Can we accept mobile apps as a part of data communication networks.pdf
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Can we accept mobile apps as a part of data communication networks? Solution Answer: Yes, we accept mobile apps as a part of data communication because mobile computing involves mobile communication. It allows transmission of data like texts , audio , vedio and now a days VoIP calls are made through the mobile apps. There are many mobiles apps which gives the services like voice over the internet. Thus mobile apps play a very important role in data communication network..
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One brain area in which neurogenesis is especially apparent is the.pdf
One brain area in which neurogenesis is especially apparent is the:
A.amygdala
B.thalamus
c.hypothalamus
D.hippocampus
Solution
The adult neurogenesis among mammals was initially found majorly in the dentate gyrus (DG)
part of the hippocampus. Lately, the minimal contribution of amygdala and hypothalamus has
been also studied.
Therefore, the correct option is: D-hippocampus.
Ive used op amps as comparators a number of times because of the e.pdf
I\'ve used op amps as comparators a number of times because of the easy availability (they can
be purchased 4 op amps to the IC package) so one was used as a comparator while the other
three for other purposes. But there are units made that are actually called comparators. Is there
something different about these - not just ordinary op amps?
Solution
there is no difference it is just ordinary op amps.
Is there any reason that mitosis could not occur in a cell whose gen.pdf
Is there any reason that mitosis could not occur in a cell whose genome is haploid? Why or why
not? Relate your answer to the way chromosomes line up for metaphase in mitosis.
Solution
Answer:
Haploid cells can undergo mitosis. It is beneficial for some plants and protists, the end result is
two identical haploid cells (nn).
Some plants use the alternation of generations reproductive cycle. A haploid gamete is
mitotically reproduced. Mitosis is much quicker than meiosis, allowing for many haploid
gametes to be produced. Example: Gametophyte of Bryophyte plants. It is produced from mitotic
cell division of spores, which are produced by meiosis in sporophytes.
During S-phase, DNA gets duplicated with each chromosome containing 4 chromatids. The
chromosomes line up at the center of the cell during metaphase. The centromeres split and divide
each chromosome into two which migrate to the opposite poles during anaphase..
In fruit flies, the brown gene, which controls eye color, is located .pdf
In fruit flies, the brown gene, which controls eye color, is located on an autosome. Brown
mutants (bw) have brown eyes, while wild-type flies (bw^+) have red eyes. A gene on the X
chromosome, miniature, controls wing size. Miniature wings are caused by the m allele, which is
recessive to the allele for normal wings (m^+). True-breeding miniature-winged, red-eyed
females are crossed to true-breeding normal-winged, brown-eyed males. Predict the phenotypes
(including their ratios of the F1 flies. If the F1 flies are intercrossed with each other, what
phenotypes will appear in the F2 and in what proportions?
Solution
1.
Phenotype
Normal wings Red Eyes 4418
Miniature wings Brown eyes 4330
Normal wings Brown eyes 610
Miniature wings Red eyes 590
2. The F1 was self-crossed to give rise to the F2 generation. Of these offspring, 3/4 had brown
eyes while 1/4 had Red eyes. Phenotypic ratio is 3:1.
In a population of Gouldian Finches, 1200 have a white breast colora.pdf
In a population of Gouldian Finches, 1200 have a white breast coloration and 800 have purple
breast coloration. The population is at Hardy-Weinberg equilibrium and breast color is
determined by a single mutation at one locus. What is the frequency of the dominant genotype if
purple is the dominant trait?
Solution
1200 have white. Let the symbol for recessive trait be \"p\" and that for dominant trait be \"P\".
So, genotype of 1200 finches will be \"pp\".
Total finches=1200+800=2000
According to HW equilibrium, q^2=pp=1200/2000=0.6
q= 0.6
p=(1-q)=(1-.77)=0.23=frequency of dominant phenotype.
IL 0 Group II Group I Select GP I Select GP II Solution Basic P.pdf
IL 0 Group II Group I Select GP I Select GP II
Solution
Basic Pneumatic Circuitry For control and automation 2. Contents Introduction Symbols
Circuit layout Actuator control 2/2 Valve Actuator control 3/2 Valve Actuator control 5/2 Valve
Sequential control Sequence solution 5/3 Valves Poppet/spool logic Balanced spool logic
Feedback Click the section to advance directly to it 3. Introduction This module shows the
methods of application of pneumatic valves and components for control and automation The
methods of pure pneumatic sequential control are confined to simple examples The majority of
modern systems are controlled electronically and is the subject of electropneumatic modules A
message to pneumatic circuit designers: Use proven and reliable design techniques Produce
circuits and documentation that are clear to read Design for safety Do not try to be too clever, the
circuit will be difficult for others to read and maintain 4. Symbols The standard for fluid
power symbols is ISO 1219-1. This is a set of basic shapes and rules for the construction of fluid
power symbols Cylinders can be drawn to show their extreme or intermediate positions of stroke
and any length above their width Valves show all states in the one symbol. The prevailing state is
shown with the port connections Other components are single state symbols 5. Symbols single
acting actuators Single acting, sprung instroked Single acting, sprung outstroked Single acting,
sprung instroked, magnetic Single acting, sprung outstroked, magnetic 6. Symbols double acting
actuators Double acting, noncushioned Double acting, adjustable cushions Double acting,
through rod, adjustable cushions Double acting, magnetic, adjustable cushions Double acting,
rodless, magnetic, adjustable cushions 7. Symbols rotary actuators Semi-rotary double acting
Rotary motor single direction of rotation Rotary motor bidirectional 8. Symbols valves 2/2
Valve push button / spring 3/2 Valve detented lever operated 1 3/2 Valve push button / spring 2
10 12 2 10 12 3 2 10 12 3 1 1 9. Symbols valves 3/2 Valve differential pressure operated 2 12 3
10 1 5/2 Valve push button / spring 5/3 Valve double pressure operated spring centre 14 4 5 4 2 5
1 3 2 12 1 3 10. Symbols valves A valve function is known by a pair of numbers e.g. 3/2. This
indicates the valve has 3 main ports and 2 states The valve symbol shows both of the states Port
numbering is to CETOP RP68P and shows: when the valve is operated at the 12 end port 1 is
connected to port 2 when reset to the normal state at the 10 end port 1 is connected to nothing (0)
2 12 10 3 1 11. Symbols valves A valve function is known by a pair of numbers e.g. 3/2. This
indicates the valve has 3 main ports and 2 states The valve symbol shows both of the states Port
numbering is to CETOP RP68P and shows: when the valve is operated at the 12 end port 1 is
connected to port 2 when reset to the normal state at the 10 end port 1 is connected to nothing (0)
2 12 10 3 1 12. Symbols val.
If noncovalent interactions such as hydrogen bonds are so weak , how.pdf
If noncovalent interactions such as hydrogen bonds are so weak , how are they able to hold
molecules in a particular conformation and hold two molecules (such as enzyme and substrate)
together?
Solution
The non-covalent interactions are weak forces like hydrogen bonds, hydrophobic interactions,
electrostatic interactions, and Vander Waals forces. Conformations in the structure of proteins at
secondary and tertiary level are due to extensive hydrogen bonding. Though individually weak,
these non-covalent interactions are able to hold molecules in a particular conformation because,
they occur in large numbers. When a substrate binds to an enzyme, the attractive forces between
the substrate and the enzyme alter the enzyme\'s conformation. As the substrate approaches the
active site, these non-covalent interactions cause the active site to bend into such a shape that the
substrate molecule closely fits into it. As the shape of the active site changes, it produces changes
in the shape of other regions of the enzyme molecule altering the overall conformation of the
enzyme. The conformational change in the enzyme allows the formation of more non-covalent
interactions, which hold the two molecules together..
How is the address of the individual of an array foundSolution.pdf
How is the address of the individual of an array found?
Solution
An array is a type of variable in C programming, one that you can examine for its size and
address. Now you uncover the deep, dark secret of beholding an array’s address.
The source code from Where the Array Lurks shows a teensy program that declares an int array
and then displays that array’s location in memory..
Recommended
One brain area in which neurogenesis is especially apparent is the.pdf
One brain area in which neurogenesis is especially apparent is the:
A.amygdala
B.thalamus
c.hypothalamus
D.hippocampus
Solution
The adult neurogenesis among mammals was initially found majorly in the dentate gyrus (DG)
part of the hippocampus. Lately, the minimal contribution of amygdala and hypothalamus has
been also studied.
Therefore, the correct option is: D-hippocampus.
Ive used op amps as comparators a number of times because of the e.pdf
I\'ve used op amps as comparators a number of times because of the easy availability (they can
be purchased 4 op amps to the IC package) so one was used as a comparator while the other
three for other purposes. But there are units made that are actually called comparators. Is there
something different about these - not just ordinary op amps?
Solution
there is no difference it is just ordinary op amps.
Is there any reason that mitosis could not occur in a cell whose gen.pdf
Is there any reason that mitosis could not occur in a cell whose genome is haploid? Why or why
not? Relate your answer to the way chromosomes line up for metaphase in mitosis.
Solution
Answer:
Haploid cells can undergo mitosis. It is beneficial for some plants and protists, the end result is
two identical haploid cells (nn).
Some plants use the alternation of generations reproductive cycle. A haploid gamete is
mitotically reproduced. Mitosis is much quicker than meiosis, allowing for many haploid
gametes to be produced. Example: Gametophyte of Bryophyte plants. It is produced from mitotic
cell division of spores, which are produced by meiosis in sporophytes.
During S-phase, DNA gets duplicated with each chromosome containing 4 chromatids. The
chromosomes line up at the center of the cell during metaphase. The centromeres split and divide
each chromosome into two which migrate to the opposite poles during anaphase..
In fruit flies, the brown gene, which controls eye color, is located .pdf
In fruit flies, the brown gene, which controls eye color, is located on an autosome. Brown
mutants (bw) have brown eyes, while wild-type flies (bw^+) have red eyes. A gene on the X
chromosome, miniature, controls wing size. Miniature wings are caused by the m allele, which is
recessive to the allele for normal wings (m^+). True-breeding miniature-winged, red-eyed
females are crossed to true-breeding normal-winged, brown-eyed males. Predict the phenotypes
(including their ratios of the F1 flies. If the F1 flies are intercrossed with each other, what
phenotypes will appear in the F2 and in what proportions?
Solution
1.
Phenotype
Normal wings Red Eyes 4418
Miniature wings Brown eyes 4330
Normal wings Brown eyes 610
Miniature wings Red eyes 590
2. The F1 was self-crossed to give rise to the F2 generation. Of these offspring, 3/4 had brown
eyes while 1/4 had Red eyes. Phenotypic ratio is 3:1.
In a population of Gouldian Finches, 1200 have a white breast colora.pdf
In a population of Gouldian Finches, 1200 have a white breast coloration and 800 have purple
breast coloration. The population is at Hardy-Weinberg equilibrium and breast color is
determined by a single mutation at one locus. What is the frequency of the dominant genotype if
purple is the dominant trait?
Solution
1200 have white. Let the symbol for recessive trait be \"p\" and that for dominant trait be \"P\".
So, genotype of 1200 finches will be \"pp\".
Total finches=1200+800=2000
According to HW equilibrium, q^2=pp=1200/2000=0.6
q= 0.6
p=(1-q)=(1-.77)=0.23=frequency of dominant phenotype.
IL 0 Group II Group I Select GP I Select GP II Solution Basic P.pdf
IL 0 Group II Group I Select GP I Select GP II
Solution
Basic Pneumatic Circuitry For control and automation 2. Contents Introduction Symbols
Circuit layout Actuator control 2/2 Valve Actuator control 3/2 Valve Actuator control 5/2 Valve
Sequential control Sequence solution 5/3 Valves Poppet/spool logic Balanced spool logic
Feedback Click the section to advance directly to it 3. Introduction This module shows the
methods of application of pneumatic valves and components for control and automation The
methods of pure pneumatic sequential control are confined to simple examples The majority of
modern systems are controlled electronically and is the subject of electropneumatic modules A
message to pneumatic circuit designers: Use proven and reliable design techniques Produce
circuits and documentation that are clear to read Design for safety Do not try to be too clever, the
circuit will be difficult for others to read and maintain 4. Symbols The standard for fluid
power symbols is ISO 1219-1. This is a set of basic shapes and rules for the construction of fluid
power symbols Cylinders can be drawn to show their extreme or intermediate positions of stroke
and any length above their width Valves show all states in the one symbol. The prevailing state is
shown with the port connections Other components are single state symbols 5. Symbols single
acting actuators Single acting, sprung instroked Single acting, sprung outstroked Single acting,
sprung instroked, magnetic Single acting, sprung outstroked, magnetic 6. Symbols double acting
actuators Double acting, noncushioned Double acting, adjustable cushions Double acting,
through rod, adjustable cushions Double acting, magnetic, adjustable cushions Double acting,
rodless, magnetic, adjustable cushions 7. Symbols rotary actuators Semi-rotary double acting
Rotary motor single direction of rotation Rotary motor bidirectional 8. Symbols valves 2/2
Valve push button / spring 3/2 Valve detented lever operated 1 3/2 Valve push button / spring 2
10 12 2 10 12 3 2 10 12 3 1 1 9. Symbols valves 3/2 Valve differential pressure operated 2 12 3
10 1 5/2 Valve push button / spring 5/3 Valve double pressure operated spring centre 14 4 5 4 2 5
1 3 2 12 1 3 10. Symbols valves A valve function is known by a pair of numbers e.g. 3/2. This
indicates the valve has 3 main ports and 2 states The valve symbol shows both of the states Port
numbering is to CETOP RP68P and shows: when the valve is operated at the 12 end port 1 is
connected to port 2 when reset to the normal state at the 10 end port 1 is connected to nothing (0)
2 12 10 3 1 11. Symbols valves A valve function is known by a pair of numbers e.g. 3/2. This
indicates the valve has 3 main ports and 2 states The valve symbol shows both of the states Port
numbering is to CETOP RP68P and shows: when the valve is operated at the 12 end port 1 is
connected to port 2 when reset to the normal state at the 10 end port 1 is connected to nothing (0)
2 12 10 3 1 12. Symbols val.
If noncovalent interactions such as hydrogen bonds are so weak , how.pdf
If noncovalent interactions such as hydrogen bonds are so weak , how are they able to hold
molecules in a particular conformation and hold two molecules (such as enzyme and substrate)
together?
Solution
The non-covalent interactions are weak forces like hydrogen bonds, hydrophobic interactions,
electrostatic interactions, and Vander Waals forces. Conformations in the structure of proteins at
secondary and tertiary level are due to extensive hydrogen bonding. Though individually weak,
these non-covalent interactions are able to hold molecules in a particular conformation because,
they occur in large numbers. When a substrate binds to an enzyme, the attractive forces between
the substrate and the enzyme alter the enzyme\'s conformation. As the substrate approaches the
active site, these non-covalent interactions cause the active site to bend into such a shape that the
substrate molecule closely fits into it. As the shape of the active site changes, it produces changes
in the shape of other regions of the enzyme molecule altering the overall conformation of the
enzyme. The conformational change in the enzyme allows the formation of more non-covalent
interactions, which hold the two molecules together..
How is the address of the individual of an array foundSolution.pdf
How is the address of the individual of an array found?
Solution
An array is a type of variable in C programming, one that you can examine for its size and
address. Now you uncover the deep, dark secret of beholding an array’s address.
The source code from Where the Array Lurks shows a teensy program that declares an int array
and then displays that array’s location in memory..
Given the following information (ignore the top posts on the LEGO br.pdf
Given the following information (ignore the top posts on the LEGO bricks):
The short blocks have a thickness of 3.20±.01 mm
The tall blocks have a thickness of 9.60±.01 mm
When the stack is the TALLEST: What would the maximum size of these gaps be?
Solution
I a little confuse with this, but with the gaps I think that you are talking about the separation
between them? if you put the tallest form, the tolerance is 0.02 mm, so this would be, but I don´t
really sure.
Gene Splicing -1. DNA is made up of two separate strands of base s.pdf
Gene Splicing -
1. DNA is made up of two separate strands of base sequences. The same sequence is found on
both strands, but running in opposite directions. What word describes this characteristic?
2. What does the term “sticky ends” refer to in gene splicing?
3. What is a plasmid? How is a plasmid used in gene splicing?
4. What types of vectors are used to carry DNA from one species into the DNA of another
species? Give examples.
5. What is a “transgenic organism”? Give examples
6. Why is it essential that the same restriction enzyme be used to cleave (cut) the DNA of both
organisms used to create a transgenic organism?
7. Are there any factors other than technical ones that might slow — or even prevent — the use
of bioengineering?
Solution
1, The sequences present in both the strands of DNA running in opposite direction are called
ANTIPARALLEL AND COMPLEMENTARY STRANDS.
2. STICKY ENDS refer to the DNA which has some extra nucleotides in one strand and
overhangs extra than another strand. The nucleotides of the sticly ends are unpaired.
3. PLASMID : plasmid is an extrachromosomal materia present in the bacteris. These are
autonoously replication circular genetic materials, characteristic of bacteris. Plasmids are used as
vectors in recombinant DNA technology. Plasmids carry genes for antibiotic resistance and some
other colony selection traits.
4. VECTORS:
following types of vectors are used:.
For each of these functions, is it one-to-one Is it onto For each .pdf
For each of these functions, is it one-to-one? Is it onto? For each of these functions, is it one-to-
one? Is it onto? For each of these functions, is it one-to-one? Is it onto? (recall that the range [0.
infinite) is the non-negative real numbers)
Solution
all functions are onto function because all functions have square operation..
Find all roots of Re(arcsinh(z)) = 0, where z is the complex number,.pdf
Find all roots of Re(arcsinh(z)) = 0, where z is the complex number, z = x + iy.
Solution
a) Write z = x + iy and separate sin(z) into real and imaginary parts: sin(z) = sin(x + iy) = sin(x)
cos(iy) + cos(x) sin(iy) = sin(x) cosh(y) + i cos(x) sinh(y) = i ? ( sin(x) cosh(y) = 0 cos(x) sinh(y)
= 1 The first equation gives x = n?, n = 0,.
Discuss the history of operations, information, systems, and conting.pdf
Discuss the history of operations, information, systems, and contingency management.
Solution
Answer 1. History of Operation Management
History of Operation Management has undergone many stages that shows shifts in thinking with
the evolution of management.
a. Pre-Industrial Stage:
This stage was prior to Industrial revolution. It was cotton-based guild system. The production of
goods took place in homes or cottages. The craftsmen guided apprentices to perform better hand
work.
b. Post-Industrial Revolution Era
An Industrial Revolution came into existence during 18th century and its impact spread to all
European nations and United States. Industrial Revolution led to two major developments in
Operation Management. First is widespread replacement of human power with machine power.
Secondly, the factories were established everywhere for operations. Lot of inventions were made
during that period.Gasoline and electricity were invented in 1800s. Old cottage systems were
replaced by factory system. Mass production came into existence.
The operation management history can be divided into five stages
Answer 2. History of Information Management
Information management has its history, its origin dated back to post-industrial era. With mass
production all over, the people were feeling the need for information exchange in more
formalized manner. In those days the information management was only in germ phase i.e. the
records were maintained properly handwritten on papers in the files.
Not later after, the writing machines were replaced by computers. In 1970s, the computers could
all do was file management. But the employees were able to retrieve information accurately and
speedily.
During 1990s, the information management was dealt only by high qualified employees of the
companies. It maintained important documents of the company. After 2000, the information
management itself became a profession. The companies specializing in data protection are
outsourced with the file management job. During these years, the information management has
become more efficient, more accurate with high speed of process..
Dr. Butthead received a new cell line and he wants to see which cult.pdf
Dr. Butthead received a new cell line and he wants to see which culture media grows the cells
the best. He cultures them in a supplemented (s) and complete (c) media. After 2 days he dilutes
a sample from each flask 1/2 (100uL of cells + 100 uL of trypan blue) His hemacytometer counts
for S are 15,14,19,12 and C are 92,100,95,and 84.
1. Which is the concentration of cells in each flask? (cell per mL)
2. Which media works best? explain why?
3. cells must be fed if they are too concentrated (ie 1x10^6 cells/ml). Does either flask need to be
fed? if yes why and describe how to dilute the cells to 5x10^4 cell/ml in a volume of 10 mL.
Thank you for your help
Solution
1. Which is the concentration of cells in each flask? (cell per mL)
Dilution factor =2
Supplemented (s)
Cell/mL=(15+14+19+12)/4 x dilution factor x 10^ 4
=30 x 10^ 4 cells/mL
Complete (c) media
Cell/mL=(92+100+95+84)/4 x dilution factor x 10^ 4
=185.5 x 10^ 4 cells/mL
2. Which media works best? explain why?
Complete (c) media works best since it promotes cell proliferation better than supplemented
media.
3. cells must be fed if they are too concentrated (ie 1x10^6 cells/ml). Does either flask need to be
fed? if yes why and describe how to dilute the cells to 5x10^4 cell/ml in a volume of 10 mL.
Cells in the Complete (c) media needs to be fed since the it is too concentrated
Stock of cells = 185.5 x 10^ 4 cells/mL
Required = 5 x10^4 cell/ml
Volume = 10 mL
C1V1= C2V2
(stock) (required)
V1=C2V2/C1
V1 = ( 5 x10^4 X 10 mL )/ 185.5 x 10^ 4
=0.27 mL
0.27 mL of stock + 9.73 mL of media gives 10 mL of media with 5 x10^4 cell/ml concentration.
Describe or explain how SNPs in gens at evolutionary conserved break.pdf
Describe or explain how SNPs in gens at evolutionary conserved breakpoints can lead to
dramatic phenotypic differences between species like chimpanzees and humans even though
their genomes are 98% similar? Provide and describe or explain an example of one gene with
SNP differences between gorillas and humans that might have resulted changes between humans
and apes.
Solution
A DNA sequence is a linear combination of four nucleotides. The two sequences when compared
position by position, and wherever a different nucleotides at the same position occurs, that\'s a
SNP. SNPs reflect past mutations that were mostly (but not exclusively) unique events, and two
individuals sharing a variant allele are thereby marked with a common evolutionary heritage.
Analysis of shared pattern of SNP variation can identify the genes similar to our ancestors.
Although humans and chimps have many identical genes, they often use them in different ways.
Same gene can be expressed more in humans and less in chimps. The same genes are expressed
in the same brain regions in human, chimp and gorilla, but in different amounts thereby resulting
in difference in brain activity. Thousands of differences like these affect brain development and
function, and help explain why the human brain is larger and smarter.
The chimpanzee immune system is surprisingly similar to ours. Most viruses that cause diseases
like AIDS and hepatitis in human can also infect chimpanzees. In case of malaria parasite which
can be transmitted by mosquito bite in human does not infect chimps. A small DNA difference
makes human red blood cells vulnerable to this parasite, while chimp blood cells are resistant.
Also the characteristic of speech in human is the result of SNP which is absent in chimps or apes..
create a binary search tree from an empty one by adding the key valu.pdf
create a binary search tree from an empty one by adding the key value alphabetically in c++
Solution
//Binary Search Tree Program
#include
#include
using namespace std;
class BinarySearchTree
{
private:
struct tree_node //structure to hold Binary search tree
{
tree_node* left;
tree_node* right;
int data;
};
tree_node* root;
public:
BinarySearchTree() //constructor of Binary search tree
{
root = NULL;
}
//function declarations
bool isEmpty() const { return root==NULL; }
void print_inorder();
void inorder(tree_node*);
void print_preorder();
void preorder(tree_node*);
void print_postorder();
void postorder(tree_node*);
void insert(int);
void remove(int);
};
// Smaller elements go left
// larger elements go right
void BinarySearchTree::insert(int d) //insert function to BST
{
tree_node* t = new tree_node;
tree_node* parent;
t->data = d;
t->left = NULL;
t->right = NULL;
parent = NULL;
// is this a new tree?
if(isEmpty()) root = t;
else
{
//Note: ALL insertions are as leaf nodes
tree_node* curr;
curr = root;
// Find the Node\'s parent
while(curr)
{
parent = curr;
if(t->data > curr->data) curr = curr->right;
else curr = curr->left;
}
if(t->data < parent->data)
parent->left = t;
else
parent->right = t;
}
}
void BinarySearchTree::remove(int d)
{
//Locate the element
bool found = false;
if(isEmpty())
{
cout<<\" This Tree is empty! \"<data == d)
{
found = true;
break;
}
else
{
parent = curr;
if(d>curr->data) curr = curr->right;
else curr = curr->left;
}
}
if(!found)
{
cout<<\" Data not found! \"<left == NULL && curr->right != NULL)|| (curr->left !=
NULL
&& curr->right == NULL))
{
if(curr->left == NULL && curr->right != NULL)
{
if(parent->left == curr)
{
parent->left = curr->right;
delete curr;
}
else
{
parent->right = curr->right;
delete curr;
}
}
else // left child present, no right child
{
if(parent->left == curr)
{
parent->left = curr->left;
delete curr;
}
else
{
parent->right = curr->left;
delete curr;
}
}
return;
}
//We\'re looking at a leaf node
if( curr->left == NULL && curr->right == NULL)
{
if(parent->left == curr) parent->left = NULL;
else parent->right = NULL;
delete curr;
return;
}
//Node with 2 children
// replace node with smallest value in right subtree
if (curr->left != NULL && curr->right != NULL)
{
tree_node* chkr;
chkr = curr->right;
if((chkr->left == NULL) && (chkr->right == NULL))
{
curr = chkr;
delete chkr;
curr->right = NULL;
}
else // right child has children
{
//if the node\'s right child has a left child
// Move all the way down left to locate smallest element
if((curr->right)->left != NULL)
{
tree_node* lcurr;
tree_node* lcurrp;
lcurrp = curr->right;
lcurr = (curr->right)->left;
while(lcurr->left != NULL)
{
lcurrp = lcurr;
lcurr = lcurr->left;
}
curr->data = lcurr->data;
delete lcurr;
lcurrp->left = NULL;
}
else
{
tree_node* tmp;
tmp = curr->right;
curr->data = tmp->data;
curr->right = tmp->right;
delete tmp;
}
}
return;
}
}
void BinarySearchTree::print_inorder()
{
inorder(root);
}
void BinarySearchTree::inorder(tree_node* p) .
Data Analysis correlation. A correlation R in regression analysis t.pdf
Data Analysis correlation. A correlation R in regression analysis that would he generated if
only one independent variable were considered at a time. You will recall that the Multiple R is
just the square root of R square. So the correlations shown in figure 13.21 are essentially the
square roots of the amount of variance shared between each of the variables shown.
Solution
The tables shows the multicollinearity among the variables only. The table does not reveals
anything. In the sense we cannnot make any predictions. So, the correlation coefficients in this
table represents only positive or negative association between the predictor variable..
Conduct outside reading on Salla disease. Explain its underlying mole.pdf
Conduct outside reading on Salla disease. Explain its underlying molecular biology, cause,
signs, symptoms and progression of the disease. Address whether any cures have been found. If
yes describe them. If no explain why this is the case.
Solution
Salla disease is a type of allelic disorder which falls under sialic acid storage disease
characterized by a delay in development, ataxia, and some dysmorphic features. Additional signs
are hypopigmentation of skin and hair, mild skeletal dysplasias, and course facial features.There
is an accumulation of the charged monosaccharide sialic acid (N-acetylneuraminic acid) in
lysosomes as a result of which sialic acid cannot be transported out of lysosomes.The mode of
inheritence is autosomal recessive disease. The diagnosis is dependent upon the raised excretion
of sialic acid in urine. Electron microscopic studies of reveal a lot about it.
There is actually a homozygous missense mutation in SLC17A5 which codes for the transport
protein called sialin. People with salla disease have normal life span but remain differently abled.
There is no cure for this disease as its a homozygous missense mutation which is having dual
mode of action-either altering the amino acids in translated proteins or else remaining silent.
References
1. Gorlin, R,J., Cohen, M,M., & Hennekam, R,C,M. (2001). Syndromes of the head and
neck.New York: Oxford university press.
2. Barness, E,G., Barness,L,A., & Farrell, P,M. (2017). Metabolic diseases. Amsterdam:IOS
press BV..
At absolute temperature T, a black body radiates its peak intensity a.pdf
At absolute temperature T, a black body radiates its peak intensity at wavelength lambda At
absolute temperature 2T, what would be the wavelength of the peak intensity?
Solution
For a particular black body, accoring to wein law
T=Constant
When Temperature equal to 2T,
(2T)= constant . this is possible only when =/2
/2 (2T)= T= constant
Answer :=/2.
An unsaturated solution of Na2SO4 is present in equilibrium with its.pdf
An unsaturated solution of Na2SO4 is present in equilibrium with its vapor in a closed vessel. (a)
How many phases and components are present? (b) What is the number of degrees of freedom of
the system? Identify the independent variables
Solution
No. of components = 2 (Na2SO4 and water - as each phase can be described by these two
components only)
No. of phases = 3 ( solid and liquid and vapor)
degrees of freedom = 2-3+2 = 1.
a. List the three things that happen when the following statement is.pdf
a. List the three things that happen when the following statement is executed : int x = 5;
b. List five different things you would need to know to learn a new programming language.
These should parallel the approach we’ve taken in class to learning Java.
Solution
a)
int x=5;
here x is identifier .
b) Different things you would need to know to learn a new programming language like java
1. Simple and powerful languages are always easy to learn.(less coding & easy syntax)
2. Platform independent & Portable (Easy to move and run all hardware platforms)
3. Robust & Secure (Strong & without loop holes)
4. Support Object-Oriented and functional programming.
5. Multithreaded & Distributed for achieving High performance .
6. Use to develop all types real world applications belongs to different
areas(Networks,Standalone,web,Artificial Intelligence ,Mining,..etc).
a statistics professor receives an average of five email messages pe.pdf
a statistics professor receives an average of five email messages per day from students. Assume
the number of messages approximates a poisson distrivution. What is the probility that on a
randomly selected day she will have four messages?
Solution
the probability that on a randomly selected day she will have four messages=
P(X=4) in a Poisson distribution with average = 5
= 0.1755.
A researcher is convinced that the reason that people watch Jerry Sp.pdf
A researcher is convinced that the reason that people watch Jerry Springer (the talk show that
parades the strange problems of highly disturbed people with highly disturbed relationships) is
that it boosts their self-esteem temporarily because they compare their lives with those of Jerry\'s
guests. To test this hypothesis, in random order 7 people watch two television shows. One show
is an episode of Jerry Springer and the other show is a documentary on the K-12 education
system that has been shown in previous research to have no impact on self-esteem. After each
show, the participants in the study fill out a self esteem measure.
For each person, the data are presented below in form (x,y) in which x represents the self esteem
measure after watching Jerry Springer and y represents the self-esteem measure after watching
the other show.
(17,17), (21,25), (21,22), (22,21), (22,22), (17,20), (19,15)
The mean of the self esteem scores after watching Jerry Springer is 19.857, and the sample
standard deviation is 2.193.
A. What is the obtained (computed) value of the test statistic that the researcher would compute
to test her question (give the absolute value)?.
Solution
The matched pairs t test statistic is .43 with a p-value of .34. We would not reject the null
hypothesis, and there is not sufficient evidence to say that watching Jerry Springer increases self
esteem..
You connect a host to a switch. The host is capable only of half dup.pdf
You connect a host to a switch. The host is capable only of half duplex. After the connection is
negotiated, how should the interface DUPLX LED appear?
A. Amber
B. Green
C. Green/amber
D. Off
Solution
D is correct. If the LED mode is set to duplex, there are only two options: green for full duplex
and off for half duplex.
A, B, and C are incorrect. A is incorrect because amber LEDs do not appear when the mode is
set to duplex. B is incorrect because green LEDs indicate a full-duplex connection when the
mode is set to duplex. C is incorrect because green/amber LEDs do not appear when the mode is
set to duplex..
You are a game warden studying elephant ecology in a Ugandan National.pdf
You are a game warden studying elephant ecology in a Ugandan National Park. As part of your
project you have collected the following data: What is the exploitation efficiency
Solution
Explotation efficiency = amount of food ingested / amount of prey production
= 71.5 / 40.2
= 1.778 = 17.78%
Asimilation efficiency = amount of assimilation / amount of food ingested
= 31.0 / 71.5
= 0.4335 = 43.35%.
Write a C program using low-level io which will read in the files g.pdf
Write a C program using low-level i/o which will read in the files given on the command line
using a buffer of size 1024 and print them to standard out. A sample invocation:
prog2 filea fileb filec filed
Solution
#include
int main(int argc, char** argv)
// It is good take a double character pointer , so that we can handle dynamic number of file names
in input
{
FILE *fp;
int c;
for(int i=1;i.
which type of distribution of final grades would be most desirable t.pdf
which type of distribution of final grades would be most desirable to students? a) Bimodal b)
Unimodal c) Positively skewed d) Negatively skewed
Solution
The most deisrable type of distribution of final grades to the students will be bimodal.
This is because If the graph of the distribution of final grades is bimodal then it means that the
grades are more widely spread in all directions compared to other type of distribution which
implies that grades of the students do not differ much among each other which will otherwise be
found in unimodal, positively skewed and negatively distribution of grades..
Which of the following organs produce similar, branching organs (i.e.pdf
Which of the following organs produce similar, branching organs (i.e., a leaf on a leaf, a branch
on a stem, a lateral root on a root) from an internal cell layer called the pericycle?
Only stems
Only roots
Only leaves
Stems and roots
Leaves and stems
All of the above
None of the above
Only stems
Only roots
Only leaves
Stems and roots
Leaves and stems
All of the above
None of the above
Solution
Stems produce similar branching organs ie roots from the pericycle which gives rise to branch
roots.
The pericycle is a cylinder of parenchyma (vascular- storage and transport of nutrients) or
sclerenchyma cells (nonvascular- support and protection) that lies inside the endodermis and
outer side of stele of plants..
Model Attribute Check Company Auto Property
In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
1.4 modern child centered education - mahatma gandhi-2.pptx
Child centred education is an educational approach that priorities the interest, needs and abilities of the child in the learning process.
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More from erremmfab
Given the following information (ignore the top posts on the LEGO br.pdf
Given the following information (ignore the top posts on the LEGO bricks):
The short blocks have a thickness of 3.20±.01 mm
The tall blocks have a thickness of 9.60±.01 mm
When the stack is the TALLEST: What would the maximum size of these gaps be?
Solution
I a little confuse with this, but with the gaps I think that you are talking about the separation
between them? if you put the tallest form, the tolerance is 0.02 mm, so this would be, but I don´t
really sure.
Gene Splicing -1. DNA is made up of two separate strands of base s.pdf
Gene Splicing -
1. DNA is made up of two separate strands of base sequences. The same sequence is found on
both strands, but running in opposite directions. What word describes this characteristic?
2. What does the term “sticky ends” refer to in gene splicing?
3. What is a plasmid? How is a plasmid used in gene splicing?
4. What types of vectors are used to carry DNA from one species into the DNA of another
species? Give examples.
5. What is a “transgenic organism”? Give examples
6. Why is it essential that the same restriction enzyme be used to cleave (cut) the DNA of both
organisms used to create a transgenic organism?
7. Are there any factors other than technical ones that might slow — or even prevent — the use
of bioengineering?
Solution
1, The sequences present in both the strands of DNA running in opposite direction are called
ANTIPARALLEL AND COMPLEMENTARY STRANDS.
2. STICKY ENDS refer to the DNA which has some extra nucleotides in one strand and
overhangs extra than another strand. The nucleotides of the sticly ends are unpaired.
3. PLASMID : plasmid is an extrachromosomal materia present in the bacteris. These are
autonoously replication circular genetic materials, characteristic of bacteris. Plasmids are used as
vectors in recombinant DNA technology. Plasmids carry genes for antibiotic resistance and some
other colony selection traits.
4. VECTORS:
following types of vectors are used:.
For each of these functions, is it one-to-one Is it onto For each .pdf
For each of these functions, is it one-to-one? Is it onto? For each of these functions, is it one-to-
one? Is it onto? For each of these functions, is it one-to-one? Is it onto? (recall that the range [0.
infinite) is the non-negative real numbers)
Solution
all functions are onto function because all functions have square operation..
Find all roots of Re(arcsinh(z)) = 0, where z is the complex number,.pdf
Find all roots of Re(arcsinh(z)) = 0, where z is the complex number, z = x + iy.
Solution
a) Write z = x + iy and separate sin(z) into real and imaginary parts: sin(z) = sin(x + iy) = sin(x)
cos(iy) + cos(x) sin(iy) = sin(x) cosh(y) + i cos(x) sinh(y) = i ? ( sin(x) cosh(y) = 0 cos(x) sinh(y)
= 1 The first equation gives x = n?, n = 0,.
Discuss the history of operations, information, systems, and conting.pdf
Discuss the history of operations, information, systems, and contingency management.
Solution
Answer 1. History of Operation Management
History of Operation Management has undergone many stages that shows shifts in thinking with
the evolution of management.
a. Pre-Industrial Stage:
This stage was prior to Industrial revolution. It was cotton-based guild system. The production of
goods took place in homes or cottages. The craftsmen guided apprentices to perform better hand
work.
b. Post-Industrial Revolution Era
An Industrial Revolution came into existence during 18th century and its impact spread to all
European nations and United States. Industrial Revolution led to two major developments in
Operation Management. First is widespread replacement of human power with machine power.
Secondly, the factories were established everywhere for operations. Lot of inventions were made
during that period.Gasoline and electricity were invented in 1800s. Old cottage systems were
replaced by factory system. Mass production came into existence.
The operation management history can be divided into five stages
Answer 2. History of Information Management
Information management has its history, its origin dated back to post-industrial era. With mass
production all over, the people were feeling the need for information exchange in more
formalized manner. In those days the information management was only in germ phase i.e. the
records were maintained properly handwritten on papers in the files.
Not later after, the writing machines were replaced by computers. In 1970s, the computers could
all do was file management. But the employees were able to retrieve information accurately and
speedily.
During 1990s, the information management was dealt only by high qualified employees of the
companies. It maintained important documents of the company. After 2000, the information
management itself became a profession. The companies specializing in data protection are
outsourced with the file management job. During these years, the information management has
become more efficient, more accurate with high speed of process..
Dr. Butthead received a new cell line and he wants to see which cult.pdf
Dr. Butthead received a new cell line and he wants to see which culture media grows the cells
the best. He cultures them in a supplemented (s) and complete (c) media. After 2 days he dilutes
a sample from each flask 1/2 (100uL of cells + 100 uL of trypan blue) His hemacytometer counts
for S are 15,14,19,12 and C are 92,100,95,and 84.
1. Which is the concentration of cells in each flask? (cell per mL)
2. Which media works best? explain why?
3. cells must be fed if they are too concentrated (ie 1x10^6 cells/ml). Does either flask need to be
fed? if yes why and describe how to dilute the cells to 5x10^4 cell/ml in a volume of 10 mL.
Thank you for your help
Solution
1. Which is the concentration of cells in each flask? (cell per mL)
Dilution factor =2
Supplemented (s)
Cell/mL=(15+14+19+12)/4 x dilution factor x 10^ 4
=30 x 10^ 4 cells/mL
Complete (c) media
Cell/mL=(92+100+95+84)/4 x dilution factor x 10^ 4
=185.5 x 10^ 4 cells/mL
2. Which media works best? explain why?
Complete (c) media works best since it promotes cell proliferation better than supplemented
media.
3. cells must be fed if they are too concentrated (ie 1x10^6 cells/ml). Does either flask need to be
fed? if yes why and describe how to dilute the cells to 5x10^4 cell/ml in a volume of 10 mL.
Cells in the Complete (c) media needs to be fed since the it is too concentrated
Stock of cells = 185.5 x 10^ 4 cells/mL
Required = 5 x10^4 cell/ml
Volume = 10 mL
C1V1= C2V2
(stock) (required)
V1=C2V2/C1
V1 = ( 5 x10^4 X 10 mL )/ 185.5 x 10^ 4
=0.27 mL
0.27 mL of stock + 9.73 mL of media gives 10 mL of media with 5 x10^4 cell/ml concentration.
Describe or explain how SNPs in gens at evolutionary conserved break.pdf
Describe or explain how SNPs in gens at evolutionary conserved breakpoints can lead to
dramatic phenotypic differences between species like chimpanzees and humans even though
their genomes are 98% similar? Provide and describe or explain an example of one gene with
SNP differences between gorillas and humans that might have resulted changes between humans
and apes.
Solution
A DNA sequence is a linear combination of four nucleotides. The two sequences when compared
position by position, and wherever a different nucleotides at the same position occurs, that\'s a
SNP. SNPs reflect past mutations that were mostly (but not exclusively) unique events, and two
individuals sharing a variant allele are thereby marked with a common evolutionary heritage.
Analysis of shared pattern of SNP variation can identify the genes similar to our ancestors.
Although humans and chimps have many identical genes, they often use them in different ways.
Same gene can be expressed more in humans and less in chimps. The same genes are expressed
in the same brain regions in human, chimp and gorilla, but in different amounts thereby resulting
in difference in brain activity. Thousands of differences like these affect brain development and
function, and help explain why the human brain is larger and smarter.
The chimpanzee immune system is surprisingly similar to ours. Most viruses that cause diseases
like AIDS and hepatitis in human can also infect chimpanzees. In case of malaria parasite which
can be transmitted by mosquito bite in human does not infect chimps. A small DNA difference
makes human red blood cells vulnerable to this parasite, while chimp blood cells are resistant.
Also the characteristic of speech in human is the result of SNP which is absent in chimps or apes..
create a binary search tree from an empty one by adding the key valu.pdf
create a binary search tree from an empty one by adding the key value alphabetically in c++
Solution
//Binary Search Tree Program
#include
#include
using namespace std;
class BinarySearchTree
{
private:
struct tree_node //structure to hold Binary search tree
{
tree_node* left;
tree_node* right;
int data;
};
tree_node* root;
public:
BinarySearchTree() //constructor of Binary search tree
{
root = NULL;
}
//function declarations
bool isEmpty() const { return root==NULL; }
void print_inorder();
void inorder(tree_node*);
void print_preorder();
void preorder(tree_node*);
void print_postorder();
void postorder(tree_node*);
void insert(int);
void remove(int);
};
// Smaller elements go left
// larger elements go right
void BinarySearchTree::insert(int d) //insert function to BST
{
tree_node* t = new tree_node;
tree_node* parent;
t->data = d;
t->left = NULL;
t->right = NULL;
parent = NULL;
// is this a new tree?
if(isEmpty()) root = t;
else
{
//Note: ALL insertions are as leaf nodes
tree_node* curr;
curr = root;
// Find the Node\'s parent
while(curr)
{
parent = curr;
if(t->data > curr->data) curr = curr->right;
else curr = curr->left;
}
if(t->data < parent->data)
parent->left = t;
else
parent->right = t;
}
}
void BinarySearchTree::remove(int d)
{
//Locate the element
bool found = false;
if(isEmpty())
{
cout<<\" This Tree is empty! \"<data == d)
{
found = true;
break;
}
else
{
parent = curr;
if(d>curr->data) curr = curr->right;
else curr = curr->left;
}
}
if(!found)
{
cout<<\" Data not found! \"<left == NULL && curr->right != NULL)|| (curr->left !=
NULL
&& curr->right == NULL))
{
if(curr->left == NULL && curr->right != NULL)
{
if(parent->left == curr)
{
parent->left = curr->right;
delete curr;
}
else
{
parent->right = curr->right;
delete curr;
}
}
else // left child present, no right child
{
if(parent->left == curr)
{
parent->left = curr->left;
delete curr;
}
else
{
parent->right = curr->left;
delete curr;
}
}
return;
}
//We\'re looking at a leaf node
if( curr->left == NULL && curr->right == NULL)
{
if(parent->left == curr) parent->left = NULL;
else parent->right = NULL;
delete curr;
return;
}
//Node with 2 children
// replace node with smallest value in right subtree
if (curr->left != NULL && curr->right != NULL)
{
tree_node* chkr;
chkr = curr->right;
if((chkr->left == NULL) && (chkr->right == NULL))
{
curr = chkr;
delete chkr;
curr->right = NULL;
}
else // right child has children
{
//if the node\'s right child has a left child
// Move all the way down left to locate smallest element
if((curr->right)->left != NULL)
{
tree_node* lcurr;
tree_node* lcurrp;
lcurrp = curr->right;
lcurr = (curr->right)->left;
while(lcurr->left != NULL)
{
lcurrp = lcurr;
lcurr = lcurr->left;
}
curr->data = lcurr->data;
delete lcurr;
lcurrp->left = NULL;
}
else
{
tree_node* tmp;
tmp = curr->right;
curr->data = tmp->data;
curr->right = tmp->right;
delete tmp;
}
}
return;
}
}
void BinarySearchTree::print_inorder()
{
inorder(root);
}
void BinarySearchTree::inorder(tree_node* p) .
Data Analysis correlation. A correlation R in regression analysis t.pdf
Data Analysis correlation. A correlation R in regression analysis that would he generated if
only one independent variable were considered at a time. You will recall that the Multiple R is
just the square root of R square. So the correlations shown in figure 13.21 are essentially the
square roots of the amount of variance shared between each of the variables shown.
Solution
The tables shows the multicollinearity among the variables only. The table does not reveals
anything. In the sense we cannnot make any predictions. So, the correlation coefficients in this
table represents only positive or negative association between the predictor variable..
Conduct outside reading on Salla disease. Explain its underlying mole.pdf
Conduct outside reading on Salla disease. Explain its underlying molecular biology, cause,
signs, symptoms and progression of the disease. Address whether any cures have been found. If
yes describe them. If no explain why this is the case.
Solution
Salla disease is a type of allelic disorder which falls under sialic acid storage disease
characterized by a delay in development, ataxia, and some dysmorphic features. Additional signs
are hypopigmentation of skin and hair, mild skeletal dysplasias, and course facial features.There
is an accumulation of the charged monosaccharide sialic acid (N-acetylneuraminic acid) in
lysosomes as a result of which sialic acid cannot be transported out of lysosomes.The mode of
inheritence is autosomal recessive disease. The diagnosis is dependent upon the raised excretion
of sialic acid in urine. Electron microscopic studies of reveal a lot about it.
There is actually a homozygous missense mutation in SLC17A5 which codes for the transport
protein called sialin. People with salla disease have normal life span but remain differently abled.
There is no cure for this disease as its a homozygous missense mutation which is having dual
mode of action-either altering the amino acids in translated proteins or else remaining silent.
References
1. Gorlin, R,J., Cohen, M,M., & Hennekam, R,C,M. (2001). Syndromes of the head and
neck.New York: Oxford university press.
2. Barness, E,G., Barness,L,A., & Farrell, P,M. (2017). Metabolic diseases. Amsterdam:IOS
press BV..
At absolute temperature T, a black body radiates its peak intensity a.pdf
At absolute temperature T, a black body radiates its peak intensity at wavelength lambda At
absolute temperature 2T, what would be the wavelength of the peak intensity?
Solution
For a particular black body, accoring to wein law
T=Constant
When Temperature equal to 2T,
(2T)= constant . this is possible only when =/2
/2 (2T)= T= constant
Answer :=/2.
An unsaturated solution of Na2SO4 is present in equilibrium with its.pdf
An unsaturated solution of Na2SO4 is present in equilibrium with its vapor in a closed vessel. (a)
How many phases and components are present? (b) What is the number of degrees of freedom of
the system? Identify the independent variables
Solution
No. of components = 2 (Na2SO4 and water - as each phase can be described by these two
components only)
No. of phases = 3 ( solid and liquid and vapor)
degrees of freedom = 2-3+2 = 1.
a. List the three things that happen when the following statement is.pdf
a. List the three things that happen when the following statement is executed : int x = 5;
b. List five different things you would need to know to learn a new programming language.
These should parallel the approach we’ve taken in class to learning Java.
Solution
a)
int x=5;
here x is identifier .
b) Different things you would need to know to learn a new programming language like java
1. Simple and powerful languages are always easy to learn.(less coding & easy syntax)
2. Platform independent & Portable (Easy to move and run all hardware platforms)
3. Robust & Secure (Strong & without loop holes)
4. Support Object-Oriented and functional programming.
5. Multithreaded & Distributed for achieving High performance .
6. Use to develop all types real world applications belongs to different
areas(Networks,Standalone,web,Artificial Intelligence ,Mining,..etc).
a statistics professor receives an average of five email messages pe.pdf
a statistics professor receives an average of five email messages per day from students. Assume
the number of messages approximates a poisson distrivution. What is the probility that on a
randomly selected day she will have four messages?
Solution
the probability that on a randomly selected day she will have four messages=
P(X=4) in a Poisson distribution with average = 5
= 0.1755.
A researcher is convinced that the reason that people watch Jerry Sp.pdf
A researcher is convinced that the reason that people watch Jerry Springer (the talk show that
parades the strange problems of highly disturbed people with highly disturbed relationships) is
that it boosts their self-esteem temporarily because they compare their lives with those of Jerry\'s
guests. To test this hypothesis, in random order 7 people watch two television shows. One show
is an episode of Jerry Springer and the other show is a documentary on the K-12 education
system that has been shown in previous research to have no impact on self-esteem. After each
show, the participants in the study fill out a self esteem measure.
For each person, the data are presented below in form (x,y) in which x represents the self esteem
measure after watching Jerry Springer and y represents the self-esteem measure after watching
the other show.
(17,17), (21,25), (21,22), (22,21), (22,22), (17,20), (19,15)
The mean of the self esteem scores after watching Jerry Springer is 19.857, and the sample
standard deviation is 2.193.
A. What is the obtained (computed) value of the test statistic that the researcher would compute
to test her question (give the absolute value)?.
Solution
The matched pairs t test statistic is .43 with a p-value of .34. We would not reject the null
hypothesis, and there is not sufficient evidence to say that watching Jerry Springer increases self
esteem..
You connect a host to a switch. The host is capable only of half dup.pdf
You connect a host to a switch. The host is capable only of half duplex. After the connection is
negotiated, how should the interface DUPLX LED appear?
A. Amber
B. Green
C. Green/amber
D. Off
Solution
D is correct. If the LED mode is set to duplex, there are only two options: green for full duplex
and off for half duplex.
A, B, and C are incorrect. A is incorrect because amber LEDs do not appear when the mode is
set to duplex. B is incorrect because green LEDs indicate a full-duplex connection when the
mode is set to duplex. C is incorrect because green/amber LEDs do not appear when the mode is
set to duplex..
You are a game warden studying elephant ecology in a Ugandan National.pdf
You are a game warden studying elephant ecology in a Ugandan National Park. As part of your
project you have collected the following data: What is the exploitation efficiency
Solution
Explotation efficiency = amount of food ingested / amount of prey production
= 71.5 / 40.2
= 1.778 = 17.78%
Asimilation efficiency = amount of assimilation / amount of food ingested
= 31.0 / 71.5
= 0.4335 = 43.35%.
Write a C program using low-level io which will read in the files g.pdf
Write a C program using low-level i/o which will read in the files given on the command line
using a buffer of size 1024 and print them to standard out. A sample invocation:
prog2 filea fileb filec filed
Solution
#include
int main(int argc, char** argv)
// It is good take a double character pointer , so that we can handle dynamic number of file names
in input
{
FILE *fp;
int c;
for(int i=1;i.
which type of distribution of final grades would be most desirable t.pdf
which type of distribution of final grades would be most desirable to students? a) Bimodal b)
Unimodal c) Positively skewed d) Negatively skewed
Solution
The most deisrable type of distribution of final grades to the students will be bimodal.
This is because If the graph of the distribution of final grades is bimodal then it means that the
grades are more widely spread in all directions compared to other type of distribution which
implies that grades of the students do not differ much among each other which will otherwise be
found in unimodal, positively skewed and negatively distribution of grades..
Which of the following organs produce similar, branching organs (i.e.pdf
Which of the following organs produce similar, branching organs (i.e., a leaf on a leaf, a branch
on a stem, a lateral root on a root) from an internal cell layer called the pericycle?
Only stems
Only roots
Only leaves
Stems and roots
Leaves and stems
All of the above
None of the above
Only stems
Only roots
Only leaves
Stems and roots
Leaves and stems
All of the above
None of the above
Solution
Stems produce similar branching organs ie roots from the pericycle which gives rise to branch
roots.
The pericycle is a cylinder of parenchyma (vascular- storage and transport of nutrients) or
sclerenchyma cells (nonvascular- support and protection) that lies inside the endodermis and
outer side of stele of plants..
More from erremmfab (20)
Given the following information (ignore the top posts on the LEGO br.pdf
Given the following information (ignore the top posts on the LEGO br.pdf
Gene Splicing -1. DNA is made up of two separate strands of base s.pdf
Gene Splicing -1. DNA is made up of two separate strands of base s.pdf
For each of these functions, is it one-to-one Is it onto For each .pdf
For each of these functions, is it one-to-one Is it onto For each .pdf
Find all roots of Re(arcsinh(z)) = 0, where z is the complex number,.pdf
Find all roots of Re(arcsinh(z)) = 0, where z is the complex number,.pdf
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Discuss the history of operations, information, systems, and conting.pdf
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Dr. Butthead received a new cell line and he wants to see which cult.pdf
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Describe or explain how SNPs in gens at evolutionary conserved break.pdf
create a binary search tree from an empty one by adding the key valu.pdf
create a binary search tree from an empty one by adding the key valu.pdf
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Data Analysis correlation. A correlation R in regression analysis t.pdf
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Conduct outside reading on Salla disease. Explain its underlying mole.pdf
At absolute temperature T, a black body radiates its peak intensity a.pdf
At absolute temperature T, a black body radiates its peak intensity a.pdf
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An unsaturated solution of Na2SO4 is present in equilibrium with its.pdf
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a. List the three things that happen when the following statement is.pdf
a statistics professor receives an average of five email messages pe.pdf
a statistics professor receives an average of five email messages pe.pdf
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A researcher is convinced that the reason that people watch Jerry Sp.pdf
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You connect a host to a switch. The host is capable only of half dup.pdf
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You are a game warden studying elephant ecology in a Ugandan National.pdf
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Write a C program using low-level io which will read in the files g.pdf
which type of distribution of final grades would be most desirable t.pdf
which type of distribution of final grades would be most desirable t.pdf
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Which of the following organs produce similar, branching organs (i.e.pdf
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Model Attribute Check Company Auto Property
In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
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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.
Instructions for Submissions thorugh G- Classroom.pptx
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
678020731-Sumas-y-Restas-Para-Colorear.pdf
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How libraries can support authors with open access requirements for UKRI fund...
How libraries can support authors with open access requirements for UKRI funded books
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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.
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Presentation required for the master in Education.
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1.4 modern child centered education - mahatma gandhi-2.pptx
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Adversarial Attention Modeling for Multi-dimensional Emotion Regression.pdf
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Instructions for Submissions thorugh G- Classroom.pptx
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How libraries can support authors with open access requirements for UKRI fund...
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Home assignment II on Spectroscopy 2024 Answers.pdf
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Unit 8 - Information and Communication Technology (Paper I).pdf
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Francesca Gottschalk - How can education support child empowerment.pptx
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Thesis Statement for students diagnonsed withADHD.ppt
Can we accept mobile apps as a part of data communication networks.pdf
- 1. Can we accept mobile apps as a part of data communication networks? Solution Answer: Yes, we accept mobile apps as a part of data communication because mobile computing involves mobile communication. It allows transmission of data like texts , audio , vedio and now a days VoIP calls are made through the mobile apps. There are many mobiles apps which gives the services like voice over the internet. Thus mobile apps play a very important role in data communication network.