes diarrhea.
Bacillary dysentery: - Also known as shigellosis, is a disease caused by poor hygiene. Bacillary
dysentery is caused by a group of bacteria of the genus, Shigella which can cause intestinal
inflammation and symptoms of gastroenteritis without dysentery.
Travelers diarrhea: - A disease often caused due to a new diet, traveling abroad, a change in
climate, stress or due to lack of sleep. It is spread by contaminated food and drinks and is
characterized by the passage of three or more loose motions in a day, accompanied by fever,
abdominal cramps, nausea or vomiting..
How do I find the repeated ration number 0.415, where 15 is repeat.pdf
How do I find the repeated ration number 0.415, where 15 is repeating?
How do I find the repeated ration number 0.415, where 15 is repeating?
Solution
A repeating decimal, also called a recurring decimal, is a number whose decimal representation
eventually becomes periodic (i.e., the same sequence of digits repeats indefinitely).
0.415 15 1515 =8303 / 20000.
Describe three early signs of shock and the rationale for each.So.pdf
Describe three early signs of shock and the rationale for each.
Solution
shock is defined as an acute clinical syndrome characterized by hypoperfusion and severe
dysfunction of vital organs. it is a result of circulatory collapse.
early signs/first signs
paleness: and excessive sweating due to reduced blood supply to perpheries like skin due to
perpheral pooling of blood ( moves into central circulation)
weak wide pulse due to initially due to increase adrenaline release and subsequent peripheral
vasoconstriction and increased diastolic pressure
tachycardia due to more work of heart . in shock vital organs need blood supply to survice so
heart work more to maintain the perfusion to vital organs
dilated pupils due to sympathetic stimulation
increased breathing frequency ( tachypnea) due to more wok of lung to compensate metabolic
disturbence ( production of more co2 to which lung work more to send out and provide more
oxygen to tissue aganist muscle impairment.
Describe a real world application for right angle trigonometry. Expl.pdf
Describe a real world application for right angle trigonometry. Explain in 3-4 sentences.
Solution
Trigonometry has a wide range of applications in many fields and one of the application is that it
is used to calculate distance between planets and even astronomers use trigonometry to observe
the path of planets in which they are moving and prepare the astrological survey..
Define a class template for an AVL tree. Create an object of su.pdf
Define a class template for an AVL tree.
Create an object of such tree.
Create a menu where you can insert, delete, and find elements.
The menu must also include a print function that will list the preorder, inorder, and postorder
visualization of the tree (Your print function must call all tree traversal types at the same time).
The function insert must have 2 choices: insert_one() and insert_k().
insert_one() will ask the user for one value to get inserted.
insert_k() will ask the user for the size of input (k) and a list of k elements will be read from the
console/file.
Both functions must give the option to populate it with: Int, Float or Character values.
(Character values will be ordered based on their ASCII value).
Once a tree of any type is created you must be able to manipulate it at all times.
If a new tree of a different type is created, don\'t replace your current tree with it.
You can potentially have one tree of each type at any given time.
Write a function to find the depth of a given node and the depth of the tree.
Solution
#include #include #include #define FALSE 0 #define TRUE 1 struct AVLNode { int data
; int balfact ; AVLNode *left ; AVLNode *right ; } ; class avltree { private
: AVLNode *root ; public : avltree( ) ; AVLNode* insert ( int
data, int *h ) ; static AVLNode* buildtree ( AVLNode *root, int data, int *h ) ;
void display( AVLNode *root ) ; AVLNode* deldata ( AVLNode* root, int data, int
*h ) ; static AVLNode* del ( AVLNode *node, AVLNode* root, int *h ) ;
static AVLNode* balright ( AVLNode *root, int *h ) ; static AVLNode* balleft (
AVLNode* root, int *h ) ; void setroot ( AVLNode *avl ) ; ~avltree( ) ;
static void deltree ( AVLNode *root ) ; } ; avltree :: avltree( ) { root = NULL ; }
AVLNode* avltree :: insert ( int data, int *h ) { root = buildtree ( root, data, h ) ; return
root ; } AVLNode* avltree :: buildtree ( AVLNode *root, int data, int *h ) { AVLNode
*node1, *node2 ; if ( root == NULL ) { root = new AVLNode ;
root -> data = data ; root -> left = NULL ; root -> right = NULL ;
root -> balfact = 0 ; *h = TRUE ; return ( root ) ; } if ( data < root -
> data ) { root -> left = buildtree ( root -> left, data, h ) ; // If left
subtree is higher if ( *h ) { switch ( root -> balfact )
{ case 1 : node1 = root -> left ;
if ( node1 -> balfact == 1 ) { cout
<< \"\ Right rotation.\" ; root -> left = node1 -> right ;
node1 -> right = root ; root -> balfact = 0 ;
root = node1 ; } else
{ cout << \"\ Double rotation, left then right.\" ;
node2 = node1 -> right ; node1 ->
right = node2 -> left ; node2 -> left = node1 ;
root -> left = node2 -> right ; node2 -> right = root ;
if ( node2 -> balfact == 1 ) root ->
balfact = -1 ; else root -> balfact
= 0 ; if ( node2 -> balfact == -1 )
node1 -> balfact = 1 ; else
node1 -> balfact = 0 ; root = node2 ; }
root -> balfact = 0 ; *h = FALSE ;
break ; case 0 : root -> balfact = 1 ;
break ; case -1 : root -> balfact = 0 ;
*h = FALSE ; } } } if ( data > root -
> data ) { root -> right.
Consider a collection of identical boards, each with mass m, length L.pdf
Consider a collection of identical boards, each with mass m, length L, and uniform mass density.
Clearly lacking other forms of entertainment, we decide place a board on a ledge, then a second
board on top of the first board, then a third board on top of the second board, and so on. Let x_n
represent the greatest possible horizontal distance between far edge of the uppermost board and
the ledge, such that the assembly of boards may remain stationary. Prove the recursive relation
for x_n, x_n= X_n-1 + L/2n Prove the explicit licit form of x_n Imagine Romeo is attempting
to sneak into Juliet\'s nd story room. Lord Capulet has coated the ding wall with a material so
slippery that is has no
Solution
a) Length of the identical Blocks is L and Mass of of the Blocks is M
We have to calulate the center of mass of the structure . The distance between the centre of mass
of the lower blockand right hand edge of the bottom blockwe\'ll define as L/2
From this we can write the equation
X2(2M)=(X1)M
Here X1=L/2
X2(2M)=(L/2)M
X2=L/4
at this point over hang of the block is X1+X2=L/2+L/4=3L/4
For third block
X3(3M)=(L/2)M
X3=L/6
at this point over hang of the block is
X1+X2+X3=L/2+L/4+L/6=11L/12
This value is close to L.
For fourth Block
X4(4M)=(L/2)M
X4=L/8
At this point over hang of the Block X1+X2+X3+X4=L/2+L/4+L/6+L/8=25L/24. This is greater
than one
Xn(nM)=(L/2)M
The displacement for each block is L/2n
Hence we can write The recursive relation
Xn=Xn-1+L/2n
b) Over hang =L/2 (Summation of (k =1 to n)).
CH27Quiz aect.pdf
CH27Quiz aecture 4) Help Save Exit submit The new model of evolution that proposes that
organisms spend most of their tme in fairly unchanging form and undergo occasional bursts of
speciation is called Multiple Choice genetic drift. phyletic gradualism. sympatric speciation
punctuated equilibrium Prev 25 of 25 t
Solution
25) punctuated equilibrium
The theory of evolution based on patters of first appearance and subsequent histories of species
in the fossil record is called as punctuated equilibrium.
24) recent common ancestor and allits descendents
A cladogram is a diagramatic representation to show relations among organisms witha a recent
common ancestor. It differs from an evolutionary tree as it does not show how ancestors are
related to descendants, nor does it show how much they have changed thus many evolutionary
trees can be inferred from a single cladogram.
26) Originally one species but now two species.
The insect originally same species but due to geographic barriers over time a new species
develoved.
27) a geographic separation occurs between subpopulation.
Allopatric speciation is also called as geographic speciation. This type of speciation occurs when
a biologically same species become a victim of geographically barrier to an extent that interferes
with genetic interchange.
28)reproductive isolation between certain subpopulation occurs.
Evolution of a new species from a single ancestral species while inhabiting the same geographic
region is called as Sympatric speciation..
Based on the given abstract and introduction to this journal article.pdf
Based on the given abstract and introduction to this journal article, what is/are:
a.) The goals and need of the study?
b.) The theory/theories used to frame the research? MOLECULAR ECOLOGY Molecular
Eoology (20ts) 24,5813-SE28 Genetic diversity loss in a biodiversity hotspot: ancient DNA
quantifies genetic decline and former connectivity in a critically endangered marsupial NTOFT, t
TIMOTHY G. VAUGHAN MORTEN E. ALLE CARLO PACIONI HELEN HUNT ADRIAN
F. WAYNE, ALEXANDER BAYNES, DALAL HAOUCHAR, JOE DORTCH\" and
MICHAEL BUNCE tt wA 62ss, Australie, sniedem Antralinn Masnam. Lacked Beg 49,
Weldrpool DC, WA 6986, Parker and Willlife, Marjamap, M257, unirersiyof western Auestralia.
35 Stirling Highnny, Nindlamas, WA 6009, Australia. DTartwamt Eneinartment and
Agricalturer. Tror and Environmental DNA Laboratory, Kent Street. Bently, Perti, WA Abstract
The extent of genetic diversity loss and former connectivity between fragmented popu- lations
are often unknown factors when studying endangered species. While genetic commonly applied
in extant Populations to assess temporal and spatial it is no substitute for directly measuring past
diversity using emographic changes, ancient DNA (aDNA. We analysed both mitochondrial
DNA (mtDNA) and nuclear microsatellite loci from 64 historical fossil and skin samples of the
critically endan- gered Western Australian woylie (Bettongia penicillata ogilbyi, and compared
them with 231 152 for mtDNA) modem samples. In modern woylie Populations 15 mito-
chondrial control region (CR) haplotypes were identified. Interestingly, mtDNA CR data from
only 29 historical samples demonstrated 15 previously unknown haplotypes and detected extinct
divergent clade. Through modelling we estimated the loss of CR mtDNA diversity to be between
46% and 91% and estimated this to have occurred in the past 2000-4000 years in association
with a dramatic population decline. In addi tion, we obtained near-com plete 11-oci
microsatellite profiles from 21 historical sam- ples. In agreement with the mtDNA data, a
number of new\" microsatellite alleles was only detected in the historical populations despite
extensive modem sampling indicat. ing a nuclear genetic diversity loss 20%. Calculations of
genetic diversity 0heterozy gosity and alleli rarefaction) showed that these were significantly
higher in the past and that there was a high degree of gene flow across the woylie\'s historical
range. These findings have an immediate impact on how the extant populations are managed and
we recommend the implementation of an assisted migration programme further of genetic
diversity. our study demonstrates the value of integrating data conservation strategies. ancient
DNA Bettorgia, biodiversity loss, genetic diversity, population botdeneck Racinad 22 May 2014
reision receioed 7 October 2015. accepted 13 october 2015 Introduction spedes and populations
that are adversely affected by Correspondence Michael Bunce, Fax: +él 8926 2495; E-mail
anthropogenic activit.
9-42 The hospital administrator at St. Charles General must appoin.pdf
9-42 The hospital administrator at St. Charles General
must appoint head nurses to four newly established
departments: urology, cardiology, orthopedics, and
obstetrics. In anticipation of this staffing problem,
she had hired four nurses: Hawkins, Condriac, Bardot, and Hoolihan. Believing in the
quantitative
analysis approach to problem solving, the adminis-
trator has interviewed each nurse, considered his or
her background, personality, and talents, and devel-
oped a cost scale ranging from 0 to 100 to be used in
the assignment. A 0 for Nurse Bardot being assigned
to the cardiology unit implies that she would be per-
fectly suited to that task. A value close to 100, on the
other hand, would imply that she is not at all suited
to head that unit. The accompanying table gives the
complete set of cost figures that the hospital admin-
istrator felt represented all possible assignments.
Which nurse should be assigned to which unit?
What is the solution?
NURSE UROLOGY CARDIOLOGY ORTHOPEDICS OBSTETRICS
Hawkins 28 18 15 75
Condriac 32 48 23 38
Bardot 51 36 24 36
Hoolihan 25 38 55 1
Solution
Hawkins will head Cardiology
Condriac wil head Urology
ON the cost scale nurse Bardot will cost the hospital 24 if he/she heads the orthopedics
department , and this cost is the least for bardot against the remaining departments(if she/he\'
assigned to them)
=>Bardot will head orthopedics
Likewise Hoolihan will head the obstetrics as he/she\'ll cost just 1 for this department
on the cost scale
=> Hoolihan will head Obstetrics.
624 Chapter Twenty-Two The Candiovarcular system. Vessels and Cimalat.pdf
624 Chapter Twenty-Two The Candiovarcular system. Vessels and Cimalation 13. Label the
diagram with the appropriate vein namas. O-G 13 supersoial wins Deep veins (b) Veins, anterior
view 30
Solution
1.Subclavian
2. Branchiocephalic
3. INternal thoracic vein
4.Axillary vein
5. Brachial veins
6. Hepatic veins
7. inferior vena cava
8. Renal veins.
9. Superior mesentric vein
10. Ulnar
11. Radial
12. femoral vein
13. popliteal vein
14. Anterior - posterior tibial veins
15. sphenous veins.
16. facial vein
17. Jugular vein
18. Subclavian vein
19. brachiocphalic vein
20. Superior Vena Cava
21. Basilic vein
22. Cephalic veins
23. Median cubital vein
24. Gonadal vein
25. Common Iliac vein
26. External iliac vein
27. Internal iliac vein
28. Deep femoral vein
29. Great sephanous vein
30. dorsal digital vein..
4. What type up bond would form between two atoms of chlorine (Cl), .pdf
4. What type up bond would form between two atoms of chlorine (Cl), a nonmetal?
A. metallic bond
B. nonpolar covalent bond
C. polar covalent bond
D. ionic bond
E. none of these
5. If neutral atoms become positive ions, they
A. gain electrons.
B. lose electrons.
C. gain protons.
D. lose neutrons.
E. gain neutrons.
6. What is the name of the functional group --OH?
A. amino
B. methyl
C. carboxyl
D. hydroxyl
E. oxyhydry
Solution.
• Leo Williams is single and 45 years old• Leo works as an IT mana.pdf
• Leo Williams is single and 45 years old
• Leo works as an IT manager and his Form W-2 shows wages of $47,250
• Leo participated in his employer’s self-only coverage High Deductible Health Plan (HDHP) all
year
• Leo does not have any other health coverage
• Leo has had an HSA for two years
• Leo’s employer contributed $1,500 in 2017 to Leo’s HSA
• In 2017, Leo’s Aunt contributed $2,000 to Leo’s HSA
• Leo is a U S citizen and has a valid Social Security number
1. For Leo to be an eligible individual and qualify for an HSA, which of the following
must be true?
a He cannot be claimed as a dependent on someone else’s return
b He must have an HDHP and cannot be enrolled in Medicare
c He cannot have any other health coverage
d All of the above
2. Where on Form 1040 would Leo report his HSA deduction?
a Payments section
b Income section
c Adjusted Gross Income section
d None of the above
3. Employer contributions to an HSA are reported on an employee’s Form W-2
a True
b False
Solution
1) he can not be claimed as a dependent on someone else\'s return .
2) income section
3) True.
Write assembly or C software to implement the following Mealy FSM. I.pdf
Write assembly or C software to implement the following Mealy FSM. Include the FSM state
machine, port initialization, timer initialization, and the FSM controller. The command sequence
will be input, output, wait 10 ms, input, then branch to next state. The 1-bit input is on Port P
(PP0), and the 3-bit output is on Port P (PP3, PP2, PP1). Assume the E clock is 8 MHz.
Solution
The header files for TM4C123 and TM4C1294. Defines all the names and addresses of the
microcontrollers in C99 syntax. These assembly files contain all the port addresses for the
microcontroller. To use this file, you open this file copy the lines you need and paste the lines
into your program. You cannot include the .s files, like you can include the header files in C.
1a- What are three ways that bacteria can exchange genetic informati.pdf
1a- What are three ways that bacteria can exchange genetic information? How are these similar
and how are they different? How can they be used to map bacterial genomes?
b) Explain how genes in phages are mapped. (This is different from the above question)
Solution
Ques-1: What are three ways that bacteria can exchange genetic information? How are these
similar and how are they different? How can they be used to map bacterial genomes?
Answer:
Horizontal gene transfer (HGT) or lateral gene transfer is the transfer of genes between two
organisms that are phylogenetically unrelated but can readily exchange genes by conjugation or
transduction or transformation. HGT commonly occurs in prokaryotes and it plays a key role in
their evolution. HGT accelerates evolution.
Horizontal gene transfer: 1. conjugation, 2. transformation; 3. transduction
Similarities & differences:
Horizontal gene transfer is the transfer of genomes by between bacteria by the routes that does
not involve parent-offspring transmission. Conjugation, transduction and transformation are the
three mechanism of horizontal gene transfer.
In the above case, it has clearly illustrated that the conditions, which can successfully prevents
transduction and abolish transfer of chromosomal DNA from one bacterial cell to another by
generalized transduction with P22 bacteriophage. Initially recipient cell- surface receptors are
important for the attachment of the bacteriophage (P22) and finally spikes of phage enable
bacterial membrane to lyse finally incorporate phage genome into the bacterial cell for
integration. Therefore, for efficient transduction, presence of recipient cell- surface receptors is
essential in the culture otherwise, bacteria exert resistant effects to the phages.
Transformation type of gene transfer occurs between bacteria. The genes released by dead
bacteria can be transferred to the live bacteria, and get integrated into its genome, which results
in genetic recombination of the host.
Transduction is also involving the genetic material transfer, but this needs bacteriophage as
media. The bacteriophage infects the bacteria in lytic cycle and releases the new virions, the
capsids of some of these virions carry the bacterial DNA fragments into the new host bacteria,
followed by genetic recombination of the host bacteria.
Conjugation: It is the transfer of genetic material from cell to cell by means of direct contact or a
bridge like structure. Bacterial conjugation needs the conjugation tube for genome transfer, both
transduction and transformation does not need this.
Mapping the genome -mechanisms
F factor is mainly pertaining to a class of conjugate plasmids, which meticulously originated to
control sexual functions in prokaryotes via fertility inhibition. The origin of F-factor is mainly
with most common functional segments such as gene traJ, Ori, OriC. The structure of this F
factor contains OriT for origin of transfer and acts as a starting point for gene transmission.
Why has the share of healthcare output produced by hospitals fallen.pdf
Why has the share of healthcare output produced by hospitals fallen? Will this trend continue?
What implications does this have for the careers of healthcare managers?
Solution
Mainly, Hospital operating cost such as Labor Wages, doctor salaries, admin salaries, electricity
charges and maintenance charge are go for high, Second major reason is Drug price risen .
Hospital are hike the charge but health care giving nominal amount covered by insurance plan,
So hospitals are fallen. Mostly profit earning hospital having huge fall.
Temporarily it will continue but not for long time i) Hospitals realised its fallen and started to
reduce the excess expenditure and situation has been under control
Careers of healthcare managers is potential and growth in their path, They have to motivate the
patient and co operative with them, They are bridge between inside and outside person
Inside people : Doctor, Staff, Nurse, Light men/Women and some labors
outside people: Patient and visitors
They need knowledge about technology advance such in medical devices and they have to lead
hospital finance in right way with positive..
What three types of inventories are reported on the balance sheet.pdf
What three types of inventories are reported on the balance sheet of a manufacturing company?
What three types of inventories are reported on the balance sheet of a manufacturing company?
Solution
Answer is
Raw Material Inventory = Contains invetoru of Materials required for Manufacturing
Work in Progress Inventory= Contains inventory of goods in Process or semi finished goods
Finished Goods Inventory = Contains Inventory of goods Finished goods.
What are the major components of a CAN 2.0A frameSolutionThe .pdf
What are the major components of a CAN 2.0A frame?
Solution
The CAN standard frame is known as CAN A 2.0 which it supports 11 bits in the identifier.
The following four of them which are indicating below are the major components of the CAN A
2.0
- Agent
- Coverage
- Configuration
- Checkers ( data and protocol)..
What is meant by anthropogenic greenhouse gasesSolutionant.pdf
What is meant by \"anthropogenic greenhouse gases\"?
Solution
anthropogenic means man-made
Since from a long time human activities have greatly increased the concentrations of greenhouse
gases in the atmosphere. Consequently, measured atmospheric concentrations of carbon dioxide
are many times higher than pre-industrial levels.
The main sources of greenhouse gases due to human activity are:.
What country specific standard was developed to evaluate integrity o.pdf
What country specific standard was developed to evaluate integrity of individual systems and is
broken into four categories?
A. ITSEC
B. TCSEC
C. Common Criteria
D. CTCPEC
Solution
B. TCSEC (or the Orange Book) was developed to evaluate the integrity of standalone systems.
Answer A is incorrect because the ITSEC is an international standard developed in Europe.
Answer C is incorrect because Common Criteria is a global standard that built on TCSEC,
ITSEC, and the CTCPEC. Answer D is incorrect because the CTCPEC is the Canadian version
of the Orange Book..
Hindi varnamala | hindi alphabet PPT.pdf
हिंदी वर्णमाला पीपीटी, hindi alphabet PPT presentation, hindi varnamala PPT, Hindi Varnamala pdf, हिंदी स्वर, हिंदी व्यंजन, sikhiye hindi varnmala, dr. mulla adam ali, hindi language and literature, hindi alphabet with drawing, hindi alphabet pdf, hindi varnamala for childrens, hindi language, hindi varnamala practice for kids, https://www.drmullaadamali.com
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If a concentration gradient has been established across a membrane, t.pdf
If a concentration gradient has been established across a membrane, the permeable solute will
diffuse across the membrane from the side with the higher concentration to the side with the
lower concentration so slowly that there will be no movement of these molecules, even though
the membrane is permeable to them. from the side with the lower concentration to the side with
the higher concentration. In both directions, but with no \"net\" movement In any particular
direction.
Solution
Answer: (a)
If a concentration gradient has been established across a membrane, the permeable solute will
diffuse across the membrane from the side with the higher concentration to the side with the
lower concentration.
Passive transport is the movement of solutes and molecular substances across cell membranes
without need of energy input. It is the net movement of solutes across the cell membrane along
the concentration gradient and continues until an equilibrium has been reached.
Simple diffusion and osmosis are in someway similar. Simple diffusion is the passive movement
of solute from a high concentration to a lower concentration until the concentration of the solute
is uniform. Osmosis describes the movement of water (not the solute) across a selectively
permeable membrane until there is an equal concentration of water and solute on both sides of
the membrane..
I need help to define each one. thank you. Define Acute enterocoli.pdf
I need help to define each one. thank you. Define: Acute enterocolitis Enterohemorrhagic
Escherichia coli Typhoid fever Bacillary dysentery Travellers\' diarrhea
Solution
Acute enterocolitis: - It is a disease caused by various bacteria, viruses, fungi, and parasites. It is
an inflammation of the colon and small intestine caused by eating or drinking substances that are
contaminated with the disease-causing organism. It may result in fever, nausea, vomiting,
diarrhea, and abdominal pain.
Enterohemorrhagic E.coli: - Enterohemorrhagic Escherichia coli is a subdivision of disease
causing E. coli. which can cause diarrhea or hemorrhagic colitis in humans. It produces a toxin
that is capable of killing intestinal cells by disrupting their protein synthesis and can ultimately
lead to kidney damage, platelet aggregation, anemia, and death.
Typhoid fever: - It is an infection that is spread by contaminated food and drinks. Disease
causing organism is Salmonella typhi. The illness is characterized by a headache, nausea, fever,
and constipation or sometimes diarrhea.
Bacillary dysentery: - Also known as shigellosis, is a disease caused by poor hygiene. Bacillary
dysentery is caused by a group of bacteria of the genus, Shigella which can cause intestinal
inflammation and symptoms of gastroenteritis without dysentery.
Travelers diarrhea: - A disease often caused due to a new diet, traveling abroad, a change in
climate, stress or due to lack of sleep. It is spread by contaminated food and drinks and is
characterized by the passage of three or more loose motions in a day, accompanied by fever,
abdominal cramps, nausea or vomiting..
How do I find the repeated ration number 0.415, where 15 is repeat.pdf
How do I find the repeated ration number 0.415, where 15 is repeating?
How do I find the repeated ration number 0.415, where 15 is repeating?
Solution
A repeating decimal, also called a recurring decimal, is a number whose decimal representation
eventually becomes periodic (i.e., the same sequence of digits repeats indefinitely).
0.415 15 1515 =8303 / 20000.
Describe three early signs of shock and the rationale for each.So.pdf
Describe three early signs of shock and the rationale for each.
Solution
shock is defined as an acute clinical syndrome characterized by hypoperfusion and severe
dysfunction of vital organs. it is a result of circulatory collapse.
early signs/first signs
paleness: and excessive sweating due to reduced blood supply to perpheries like skin due to
perpheral pooling of blood ( moves into central circulation)
weak wide pulse due to initially due to increase adrenaline release and subsequent peripheral
vasoconstriction and increased diastolic pressure
tachycardia due to more work of heart . in shock vital organs need blood supply to survice so
heart work more to maintain the perfusion to vital organs
dilated pupils due to sympathetic stimulation
increased breathing frequency ( tachypnea) due to more wok of lung to compensate metabolic
disturbence ( production of more co2 to which lung work more to send out and provide more
oxygen to tissue aganist muscle impairment.
Describe a real world application for right angle trigonometry. Expl.pdf
Describe a real world application for right angle trigonometry. Explain in 3-4 sentences.
Solution
Trigonometry has a wide range of applications in many fields and one of the application is that it
is used to calculate distance between planets and even astronomers use trigonometry to observe
the path of planets in which they are moving and prepare the astrological survey..
Define a class template for an AVL tree. Create an object of su.pdf
Define a class template for an AVL tree.
Create an object of such tree.
Create a menu where you can insert, delete, and find elements.
The menu must also include a print function that will list the preorder, inorder, and postorder
visualization of the tree (Your print function must call all tree traversal types at the same time).
The function insert must have 2 choices: insert_one() and insert_k().
insert_one() will ask the user for one value to get inserted.
insert_k() will ask the user for the size of input (k) and a list of k elements will be read from the
console/file.
Both functions must give the option to populate it with: Int, Float or Character values.
(Character values will be ordered based on their ASCII value).
Once a tree of any type is created you must be able to manipulate it at all times.
If a new tree of a different type is created, don\'t replace your current tree with it.
You can potentially have one tree of each type at any given time.
Write a function to find the depth of a given node and the depth of the tree.
Solution
#include #include #include #define FALSE 0 #define TRUE 1 struct AVLNode { int data
; int balfact ; AVLNode *left ; AVLNode *right ; } ; class avltree { private
: AVLNode *root ; public : avltree( ) ; AVLNode* insert ( int
data, int *h ) ; static AVLNode* buildtree ( AVLNode *root, int data, int *h ) ;
void display( AVLNode *root ) ; AVLNode* deldata ( AVLNode* root, int data, int
*h ) ; static AVLNode* del ( AVLNode *node, AVLNode* root, int *h ) ;
static AVLNode* balright ( AVLNode *root, int *h ) ; static AVLNode* balleft (
AVLNode* root, int *h ) ; void setroot ( AVLNode *avl ) ; ~avltree( ) ;
static void deltree ( AVLNode *root ) ; } ; avltree :: avltree( ) { root = NULL ; }
AVLNode* avltree :: insert ( int data, int *h ) { root = buildtree ( root, data, h ) ; return
root ; } AVLNode* avltree :: buildtree ( AVLNode *root, int data, int *h ) { AVLNode
*node1, *node2 ; if ( root == NULL ) { root = new AVLNode ;
root -> data = data ; root -> left = NULL ; root -> right = NULL ;
root -> balfact = 0 ; *h = TRUE ; return ( root ) ; } if ( data < root -
> data ) { root -> left = buildtree ( root -> left, data, h ) ; // If left
subtree is higher if ( *h ) { switch ( root -> balfact )
{ case 1 : node1 = root -> left ;
if ( node1 -> balfact == 1 ) { cout
<< \"\ Right rotation.\" ; root -> left = node1 -> right ;
node1 -> right = root ; root -> balfact = 0 ;
root = node1 ; } else
{ cout << \"\ Double rotation, left then right.\" ;
node2 = node1 -> right ; node1 ->
right = node2 -> left ; node2 -> left = node1 ;
root -> left = node2 -> right ; node2 -> right = root ;
if ( node2 -> balfact == 1 ) root ->
balfact = -1 ; else root -> balfact
= 0 ; if ( node2 -> balfact == -1 )
node1 -> balfact = 1 ; else
node1 -> balfact = 0 ; root = node2 ; }
root -> balfact = 0 ; *h = FALSE ;
break ; case 0 : root -> balfact = 1 ;
break ; case -1 : root -> balfact = 0 ;
*h = FALSE ; } } } if ( data > root -
> data ) { root -> right.
Consider a collection of identical boards, each with mass m, length L.pdf
Consider a collection of identical boards, each with mass m, length L, and uniform mass density.
Clearly lacking other forms of entertainment, we decide place a board on a ledge, then a second
board on top of the first board, then a third board on top of the second board, and so on. Let x_n
represent the greatest possible horizontal distance between far edge of the uppermost board and
the ledge, such that the assembly of boards may remain stationary. Prove the recursive relation
for x_n, x_n= X_n-1 + L/2n Prove the explicit licit form of x_n Imagine Romeo is attempting
to sneak into Juliet\'s nd story room. Lord Capulet has coated the ding wall with a material so
slippery that is has no
Solution
a) Length of the identical Blocks is L and Mass of of the Blocks is M
We have to calulate the center of mass of the structure . The distance between the centre of mass
of the lower blockand right hand edge of the bottom blockwe\'ll define as L/2
From this we can write the equation
X2(2M)=(X1)M
Here X1=L/2
X2(2M)=(L/2)M
X2=L/4
at this point over hang of the block is X1+X2=L/2+L/4=3L/4
For third block
X3(3M)=(L/2)M
X3=L/6
at this point over hang of the block is
X1+X2+X3=L/2+L/4+L/6=11L/12
This value is close to L.
For fourth Block
X4(4M)=(L/2)M
X4=L/8
At this point over hang of the Block X1+X2+X3+X4=L/2+L/4+L/6+L/8=25L/24. This is greater
than one
Xn(nM)=(L/2)M
The displacement for each block is L/2n
Hence we can write The recursive relation
Xn=Xn-1+L/2n
b) Over hang =L/2 (Summation of (k =1 to n)).
CH27Quiz aect.pdf
CH27Quiz aecture 4) Help Save Exit submit The new model of evolution that proposes that
organisms spend most of their tme in fairly unchanging form and undergo occasional bursts of
speciation is called Multiple Choice genetic drift. phyletic gradualism. sympatric speciation
punctuated equilibrium Prev 25 of 25 t
Solution
25) punctuated equilibrium
The theory of evolution based on patters of first appearance and subsequent histories of species
in the fossil record is called as punctuated equilibrium.
24) recent common ancestor and allits descendents
A cladogram is a diagramatic representation to show relations among organisms witha a recent
common ancestor. It differs from an evolutionary tree as it does not show how ancestors are
related to descendants, nor does it show how much they have changed thus many evolutionary
trees can be inferred from a single cladogram.
26) Originally one species but now two species.
The insect originally same species but due to geographic barriers over time a new species
develoved.
27) a geographic separation occurs between subpopulation.
Allopatric speciation is also called as geographic speciation. This type of speciation occurs when
a biologically same species become a victim of geographically barrier to an extent that interferes
with genetic interchange.
28)reproductive isolation between certain subpopulation occurs.
Evolution of a new species from a single ancestral species while inhabiting the same geographic
region is called as Sympatric speciation..
Based on the given abstract and introduction to this journal article.pdf
Based on the given abstract and introduction to this journal article, what is/are:
a.) The goals and need of the study?
b.) The theory/theories used to frame the research? MOLECULAR ECOLOGY Molecular
Eoology (20ts) 24,5813-SE28 Genetic diversity loss in a biodiversity hotspot: ancient DNA
quantifies genetic decline and former connectivity in a critically endangered marsupial NTOFT, t
TIMOTHY G. VAUGHAN MORTEN E. ALLE CARLO PACIONI HELEN HUNT ADRIAN
F. WAYNE, ALEXANDER BAYNES, DALAL HAOUCHAR, JOE DORTCH\" and
MICHAEL BUNCE tt wA 62ss, Australie, sniedem Antralinn Masnam. Lacked Beg 49,
Weldrpool DC, WA 6986, Parker and Willlife, Marjamap, M257, unirersiyof western Auestralia.
35 Stirling Highnny, Nindlamas, WA 6009, Australia. DTartwamt Eneinartment and
Agricalturer. Tror and Environmental DNA Laboratory, Kent Street. Bently, Perti, WA Abstract
The extent of genetic diversity loss and former connectivity between fragmented popu- lations
are often unknown factors when studying endangered species. While genetic commonly applied
in extant Populations to assess temporal and spatial it is no substitute for directly measuring past
diversity using emographic changes, ancient DNA (aDNA. We analysed both mitochondrial
DNA (mtDNA) and nuclear microsatellite loci from 64 historical fossil and skin samples of the
critically endan- gered Western Australian woylie (Bettongia penicillata ogilbyi, and compared
them with 231 152 for mtDNA) modem samples. In modern woylie Populations 15 mito-
chondrial control region (CR) haplotypes were identified. Interestingly, mtDNA CR data from
only 29 historical samples demonstrated 15 previously unknown haplotypes and detected extinct
divergent clade. Through modelling we estimated the loss of CR mtDNA diversity to be between
46% and 91% and estimated this to have occurred in the past 2000-4000 years in association
with a dramatic population decline. In addi tion, we obtained near-com plete 11-oci
microsatellite profiles from 21 historical sam- ples. In agreement with the mtDNA data, a
number of new\" microsatellite alleles was only detected in the historical populations despite
extensive modem sampling indicat. ing a nuclear genetic diversity loss 20%. Calculations of
genetic diversity 0heterozy gosity and alleli rarefaction) showed that these were significantly
higher in the past and that there was a high degree of gene flow across the woylie\'s historical
range. These findings have an immediate impact on how the extant populations are managed and
we recommend the implementation of an assisted migration programme further of genetic
diversity. our study demonstrates the value of integrating data conservation strategies. ancient
DNA Bettorgia, biodiversity loss, genetic diversity, population botdeneck Racinad 22 May 2014
reision receioed 7 October 2015. accepted 13 october 2015 Introduction spedes and populations
that are adversely affected by Correspondence Michael Bunce, Fax: +él 8926 2495; E-mail
anthropogenic activit.
9-42 The hospital administrator at St. Charles General must appoin.pdf
9-42 The hospital administrator at St. Charles General
must appoint head nurses to four newly established
departments: urology, cardiology, orthopedics, and
obstetrics. In anticipation of this staffing problem,
she had hired four nurses: Hawkins, Condriac, Bardot, and Hoolihan. Believing in the
quantitative
analysis approach to problem solving, the adminis-
trator has interviewed each nurse, considered his or
her background, personality, and talents, and devel-
oped a cost scale ranging from 0 to 100 to be used in
the assignment. A 0 for Nurse Bardot being assigned
to the cardiology unit implies that she would be per-
fectly suited to that task. A value close to 100, on the
other hand, would imply that she is not at all suited
to head that unit. The accompanying table gives the
complete set of cost figures that the hospital admin-
istrator felt represented all possible assignments.
Which nurse should be assigned to which unit?
What is the solution?
NURSE UROLOGY CARDIOLOGY ORTHOPEDICS OBSTETRICS
Hawkins 28 18 15 75
Condriac 32 48 23 38
Bardot 51 36 24 36
Hoolihan 25 38 55 1
Solution
Hawkins will head Cardiology
Condriac wil head Urology
ON the cost scale nurse Bardot will cost the hospital 24 if he/she heads the orthopedics
department , and this cost is the least for bardot against the remaining departments(if she/he\'
assigned to them)
=>Bardot will head orthopedics
Likewise Hoolihan will head the obstetrics as he/she\'ll cost just 1 for this department
on the cost scale
=> Hoolihan will head Obstetrics.
624 Chapter Twenty-Two The Candiovarcular system. Vessels and Cimalat.pdf
624 Chapter Twenty-Two The Candiovarcular system. Vessels and Cimalation 13. Label the
diagram with the appropriate vein namas. O-G 13 supersoial wins Deep veins (b) Veins, anterior
view 30
Solution
1.Subclavian
2. Branchiocephalic
3. INternal thoracic vein
4.Axillary vein
5. Brachial veins
6. Hepatic veins
7. inferior vena cava
8. Renal veins.
9. Superior mesentric vein
10. Ulnar
11. Radial
12. femoral vein
13. popliteal vein
14. Anterior - posterior tibial veins
15. sphenous veins.
16. facial vein
17. Jugular vein
18. Subclavian vein
19. brachiocphalic vein
20. Superior Vena Cava
21. Basilic vein
22. Cephalic veins
23. Median cubital vein
24. Gonadal vein
25. Common Iliac vein
26. External iliac vein
27. Internal iliac vein
28. Deep femoral vein
29. Great sephanous vein
30. dorsal digital vein..
4. What type up bond would form between two atoms of chlorine (Cl), .pdf
4. What type up bond would form between two atoms of chlorine (Cl), a nonmetal?
A. metallic bond
B. nonpolar covalent bond
C. polar covalent bond
D. ionic bond
E. none of these
5. If neutral atoms become positive ions, they
A. gain electrons.
B. lose electrons.
C. gain protons.
D. lose neutrons.
E. gain neutrons.
6. What is the name of the functional group --OH?
A. amino
B. methyl
C. carboxyl
D. hydroxyl
E. oxyhydry
Solution.
• Leo Williams is single and 45 years old• Leo works as an IT mana.pdf
• Leo Williams is single and 45 years old
• Leo works as an IT manager and his Form W-2 shows wages of $47,250
• Leo participated in his employer’s self-only coverage High Deductible Health Plan (HDHP) all
year
• Leo does not have any other health coverage
• Leo has had an HSA for two years
• Leo’s employer contributed $1,500 in 2017 to Leo’s HSA
• In 2017, Leo’s Aunt contributed $2,000 to Leo’s HSA
• Leo is a U S citizen and has a valid Social Security number
1. For Leo to be an eligible individual and qualify for an HSA, which of the following
must be true?
a He cannot be claimed as a dependent on someone else’s return
b He must have an HDHP and cannot be enrolled in Medicare
c He cannot have any other health coverage
d All of the above
2. Where on Form 1040 would Leo report his HSA deduction?
a Payments section
b Income section
c Adjusted Gross Income section
d None of the above
3. Employer contributions to an HSA are reported on an employee’s Form W-2
a True
b False
Solution
1) he can not be claimed as a dependent on someone else\'s return .
2) income section
3) True.
Write assembly or C software to implement the following Mealy FSM. I.pdf
Write assembly or C software to implement the following Mealy FSM. Include the FSM state
machine, port initialization, timer initialization, and the FSM controller. The command sequence
will be input, output, wait 10 ms, input, then branch to next state. The 1-bit input is on Port P
(PP0), and the 3-bit output is on Port P (PP3, PP2, PP1). Assume the E clock is 8 MHz.
Solution
The header files for TM4C123 and TM4C1294. Defines all the names and addresses of the
microcontrollers in C99 syntax. These assembly files contain all the port addresses for the
microcontroller. To use this file, you open this file copy the lines you need and paste the lines
into your program. You cannot include the .s files, like you can include the header files in C.
1a- What are three ways that bacteria can exchange genetic informati.pdf
1a- What are three ways that bacteria can exchange genetic information? How are these similar
and how are they different? How can they be used to map bacterial genomes?
b) Explain how genes in phages are mapped. (This is different from the above question)
Solution
Ques-1: What are three ways that bacteria can exchange genetic information? How are these
similar and how are they different? How can they be used to map bacterial genomes?
Answer:
Horizontal gene transfer (HGT) or lateral gene transfer is the transfer of genes between two
organisms that are phylogenetically unrelated but can readily exchange genes by conjugation or
transduction or transformation. HGT commonly occurs in prokaryotes and it plays a key role in
their evolution. HGT accelerates evolution.
Horizontal gene transfer: 1. conjugation, 2. transformation; 3. transduction
Similarities & differences:
Horizontal gene transfer is the transfer of genomes by between bacteria by the routes that does
not involve parent-offspring transmission. Conjugation, transduction and transformation are the
three mechanism of horizontal gene transfer.
In the above case, it has clearly illustrated that the conditions, which can successfully prevents
transduction and abolish transfer of chromosomal DNA from one bacterial cell to another by
generalized transduction with P22 bacteriophage. Initially recipient cell- surface receptors are
important for the attachment of the bacteriophage (P22) and finally spikes of phage enable
bacterial membrane to lyse finally incorporate phage genome into the bacterial cell for
integration. Therefore, for efficient transduction, presence of recipient cell- surface receptors is
essential in the culture otherwise, bacteria exert resistant effects to the phages.
Transformation type of gene transfer occurs between bacteria. The genes released by dead
bacteria can be transferred to the live bacteria, and get integrated into its genome, which results
in genetic recombination of the host.
Transduction is also involving the genetic material transfer, but this needs bacteriophage as
media. The bacteriophage infects the bacteria in lytic cycle and releases the new virions, the
capsids of some of these virions carry the bacterial DNA fragments into the new host bacteria,
followed by genetic recombination of the host bacteria.
Conjugation: It is the transfer of genetic material from cell to cell by means of direct contact or a
bridge like structure. Bacterial conjugation needs the conjugation tube for genome transfer, both
transduction and transformation does not need this.
Mapping the genome -mechanisms
F factor is mainly pertaining to a class of conjugate plasmids, which meticulously originated to
control sexual functions in prokaryotes via fertility inhibition. The origin of F-factor is mainly
with most common functional segments such as gene traJ, Ori, OriC. The structure of this F
factor contains OriT for origin of transfer and acts as a starting point for gene transmission.
Why has the share of healthcare output produced by hospitals fallen.pdf
Why has the share of healthcare output produced by hospitals fallen? Will this trend continue?
What implications does this have for the careers of healthcare managers?
Solution
Mainly, Hospital operating cost such as Labor Wages, doctor salaries, admin salaries, electricity
charges and maintenance charge are go for high, Second major reason is Drug price risen .
Hospital are hike the charge but health care giving nominal amount covered by insurance plan,
So hospitals are fallen. Mostly profit earning hospital having huge fall.
Temporarily it will continue but not for long time i) Hospitals realised its fallen and started to
reduce the excess expenditure and situation has been under control
Careers of healthcare managers is potential and growth in their path, They have to motivate the
patient and co operative with them, They are bridge between inside and outside person
Inside people : Doctor, Staff, Nurse, Light men/Women and some labors
outside people: Patient and visitors
They need knowledge about technology advance such in medical devices and they have to lead
hospital finance in right way with positive..
What three types of inventories are reported on the balance sheet.pdf
What three types of inventories are reported on the balance sheet of a manufacturing company?
What three types of inventories are reported on the balance sheet of a manufacturing company?
Solution
Answer is
Raw Material Inventory = Contains invetoru of Materials required for Manufacturing
Work in Progress Inventory= Contains inventory of goods in Process or semi finished goods
Finished Goods Inventory = Contains Inventory of goods Finished goods.
What are the major components of a CAN 2.0A frameSolutionThe .pdf
What are the major components of a CAN 2.0A frame?
Solution
The CAN standard frame is known as CAN A 2.0 which it supports 11 bits in the identifier.
The following four of them which are indicating below are the major components of the CAN A
2.0
- Agent
- Coverage
- Configuration
- Checkers ( data and protocol)..
What is meant by anthropogenic greenhouse gasesSolutionant.pdf
What is meant by \"anthropogenic greenhouse gases\"?
Solution
anthropogenic means man-made
Since from a long time human activities have greatly increased the concentrations of greenhouse
gases in the atmosphere. Consequently, measured atmospheric concentrations of carbon dioxide
are many times higher than pre-industrial levels.
The main sources of greenhouse gases due to human activity are:.
What country specific standard was developed to evaluate integrity o.pdf
What country specific standard was developed to evaluate integrity of individual systems and is
broken into four categories?
A. ITSEC
B. TCSEC
C. Common Criteria
D. CTCPEC
Solution
B. TCSEC (or the Orange Book) was developed to evaluate the integrity of standalone systems.
Answer A is incorrect because the ITSEC is an international standard developed in Europe.
Answer C is incorrect because Common Criteria is a global standard that built on TCSEC,
ITSEC, and the CTCPEC. Answer D is incorrect because the CTCPEC is the Canadian version
of the Orange Book..
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