Find the probability mass function (PMF) of L A source of informatio.pdf
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Find the probability mass function (PMF) of L A source of information randomly generates symbols from a five letter alphabet {a, b, c. d, e}. The probability of each symbol is as follows: P(a) = 1/2; P(b) = 1/4; P(c) = 1/8; P(d) = 1/16; and P(e) = 1/16. These symbols are now encoded into binary codes using the scheme shown below. Let the random variable L denote the length of the binary code (for example, the length of the binary code for the symbol c is 3 bits). Find the probability mass function (PMF) of L. Solution PMF of L: L can take value 1, 2, 3, 4. For L=0, P(L=0)=1/2 For L=1, P(L=1)=1/4 For L=2, P(L=2)=1/8 For L=4, P(L=4)=(1/16)+(1/16)=1/8.
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Consider a binary search tree T with nodes containing the four fields.pdf
Consider a binary search tree T with nodes containing the four fields key, left, right, p. and ht,
where key is an integer, left, and right are pointers to the two subtrees, p is a pointer to the parent
of the node, and ht is the height of the node (the maximum length of a path from this node to a
leaf). Design an algorithm for a procedure insert (T, x) that inserts to a binary searches tree T a
new node x (whose key is not in the tree yet), and updates lit fields accordingly. The function
should work in O(h) time, where h is the length of the access path.
Solution
Algorithm is implemented in C:
struct node
{
int key;
struct node *left, *right;
};
//function to create a new BST node
struct node *new_Node(int item)
{
struct node *tem = (struct node *)malloc(sizeof(struct node));
tem->key = item;
tem->left = tem->right = NULL;
return tem;
}
/INSERT PROCEDURE
/* Given key in BST,function to insert a new node. */
struct node* insert(struct node* T, int x)
{
/* If tree is empty then return a new node */
if (T == NULL) return new_Node(x);
/* else, repeat down the tree */
if (x < T->x)
T->left = insert(T->left, x);
else if (key > node->key)
T->right = insert(T->right, x);
return T;
}.
Based on the articleWhy It Is Time to Look Beyond Algal Genes in P.pdf
Based on the article
Why It Is Time to Look Beyond Algal Genes in Photosynthetic Slugs
By: Cessa Rauch1, Jan de Vries 1, Sophie Rommel2, Laura E. Rose2, Christian Woehle3,
Gregor Christa1, Elise M. Laetz4, Heike Wa ¨gele4, Aloysius G.M. Tielens5,6, Jo ¨rg
Nickelsen7, Tobias Schumann8, Peter Jahns8, and Sven B. Gould1,*
Answer this:
--> It has been argued that the reason for detection of such a small amount of mRNA (the read
count) of lateral gene transfer origin in the slug cells is due to only very few cells containing
plastids. Do you agree with this argument? Why or why not?
--> The authors present several alternative reasons for the stability of plastids within the slug
cells. Describe two of these reasons.
Solution
1. No, we donot agree with the argument.
Through LGT from algae to slugs has similar RNA sequence,supported by genome sequence of
slug DNA, failed to give an evidence for such LGT events later on. So by this, it is clear that
there is no proof that the evolution of kleptoplasty in photosynthetic slugs involves LGT events,
where they depend purely on nuclear genes.
In some of the photosynthetic slugs, plastids sequestered from algae can continue to
photosynthesize for weeks, or even months, in the absence of algal nuclei. Therefore, in
conclusion, some isolated algal plastids are by nature more robust than those of land plants and
there is no evidence that any of these require the expression of slug genes of algal origin.
2. Plastids have effective photoprotection mechanisms.
Slugs acquiring robust plastids will not automatically retain them long-term and endure
starvation.
Slugs are physiologically adapted and require to retain them functionally..
Comparecontrast the following classes of hormones by identifying thr.pdf
Compare/contrast the following classes of hormones by identifying three similarities and four
differences among/between them. In addition, be able to list several specific examples of each.
AMINE vs. PEPTIDE vs. STEROID
Solution
Three similarities between the three hormones-
1. Being hormones, the site of synthesis and the site of action of all the three hormones are
different.
2. They are all organic molecules and rich in carbon and hydrogen.
3. They are all released directly into the blood stream.
Differences
Peptide
Steroid
Amine
Composition
Made of few to many amino acids(3-200), synthesized from mRNA
Derived from cholesterol (from pregnenolone, a steroid precursor)
Derived from amino acids tyrosine or tryptophan
Solubility in lipid bilayer
Not soluble
soluble
Not soluble
Nature
Hydrophillic with cell surface receptors( except- thyroid hormones are lipid soluble)
Lipophillic with intracellular receptors
Hydrophillic with cell surface receptors
Mode of Action
Through secondary messengers like cAMP, cGMP etc by cascading effect
Once inside the cell it binds DNA and modifies transcription
Through binding to beta adrenergic receptors, by secondary messengers (signal transduction) or
calcium or phosphoinositides
Examples
ADH, calcitonin, Insulin, Pituitary Hormones (LH,FSH, ACTH, TSH,GH)
Estrogen, Progesterone, testosterone, cortisol
Epinephrine, nor epinephrine, melatonin, dopamine
Peptide
Steroid
Amine
Composition
Made of few to many amino acids(3-200), synthesized from mRNA
Derived from cholesterol (from pregnenolone, a steroid precursor)
Derived from amino acids tyrosine or tryptophan
Solubility in lipid bilayer
Not soluble
soluble
Not soluble
Nature
Hydrophillic with cell surface receptors( except- thyroid hormones are lipid soluble)
Lipophillic with intracellular receptors
Hydrophillic with cell surface receptors
Mode of Action
Through secondary messengers like cAMP, cGMP etc by cascading effect
Once inside the cell it binds DNA and modifies transcription
Through binding to beta adrenergic receptors, by secondary messengers (signal transduction) or
calcium or phosphoinositides
Examples
ADH, calcitonin, Insulin, Pituitary Hormones (LH,FSH, ACTH, TSH,GH)
Estrogen, Progesterone, testosterone, cortisol
Epinephrine, nor epinephrine, melatonin, dopamine.
Why does the incidence of acute disease decrease with ageSoluti.pdf
Why does the incidence of acute disease decrease with age?
Solution
Incidence of acute diseases decrease with age because the body which is through all the worst
times have gained enough resistance though the immunity decreases with age still the body will
be ready to fight with acute infections and go over that due to the resistance developed in course
of lifetime. i hope this is a brief answer and would help you. Thank you.
White eye color in Drosophila was discovered to be sex linked by Tho.pdf
White eye color in Drosophila was discovered to be sex linked by Thomas Hunt Morgan, who
observed that:
White eye color in Drosophila was discovered to be sex linked by Thomas Hunt Morgan, who
observed that:A. reciprocal crosses produced different results, which is not true for most
autosomal traits.B. the inheritance pattern depended on the sex of the individuals.C. the white-
eyed flies were sterile, in contrast to the red-eyed flies.D. A and B.E. All of the above
statements apply.
Solution
The answer is D. A and B.
A. reciprocal crosses produced different results, which is not true for most autosomal traits.
B. the inheritance pattern depended on the sex of the individuals..
Which of the following protons would you NOT expect to find embedded .pdf
Which of the following protons would you NOT expect to find embedded in the cytoplasmic
membrane of any bacteria? Primary active transport proteins. ATP synthase. Enzyme I complex
of the Group translocation system. Basel body protein rings of flagella. Which of the following is
true? Both photosystem I and photosystem II rely on light bombarding chlorophyll. Photosystem
II donates electrons to photosystem I in a linear electron transport chain. Photosystem I only (not
photosystem II) directly produces NADPH utilized for the Calvin cycle. All of the above are
true. None of the above are true. Which one of these prokaryotic characteristics can\'t co-exist in
a single cell? Mesospheric and acidophil. Aero-tolerant and basophilic. Thermophiles and
psychrophilic. Fermentative and Anaerobic. If a population of 4 bacterial cells in log phase, how
many cells can we expect to see in an hour\'s time if the cell has a generation time of 30 minutes?
8 cells. 16 cells. 64 cells. 128 cells. The classification of ribosomes in bacteria would be:
Ribosomes are proteins and bacteria contain 70s versions. Ribosomes are proteins and bacteria
contain 80s versions. Ribosomes are membrane lipids and bacteria contain 70s versions.
Ribosomes are membrane lipids and bacteria contain 80s versions. Where do The Calvin Cycle
and Glycolysis converge? Only after the \"prep step\" once acetyl CoA is formed, leading to the
Calvin Cycle Only when NADH is utilized in the electron transport chain At Glyceraldehyde 3
Phosphate. None of the above are correct.
Solution
8.a primary active transport proteins
The basal body of a bacterial flagellum is a rod and a system of rings embedded in the cell
envelope.Gram-negative flagella usually have an L ring in the plane of the lipopolysaccharide in
the outer membrane,a periplasmic P ring in the plane of the peptidoglycan around the flagellar
rod and a M S ring that is located within and above the cytoplasmic membrane.A ,C ring extends
into the cytoplasm.
Enzyme I transfers phosphate from PEP to HPr.PEP group translocation is a distinct method
used by bacteria for sugar uptake where the source of energy is from phosphoenolpyruvate
(PEP).It is known as multicomponent system that always involves enzymes of the plasma
membrane and those in the cytoplasm.
ATP synthase is located within the thylakoid membrane and the inner mitochondrial membrane
in bacteria.
9.d all are true.
In photosystem I energy is absorbed by a pair of P700 chlrophyll a molecules raise to an excited
energy level.From there they pass onto FeS4,then onto ferrodoxin,and finally onto ferrodoxin-
NADP reductase.After 2 electrons have reduced ferrodoxin-NADP reductase they are transferred
to NADP+ reducing it to NADPH and a H+.Photosystem II works with Photosystem I and two
series of enzymes imbedded in the thylakoid membrane to transfer energy from the form of light
to that stored in chemical bands and gradients which the plant can use in a process called
noncyclic photophosph.
Which of these characteristics distinguishes angiosperms from other .pdf
Which of these characteristics distinguishes angiosperms from other seed plants?
roots
Which of these characteristics distinguishes angiosperms from other seed
plants?flowersseedstracheids
roots
Solution
1. Only angiosperms bear true fruit and have flowers.
2.sporophyte and gametophyte generations, swimming sperm and desiccation tolerance
3. monocots (ie cereals)
4. mechanical support (in form of stem and root anchored to the soil)
5. the diploid gametophyte
6. Pollen grains contain the male gametes of gymnosperms, and seeds contain the female
gametes called ovules..
Which of the following is MOST responsible for maintaining the doubl.pdf
Which of the following is MOST responsible for maintaining the double-stranded structure of
DNA?
A. Attractive interactions between positively charged purines and negatively charged
pyrimidines.
B. Hydrogen bonding between the 3\'-OH of deoxyribose and purine and pyrimidine ring
nitrogens.
C. Hydrophobic interactions between contiguous base pairs and hydrogen bonding between
complementary bases on opposite strands.
D. Interactions between the phosphate groups in separate chains.
E. Hydrophilic interactions between the pentoses in separate chains.
Solution
A. Attractive interactions between positively charged purines and negatively charged
pyrimidines.
Explanation- Attractive interactions between partial positive charge on purines and partial
negative charge on pyrimidines is responsible for maintaining stability of DNA.
Hydrogen bonds in the DNA strands are individually weak but are collectively strong. These
bonds between complementary base pairs stabilize the helical structure of DNA.
Adenine is paired through two hydrogen bonds with thymine. Guanidine is paired through three
hydrogen bonds with cytosine..
Recommended
Consider a binary search tree T with nodes containing the four fields.pdf
Consider a binary search tree T with nodes containing the four fields key, left, right, p. and ht,
where key is an integer, left, and right are pointers to the two subtrees, p is a pointer to the parent
of the node, and ht is the height of the node (the maximum length of a path from this node to a
leaf). Design an algorithm for a procedure insert (T, x) that inserts to a binary searches tree T a
new node x (whose key is not in the tree yet), and updates lit fields accordingly. The function
should work in O(h) time, where h is the length of the access path.
Solution
Algorithm is implemented in C:
struct node
{
int key;
struct node *left, *right;
};
//function to create a new BST node
struct node *new_Node(int item)
{
struct node *tem = (struct node *)malloc(sizeof(struct node));
tem->key = item;
tem->left = tem->right = NULL;
return tem;
}
/INSERT PROCEDURE
/* Given key in BST,function to insert a new node. */
struct node* insert(struct node* T, int x)
{
/* If tree is empty then return a new node */
if (T == NULL) return new_Node(x);
/* else, repeat down the tree */
if (x < T->x)
T->left = insert(T->left, x);
else if (key > node->key)
T->right = insert(T->right, x);
return T;
}.
Based on the articleWhy It Is Time to Look Beyond Algal Genes in P.pdf
Based on the article
Why It Is Time to Look Beyond Algal Genes in Photosynthetic Slugs
By: Cessa Rauch1, Jan de Vries 1, Sophie Rommel2, Laura E. Rose2, Christian Woehle3,
Gregor Christa1, Elise M. Laetz4, Heike Wa ¨gele4, Aloysius G.M. Tielens5,6, Jo ¨rg
Nickelsen7, Tobias Schumann8, Peter Jahns8, and Sven B. Gould1,*
Answer this:
--> It has been argued that the reason for detection of such a small amount of mRNA (the read
count) of lateral gene transfer origin in the slug cells is due to only very few cells containing
plastids. Do you agree with this argument? Why or why not?
--> The authors present several alternative reasons for the stability of plastids within the slug
cells. Describe two of these reasons.
Solution
1. No, we donot agree with the argument.
Through LGT from algae to slugs has similar RNA sequence,supported by genome sequence of
slug DNA, failed to give an evidence for such LGT events later on. So by this, it is clear that
there is no proof that the evolution of kleptoplasty in photosynthetic slugs involves LGT events,
where they depend purely on nuclear genes.
In some of the photosynthetic slugs, plastids sequestered from algae can continue to
photosynthesize for weeks, or even months, in the absence of algal nuclei. Therefore, in
conclusion, some isolated algal plastids are by nature more robust than those of land plants and
there is no evidence that any of these require the expression of slug genes of algal origin.
2. Plastids have effective photoprotection mechanisms.
Slugs acquiring robust plastids will not automatically retain them long-term and endure
starvation.
Slugs are physiologically adapted and require to retain them functionally..
Comparecontrast the following classes of hormones by identifying thr.pdf
Compare/contrast the following classes of hormones by identifying three similarities and four
differences among/between them. In addition, be able to list several specific examples of each.
AMINE vs. PEPTIDE vs. STEROID
Solution
Three similarities between the three hormones-
1. Being hormones, the site of synthesis and the site of action of all the three hormones are
different.
2. They are all organic molecules and rich in carbon and hydrogen.
3. They are all released directly into the blood stream.
Differences
Peptide
Steroid
Amine
Composition
Made of few to many amino acids(3-200), synthesized from mRNA
Derived from cholesterol (from pregnenolone, a steroid precursor)
Derived from amino acids tyrosine or tryptophan
Solubility in lipid bilayer
Not soluble
soluble
Not soluble
Nature
Hydrophillic with cell surface receptors( except- thyroid hormones are lipid soluble)
Lipophillic with intracellular receptors
Hydrophillic with cell surface receptors
Mode of Action
Through secondary messengers like cAMP, cGMP etc by cascading effect
Once inside the cell it binds DNA and modifies transcription
Through binding to beta adrenergic receptors, by secondary messengers (signal transduction) or
calcium or phosphoinositides
Examples
ADH, calcitonin, Insulin, Pituitary Hormones (LH,FSH, ACTH, TSH,GH)
Estrogen, Progesterone, testosterone, cortisol
Epinephrine, nor epinephrine, melatonin, dopamine
Peptide
Steroid
Amine
Composition
Made of few to many amino acids(3-200), synthesized from mRNA
Derived from cholesterol (from pregnenolone, a steroid precursor)
Derived from amino acids tyrosine or tryptophan
Solubility in lipid bilayer
Not soluble
soluble
Not soluble
Nature
Hydrophillic with cell surface receptors( except- thyroid hormones are lipid soluble)
Lipophillic with intracellular receptors
Hydrophillic with cell surface receptors
Mode of Action
Through secondary messengers like cAMP, cGMP etc by cascading effect
Once inside the cell it binds DNA and modifies transcription
Through binding to beta adrenergic receptors, by secondary messengers (signal transduction) or
calcium or phosphoinositides
Examples
ADH, calcitonin, Insulin, Pituitary Hormones (LH,FSH, ACTH, TSH,GH)
Estrogen, Progesterone, testosterone, cortisol
Epinephrine, nor epinephrine, melatonin, dopamine.
Why does the incidence of acute disease decrease with ageSoluti.pdf
Why does the incidence of acute disease decrease with age?
Solution
Incidence of acute diseases decrease with age because the body which is through all the worst
times have gained enough resistance though the immunity decreases with age still the body will
be ready to fight with acute infections and go over that due to the resistance developed in course
of lifetime. i hope this is a brief answer and would help you. Thank you.
White eye color in Drosophila was discovered to be sex linked by Tho.pdf
White eye color in Drosophila was discovered to be sex linked by Thomas Hunt Morgan, who
observed that:
White eye color in Drosophila was discovered to be sex linked by Thomas Hunt Morgan, who
observed that:A. reciprocal crosses produced different results, which is not true for most
autosomal traits.B. the inheritance pattern depended on the sex of the individuals.C. the white-
eyed flies were sterile, in contrast to the red-eyed flies.D. A and B.E. All of the above
statements apply.
Solution
The answer is D. A and B.
A. reciprocal crosses produced different results, which is not true for most autosomal traits.
B. the inheritance pattern depended on the sex of the individuals..
Which of the following protons would you NOT expect to find embedded .pdf
Which of the following protons would you NOT expect to find embedded in the cytoplasmic
membrane of any bacteria? Primary active transport proteins. ATP synthase. Enzyme I complex
of the Group translocation system. Basel body protein rings of flagella. Which of the following is
true? Both photosystem I and photosystem II rely on light bombarding chlorophyll. Photosystem
II donates electrons to photosystem I in a linear electron transport chain. Photosystem I only (not
photosystem II) directly produces NADPH utilized for the Calvin cycle. All of the above are
true. None of the above are true. Which one of these prokaryotic characteristics can\'t co-exist in
a single cell? Mesospheric and acidophil. Aero-tolerant and basophilic. Thermophiles and
psychrophilic. Fermentative and Anaerobic. If a population of 4 bacterial cells in log phase, how
many cells can we expect to see in an hour\'s time if the cell has a generation time of 30 minutes?
8 cells. 16 cells. 64 cells. 128 cells. The classification of ribosomes in bacteria would be:
Ribosomes are proteins and bacteria contain 70s versions. Ribosomes are proteins and bacteria
contain 80s versions. Ribosomes are membrane lipids and bacteria contain 70s versions.
Ribosomes are membrane lipids and bacteria contain 80s versions. Where do The Calvin Cycle
and Glycolysis converge? Only after the \"prep step\" once acetyl CoA is formed, leading to the
Calvin Cycle Only when NADH is utilized in the electron transport chain At Glyceraldehyde 3
Phosphate. None of the above are correct.
Solution
8.a primary active transport proteins
The basal body of a bacterial flagellum is a rod and a system of rings embedded in the cell
envelope.Gram-negative flagella usually have an L ring in the plane of the lipopolysaccharide in
the outer membrane,a periplasmic P ring in the plane of the peptidoglycan around the flagellar
rod and a M S ring that is located within and above the cytoplasmic membrane.A ,C ring extends
into the cytoplasm.
Enzyme I transfers phosphate from PEP to HPr.PEP group translocation is a distinct method
used by bacteria for sugar uptake where the source of energy is from phosphoenolpyruvate
(PEP).It is known as multicomponent system that always involves enzymes of the plasma
membrane and those in the cytoplasm.
ATP synthase is located within the thylakoid membrane and the inner mitochondrial membrane
in bacteria.
9.d all are true.
In photosystem I energy is absorbed by a pair of P700 chlrophyll a molecules raise to an excited
energy level.From there they pass onto FeS4,then onto ferrodoxin,and finally onto ferrodoxin-
NADP reductase.After 2 electrons have reduced ferrodoxin-NADP reductase they are transferred
to NADP+ reducing it to NADPH and a H+.Photosystem II works with Photosystem I and two
series of enzymes imbedded in the thylakoid membrane to transfer energy from the form of light
to that stored in chemical bands and gradients which the plant can use in a process called
noncyclic photophosph.
Which of these characteristics distinguishes angiosperms from other .pdf
Which of these characteristics distinguishes angiosperms from other seed plants?
roots
Which of these characteristics distinguishes angiosperms from other seed
plants?flowersseedstracheids
roots
Solution
1. Only angiosperms bear true fruit and have flowers.
2.sporophyte and gametophyte generations, swimming sperm and desiccation tolerance
3. monocots (ie cereals)
4. mechanical support (in form of stem and root anchored to the soil)
5. the diploid gametophyte
6. Pollen grains contain the male gametes of gymnosperms, and seeds contain the female
gametes called ovules..
Which of the following is MOST responsible for maintaining the doubl.pdf
Which of the following is MOST responsible for maintaining the double-stranded structure of
DNA?
A. Attractive interactions between positively charged purines and negatively charged
pyrimidines.
B. Hydrogen bonding between the 3\'-OH of deoxyribose and purine and pyrimidine ring
nitrogens.
C. Hydrophobic interactions between contiguous base pairs and hydrogen bonding between
complementary bases on opposite strands.
D. Interactions between the phosphate groups in separate chains.
E. Hydrophilic interactions between the pentoses in separate chains.
Solution
A. Attractive interactions between positively charged purines and negatively charged
pyrimidines.
Explanation- Attractive interactions between partial positive charge on purines and partial
negative charge on pyrimidines is responsible for maintaining stability of DNA.
Hydrogen bonds in the DNA strands are individually weak but are collectively strong. These
bonds between complementary base pairs stabilize the helical structure of DNA.
Adenine is paired through two hydrogen bonds with thymine. Guanidine is paired through three
hydrogen bonds with cytosine..
What are the role of environmental factors on transpiration and h.pdf
What are the role of environmental factors on transpiration and how plants defend themselves
against water loss?
What are the role of environmental factors on transpiration and how plants defend themselves
against water loss?
Solution
Role of environmental factors on transpiration
1. Relative humidity
The loss of water vapour from stomata occurs by simple diffusion. When the atmosphere is very
humid, the rate of loss of water vapour is lowered.
Wind:
When the wind velocity is high, rate of transpiration also increases.
Temperature:
Warm air hold more water vapour so the rate of loss of water increases with increase in
temperature.
Light:
Opening and closing of stomata is regulated by light. More light result in high degree of
transpiration.
Soil water content:
When there is less water content in the soil , the leaf droops / get dehydrated making the stomata
to close resulting in less rate of transpiration.
Carbon dioxide concentration
When the concentration is high, stomata closes resulting in low rate of transpiration and vice
versa.
Plants defense mechanism against water loss
Plants require water for it\'s survival. So it will have to regulate water loss ( especially during
drought) .
Several leaf adaptations help to combat water loss.
Leaf arrangements play important role, transversely arranged leaves loss more water than
perpendicularly arranged ones.
Other anatomical features of the leaves such as sunken/vestigial stomata and closely arranged
mesophyll cells reduces transpiration.
Presence of hair, cuticle(waxy) , gums, mucilage etc on leaf surface help reducing transpiration.
During drought, plant hormone abscisic acid play an important role by closing the stomata
reducing water loss..
What will be quantization step size in numbers and in voltage for th.pdf
What will be quantization step size in numbers and in voltage for this Arduino Code? Using 5V
const int led1 = 2;
const int led2 = 3;
const int led3 = 4;
void setup() {
pinMode(led1, OUTPUT);
pinMode(led2, OUTPUT);
pinMode(led3, OUTPUT);
// initialize serial communication at 9600 bits per second:
Serial.begin(9600);
}
// the loop routine runs over and over again forever:
void loop() {
// read the input on analog pin 0:
int sensorValue = analogRead(A0);
int dataConv = sensorValue*(8.0/1024);
//write analog equvivalant data on led pins
switch(dataConv)
{
case 0: {
digitalWrite(led1,LOW);
digitalWrite(led2,LOW);
digitalWrite(led3,LOW);
break;
}
case 1: {
digitalWrite(led1,HIGH);
digitalWrite(led2,LOW);
digitalWrite(led3,LOW);
break;
}
case 2: {
digitalWrite(led1,LOW);
digitalWrite(led2,HIGH);
digitalWrite(led3,LOW);
break;
}
case 3: {
digitalWrite(led1,HIGH);
digitalWrite(led2,HIGH);
digitalWrite(led3,LOW);
break;
}
case 4: {
digitalWrite(led1,LOW);
digitalWrite(led2,LOW);
digitalWrite(led3,HIGH);
break;
}
case 5: {
digitalWrite(led1,HIGH);
digitalWrite(led2,LOW);
digitalWrite(led3,HIGH);
break;
}
case 6: {
digitalWrite(led1,LOW);
digitalWrite(led2,HIGH);
digitalWrite(led3,HIGH);
break;
}
case 7: {
digitalWrite(led1,HIGH);
digitalWrite(led2,HIGH);
digitalWrite(led3,HIGH);
break;
}
}
Serial.println(sensorValue); // print out the value you read:
Serial.println(dataConv);
delay(1000); // delay in between reads for stability
}
const int led1 = 2;
const int led2 = 3;
const int led3 = 4;
void setup() {
pinMode(led1, OUTPUT);
pinMode(led2, OUTPUT);
pinMode(led3, OUTPUT);
// initialize serial communication at 9600 bits per second:
Serial.begin(9600);
}
// the loop routine runs over and over again forever:
void loop() {
// read the input on analog pin 0:
int sensorValue = analogRead(A0);
int dataConv = sensorValue*(8.0/1024);
//write analog equvivalant data on led pins
switch(dataConv)
{
case 0: {
digitalWrite(led1,LOW);
digitalWrite(led2,LOW);
digitalWrite(led3,LOW);
break;
}
case 1: {
digitalWrite(led1,HIGH);
digitalWrite(led2,LOW);
digitalWrite(led3,LOW);
break;
}
case 2: {
digitalWrite(led1,LOW);
digitalWrite(led2,HIGH);
digitalWrite(led3,LOW);
break;
}
case 3: {
digitalWrite(led1,HIGH);
digitalWrite(led2,HIGH);
digitalWrite(led3,LOW);
break;
}
case 4: {
digitalWrite(led1,LOW);
digitalWrite(led2,LOW);
digitalWrite(led3,HIGH);
break;
}
case 5: {
digitalWrite(led1,HIGH);
digitalWrite(led2,LOW);
digitalWrite(led3,HIGH);
break;
}
case 6: {
digitalWrite(led1,LOW);
digitalWrite(led2,HIGH);
digitalWrite(led3,HIGH);
break;
}
case 7: {
digitalWrite(led1,HIGH);
digitalWrite(led2,HIGH);
digitalWrite(led3,HIGH);
break;
}
}
Serial.println(sensorValue); // print out the value you read:
Serial.println(dataConv);
delay(1000); // delay in between reads for stability
}
Solution
// Random LED Dots - from noise source
// Ed Nisley - KE4ANU - September 2015
//----------
// Pin assignments
const byte PIN_HEARTBEAT = 8; // DO - heartbeat LED
const byte PIN_SYNC = A3; // DO.
What is the need for a shadow page table (One or two brief sentences.pdf
What is the need for a shadow page table? (One or two brief sentences please)
Solution
Virtualization allows multiple operating system instances to run concurrently on a single
computer. Virtualization is the creation of a virtual rather than actual version of something, such
as an operating system, a server, a storage device or network resources.
Virtualization is where the framework divides the resource into one or more execution
environments Even something as simple as partitioning a hard drive is considered virtualization
because you take one drive and partition it to create two separate hard drives.
Types of virtualization:
Storage virtualization: is the pooling of physical storage from multiple network storage devices
into what appears to be a single storage device
Data virtualization: is abstracting the traditional technical details of data and data management,
such as location, performance or format..
Network virtualization: is a method of combining the available resources in a network by
splitting up the available bandwidth into channels, each of which is independent from the others.
Server virtualization: the partitioning a physical server into smaller virtual servers.
Shadow Tables are objects used to improve the way machines, networks and programs handle
information.
Shadow page tables are used by the hypervisor to keep track of the state in which the guest
thinks its page tables should be.
The guest can\'t be allowed access to the hardware page tables because then it would essentially
have control of the machine. So, the hypervisor keeps the \"real\" mappings guest virtual -> host
physical in the hardware when the relevant guest is executing, and keeps a representation of the
page table..
What step of protein synthesis is shown in the figure elongation st.pdf
What step of protein synthesis is shown in the figure? elongation stage of transcription
termination stage of translation mRNA processing elongation stage of translation initiation
stage of transcription
Solution
elongation stage of transcription
Explanation: newly synthesized mRNA separates from the DNA template as the RNA plymerase
continues elongation. multiple RNA plymerases transcribe simultaneously..
Water is forced through a contraction causing low pressure. The wat.pdf
Water is forced through a contraction causing low pressure. The water is observed to boil at a
pressure of -13.01 psi. If atmospheric pressure is 14.70 psi, what is the temperature, in °F, of the
water?
Solution
the water is start boiling at 212 °F
it means the temperature of water at 14.70psi is 212 °F +239.538 °F =451.53 °F.
Using a PIC 18 Write a time overflow interrupt routine that detects .pdf
Using a PIC 18 Write a time overflow interrupt routine that detects the interrupt source if it
overflows, increments the timer overflow counter, and finally clears the interrupt flag
Solution
ORG 00H
GOTO MAIN
ORG 0X08H
MOVLW 01H
ADDWF TMR0L
BCF INTCON,TMR0IF
RETIF
MAIN
ORG 100H
BCF INTCON,TMR0IF
BSF T0CON,TMR0ON
BSF INTCON,TMR0IE
BSF INTC0N,GIE
HERE GOTO HERE
END.
The three types of mixtures are and The weak bond forming a bridge .pdf
The three types of mixtures are and The weak bond forming a bridge between a hydrogen atom
and nitrogen or oxygen is a bond. A is an inorganic compound which has a bitter taste, feels
slipper and is a proton acceptor.
Solution
10. Three types of mixtures are suspensions, solutions and colloids
11.answer is hydrogen bond.
A hydrogen bond is the electrostatic attraction between two polar groups that occurs when a
hydrogen (H) atom covalently bound to a highly electronegative atom such as nitrogen (N),
oxygen (O), or fluorine (F) experiences the electrostatic field of another highly electronegative
atom nearby.
12.answer is Base..
Some commercial airplanes recirculate approximately 50 of the cabin .pdf
Some commercial airplanes recirculate approximately 50% of the cabin air in order to increase
fuel efficiency. The researchers studied 1096 airline passengers, among which some traveled on
airplanes that recirculated air and others traveled on planes that did not recirculate air. Of the 515
passengers who flew on planes that did not recirculate air, 106 reported post-flight respiratory
symptoms, while 112 of the 581 passengers on planes that did recirculate air reported such
symptoms. Is there sufficient evidence to conclude that the proportion of passengers with post-
flight respiratory symptoms differs for planes that do and do not recirculate air? Test the
appropriate hypotheses using a = 0.05. You may assume that it is reasonable to regard these two
samples as being independently selected and as representative of the two populations of interest.
(Use a statistical computer package to calculate the P-value. Use Pdo not recirculate - Pdo
recirculate- Round your test statistic to two decimal places and your P-value to four decimal
places.)
Solution
Null , There Is No Significance between them Ho: p1 = p2
Alternate , There Is Significance between them H1: p1 != p2
Test Statistic
Sample 1 : X1 =106, n1 =515, P1= X1/n1=0.206
Sample 2 : X2 =112, n2 =581, P2= X2/n2=0.193
Finding a P^ value For Proportion P^=(X1 + X2 ) / (n1+n2)
P^=0.199
Q^ Value For Proportion= 1-P^=0.801
we use Test Statistic (Z) = (P1-P2)/(P^Q^(1/n1+1/n2))
Zo =(0.206-0.193)/Sqrt((0.199*0.801(1/515+1/581))
Zo =0.54
| Zo | =0.54
Critical Value
The Value of |Z | at LOS 0.05% is 1.96
We got |Zo| =0.54 & | Z | =1.96
Make Decision
Hence Value of |Zo | < | Z | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) -Ha : ( P != 0.5403 ) = 0.589
Hence Value of P0.05 < 0.589,Here We Do not Reject Ho
No sufficient evidence to condude that the proportion of passengers
with post-flight respratory symptoms differs for planes that do and do not recirculate air
ANS:
Z = 0.54
P = 0.589.
Solve for 02 sin 2+sin 5=0. Separate your answers with commas.pdf
Solve for 0??<2?: sin 2?+sin 5?=0. Separate your answers with commas.
?= Solve for 0 lessthanorequalto theta
Solution
Use formula : sinx +siny = 2sin(x+y)/2cos(x-y)/2
So, sin 2+sin 5=0.
2sin(7)/2cos(3/2) =0
sin(7)/2 =0
= 0, 2pi/7 , 4pi/7, 6pi/7 , 8pi/7 , 10pi/7 , 12pi/7 ,
cos(3)/2 =0
= pi/3 , pi , 5pi/3.
Required 1. Briefly describe different computer controls available .pdf
Required: 1. Briefly describe different computer controls available to an organization.
a. Identify how these controls help to maintain system security and reduce risk.
b. Do companies need a formal internal policy that addresses security violations? Why/why not?
c. Consider what type of actions a company should take against employees that circumvent
computer
Solution
Security Controls:
Administrative Controls, Technical or Logical Controls, Physical Controls
The constituents of Security:
Vulnerability, Threat, Risk, Exposure, Countermeasure or Safeguard
The data vendor grants access needs or may choose to give this function to business unit
managers. And it is the data owner who will contract with security violations pertaining to the
data he is liable for defending.
Subjects that should be measured when developing a security policy:
Classification Controls:
Qualitative Risk Analysis Techniques:.
a. What is Snort [1]b. According to “Snort Users Manual” (downloa.pdf
a. What is Snort? [1]
b. According to “Snort Users Manual” (downloadable under “Docs” menu), what are the three
modes in which Snort operates and how it behaves under each mode? Show an example
command to put Snort in each mode. [6]
c. Explain the difference between the rule headers and rule options [2]
d. List 8 different types of actions Snort may take upon a rule match. Describe what each action
type does. [2]
Solution
a. Snort: Snort is an open source network capable of performing real-time traffic study and
packet-logging on IP networks. It can transmit the protocol analysis, content searching &
matching and is used to detect a range of attacks and investigates, such as buffer run over,
cautious port scans, OS fingerprinting efforts.
d. Types of snort actions:.
Proponents of decentralized forms of socialism usually supportA co.pdf
Proponents of decentralized forms of socialism usually support
A command economy
Control of the economy by a vanguard party of intellectuals
Control of the economy by various local groups of workers
State control of the economy
The Phoenix Suns
A command economy
Control of the economy by a vanguard party of intellectuals
Control of the economy by various local groups of workers
State control of the economy
The Phoenix Suns
Solution
Proponents of decentralized forms of socialism usually support the following things such a
1.A command economy.
2.Control of the economy by a vanguard party of intellectuals.
3.Control of the economy by various local groups of workers.
4.State control of the economy.
5.The phoenix suns..
Please convert the following C code to assembly Y86int i,j; ......pdf
Please convert the following C code to assembly Y86
int i,j; .....
if (i >j) {
i= i+5;
} else {
i=0;
j++;
}
Solution
Answer:
MIPS Assembly Language Code :
.zero 1
main:
push rbp
mov rbp, rsp
mov eax, DWORD PTR [rbp-4]
cmp eax, DWORD PTR [rbp-8]
jle .L2
add DWORD PTR [rbp-4], 5
jmp .L3
.L2:
mov DWORD PTR [rbp-4], 0
add DWORD PTR [rbp-8], 1
.L3:
mov eax, 0
pop rbp
ret
__static_initialization_and_destruction_0(int, int):
push rbp
mov rbp, rsp
sub rsp, 16
mov DWORD PTR [rbp-4], edi
mov DWORD PTR [rbp-8], esi
cmp DWORD PTR [rbp-4], 1
jne .L7
cmp DWORD PTR [rbp-8], 65535
jne .L7
mov edi, OFFSET FLAT:std::__ioinit
call std::ios_base::Init::Init()
mov edx, OFFSET FLAT:__dso_handle
mov esi, OFFSET FLAT:std::__ioinit
mov edi, OFFSET FLAT:std::ios_base::Init::~Init()
call __cxa_atexit
.L7:
nop
leave
ret
push rbp
mov rbp, rsp
mov esi, 65535
mov edi, 1
call __static_initialization_and_destruction_0(int, int)
pop rbp
ret.
Outline the four steps in the data collection process.Explain why .pdf
Outline the four steps in the data collection process.
Explain why it is so important to be systematic in collecting data.
Solution
Step 1: Identify issues and/or opportunities for collecting dataordata collection goals
what problem are you trying to solve by collecting this data?
Step 2:Develop operational definitions and procedures.
Here we need to be very clear as to what we are measuring, how it is to be measured, and who is
to measure it. Often times we will employ sampling in which case we need to define a sampling
plan.
. Who will the data be collected about?
. Who will the group of interest be compared to?
. What locations or geographical areas will the data be gathered from?
. What categories will be used to identify the group of interest and comparator group?
Step 3 : How should data be collected and What sources of data should be used to collect
information?
There are several methods of collecting data: online-surveys, phone interviews, focus groups and
yes, even dreaded handwritten surveys. Each type possesses its own advantages and
disadvantages. Whichever method you use, there are bound to be obstacles to overcome along
the way.
What sources of data should be used to collect information?
Pre-existing or official data
Survey data
Interviews and focus groups
Observed data
. Two types of data are used in any field of research: qualitative and quantitative.
Qualitative data:
Typically, data is called “qualitative” if it is in the form of words, but may also include any
information that is not numerical in form, such as photographs, videos and sound recordings.
Qualitative methods are aimed at describing a specific context, event, people or relationship in a
broad contextual way, by trying to understand the underlying reasons for behaviour, thoughts
and feelings.
Common qualitative research methods include observation, one-on-one interviews, focus groups
and intensive case studies
Quantitative data:
Typically, data is called “quantitative” if it is in the form of numbers.
A quantitative approach can be used to count events or the number of people who represent a
particular background.
Common quantitative tools include surveys, questionnaires and statistical data (such as Statistics
Canada census information).
It is important to note that all quantitative data is based on qualitative judgment. In other words,
numbers cannot be interpreted by themselves, without understanding the assumptions that
underlie them.
A good research effort involves the use of both types. Both approaches, while distinct, can
overlap and rely on the other to produce meaningful data, analysis and results.
Step 4 : Analyze and interpret data and Act on results
Explaining the technical steps involved in analyzing and interpreting data is beyond the scope of
this guide. An organization will have to determine whether it has the internal capacity and
expertise to analyze and interpret data itself, or whether it will need the help of an external
consultant.
. A summa.
Name the author of 1847 review article that conclusively demonstrated.pdf
Name the author of 1847 review article that conclusively demonstrated the validity of the
\"energy concept.\" Please state the \"energy concept\" that developed in Europe during the first
half of the 19^th century. Who was the first person to state the First Law in modern
mathematical terms AND in what year was his work published?
Solution
Hermann von Hermholtz was the author of validity of energy concept in 1847. The energy
concept in the first half of 19th century, i.e., the first law is the version of law of conservation of
energy, which was adopted for thermodynamic systems. It states that total energy of isolated
system is constant, energy can neither be created nor destroyed. It is often stated as the change in
the internal energy of closed system is equal to the amount of heat supplied to the system, minus
the amount of work done by the system on its surroundings and further the impossibility of
perpetual motion machines.
Rudolf clausius was the first person to state first law in modern mathematical terms and his well
esteemed work was published in 1850..
Many of our National Parks are inhabited by large numbers of black b.pdf
Many of our National Parks are inhabited by large numbers of black bears (Ursus americanus),
and their use of the Park can be a problem for humans. These creatures\' travels are governed by
the need for food, as well as a high degree of natural curiosity. As it happens, some humans
inadvertently or purposefully feed these bears, leading them to harass the human population.
When bears become troublesome, one strategy is to transplant them
Solution
Let us prepare chi square table as follows
H0: Transplant and gender are not associated
Ha: There is associaton between them
?2 Calculations
Two-way Contingency Table
?2 = 10.432, df = 2, ?2/df = 5.22 , P(?2 > 10.432) = 0.0054
As p value is less than alpha
reject null hypothesis.
It is statistically evident that transplant outcome was related to genderBlack Bears:
Transplantation Fate by Gender
MaleFemaleSuccessful32
26.85
( 0.99)17
22.15
( 1.20)49Returned34
43.29
( 1.99)45
35.71
( 2.42)79Nuisance14
9.86
( 1.74)4
8.14
( 2.10)188066146.
Implement the ADT stack by using an array stack to contain its entri.pdf
Implement the ADT stack by using an array stack to contain its entries. Expand the array
dynamically, as necessary. Maintain the stack’s bottom entry in stack[stack.length – 1].
Your Stack class must implement StackInterface (provided). You may use ArrayStack.java
(provided) as the starting point for your implementation.
You must implement a comprehensive set of unit tests using the main() method (and private
utility methods) in ArrayStack.java.
ArrayStack.java
public class ArrayStack implements StackInterface
{
private T[] stack; // Array of stack entries
private int topIndex; // Index of top entry
private boolean initialized = false;
private static final int DEFAULT_CAPACITY = 50;
private static final int MAX_CAPACITY = 10000;
public ArrayStack()
{
this(DEFAULT_CAPACITY);
} // end default constructor
public ArrayStack(int initialCapacity)
{
checkCapacity(initialCapacity);
// The cast is safe because the new array contains null entries
@SuppressWarnings(\"unchecked\")
T[] tempStack = (T[])new Object[initialCapacity];
stack = tempStack;
topIndex = -1;
initialized = true;
} // end constructor
// < Implementations of the stack operations go here. >
// < Implementations of the private methods go here; checkCapacity and
// checkInitialization are analogous to those in Chapter 2. >
// . . .
} // end ArrayStack
StackInterface.java
public interface StackInterface
{
/** Adds a new entry to the top of this stack.
@param newEntry An object to be added to the stack. */
public void push(T newEntry);
/** Removes and returns this stack\'s top entry.
@return The object at the top of the stack.
@throws EmptyStackException if the stack is empty before the operation. */
public T pop();
/** Retrieves this stack\'s top entry.
@return The object at the top of the stack.
@throws EmptyStackException if the stack is empty. */
public T peek();
/** Detects whether this stack is empty.
@return True if the stack is empty. */
public boolean isEmpty();
/** Removes all entries from this stack. */
public void clear();
} // end StackInterface
Solution
StackTester.java
public class StackTester
{
public static void main(String[] args)
{
ArrayStack aS = new ArrayStack();
aS.push(3);
aS.push(5);
aS.push(10);
aS.push(11);
System.out.println(\"\"+ aS.peek() + \", \" + aS.pop() + \", \" + aS.peek2());
aS.remove(1);
}
}
ArrayStack.java
import java.util.Arrays;
public class ArrayStack implements StackInterface
{
private T[] stack;
private int topIndex;
private boolean initialized = false;
private static final int DEFAULT_CAPACITY = 50;
private static final int MAX_CAPACITY = 10000;
public ArrayStack()
{
this(DEFAULT_CAPACITY);
}
public ArrayStack(int initialCapacity)
{
checkCapacity(initialCapacity);
@SuppressWarnings(\"unchecked\")
T[] tempStack = (T[])new Object[initialCapacity];
stack = tempStack;
topIndex = -1;
initialized = true;
}
public boolean isEmpty()
{
return topIndex < 0;
}
private void ensureCapacity()
{
if(topIndex == stack.length - 1)
{
int newLength = 2*stack.length;
.
How is the energy transferred from the source to the load in any elec.pdf
How is the energy transferred from the source to the load in any electromagnetic circuit? Does
the magnitude of the reflection coefficient increase, decrease, or stay the same as you move away
from the load on a lossless transmission line? Why?
Solution
Wave’ is a common term for a number of different ways in which energy is transferred:
Waves transfer energy but not mass
When we watch surf waves coming into shore, it’s easy to think that individual water particles
are moving towards us, but that’s not actually the case. The particles involved in waves move
back and forth perpendicularly to the way the wave is going, but don’t move significantly in the
direction of the wave. The particles ‘take part’ in the wave by bumping into one another and
transferring energy. This is why energy can be transferred, even though the average position of
the particles doesn’t change.
How does this work? It can help to think of a buoy bobbing in the ocean. The buoy is moved up
and down by the waves that pass by it, but doesn’t move directionally across the water.
You could also think about a Mexican wave at a sports match. The wave moves around the
arena, but the audience members don’t move around with it – they only stand up and sit down (a
perpendicular movement to the wave direction).
Particles in a water wave exchange kinetic energy for potential energy
When particles in water become part of a wave, they start to move up or down. This means that
kinetic energy (energy of movement) has been transferred to them. As the particles move further
away from their normal position (up towards the wave crest or down towards the trough), they
slow down. This means that some of theirkinetic energy has been converted into potential energy
– the energy of particles in a wave oscillates between kinetic and potential energy.
Thinking about potential energy can help us understand why tsunamis can be so damaging.
When a tsunami approaches the shore, it shoals (becomes much higher), so the water particles
are displaced further from equilibrium. They acquire a lot of potential energy, and this is released
when the wave interacts with land.
--------------------------------------------------------------------------------------------------------------------
-------------------------------------------------
The magnitude of reflection coefficeint decreases
that the magnitude of the reflection coefficient does not depend on the length of the line, only the
load impedance and the impedance of the transmission line. Also, note that if ZL=Z0, then the
line is \"matched\". In this case, there is no mismatch loss and all power is transferred to the load.
At this point, you should begin to understand the importance of impedance matching: grossly
mismatched impedances will lead to most of the power reflected away from the load.
Note that the reflection coefficient can be a real or a complex number..
Given Rhombus RHOB Prove HSR HSO Solution We know that .pdf
Given: Rhombus RHOB Prove: HSR HSO
Solution
We know that in traingles hsr and hso , the diagonal is equally split by other diagonal
So,
RS = SO.........1
HS is common side..........2
And, all sides of rhombus are equal, So, HR=HO.............3
Hence, since the 3 sides are equal, the triangles are congruent..
Find the real part, the imaginary part, and the absolute value of sin.pdf
Find the real part, the imaginary part, and the absolute value of sin (x - iy).
Solution
sin(x-iy)=sinx.cosiy-cosx.siniy
Since siniy=i sinhy ,cosiy=coshy
Sin(x-iy)=sinx.coshy- cosx.i sinhy
= sinx.1/2 (e^y +e^-y) -cosx.i.1/2(e^x-e^-x)
Real part=1/2 sinx(e^y+e^-y)
Imaginary part=-1/2 cosx(e^x-e^-x)
Absolute value=1/2{sinx(e^y+e^-y)+cosx(e^x-e^-x)}.
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdf
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.
CLASS 11 CBSE B.St Project AIDS TO TRADE - INSURANCE
Class 11 CBSE Business Studies Project ( AIDS TO TRADE - INSURANCE)
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What are the role of environmental factors on transpiration and h.pdf
What are the role of environmental factors on transpiration and how plants defend themselves
against water loss?
What are the role of environmental factors on transpiration and how plants defend themselves
against water loss?
Solution
Role of environmental factors on transpiration
1. Relative humidity
The loss of water vapour from stomata occurs by simple diffusion. When the atmosphere is very
humid, the rate of loss of water vapour is lowered.
Wind:
When the wind velocity is high, rate of transpiration also increases.
Temperature:
Warm air hold more water vapour so the rate of loss of water increases with increase in
temperature.
Light:
Opening and closing of stomata is regulated by light. More light result in high degree of
transpiration.
Soil water content:
When there is less water content in the soil , the leaf droops / get dehydrated making the stomata
to close resulting in less rate of transpiration.
Carbon dioxide concentration
When the concentration is high, stomata closes resulting in low rate of transpiration and vice
versa.
Plants defense mechanism against water loss
Plants require water for it\'s survival. So it will have to regulate water loss ( especially during
drought) .
Several leaf adaptations help to combat water loss.
Leaf arrangements play important role, transversely arranged leaves loss more water than
perpendicularly arranged ones.
Other anatomical features of the leaves such as sunken/vestigial stomata and closely arranged
mesophyll cells reduces transpiration.
Presence of hair, cuticle(waxy) , gums, mucilage etc on leaf surface help reducing transpiration.
During drought, plant hormone abscisic acid play an important role by closing the stomata
reducing water loss..
What will be quantization step size in numbers and in voltage for th.pdf
What will be quantization step size in numbers and in voltage for this Arduino Code? Using 5V
const int led1 = 2;
const int led2 = 3;
const int led3 = 4;
void setup() {
pinMode(led1, OUTPUT);
pinMode(led2, OUTPUT);
pinMode(led3, OUTPUT);
// initialize serial communication at 9600 bits per second:
Serial.begin(9600);
}
// the loop routine runs over and over again forever:
void loop() {
// read the input on analog pin 0:
int sensorValue = analogRead(A0);
int dataConv = sensorValue*(8.0/1024);
//write analog equvivalant data on led pins
switch(dataConv)
{
case 0: {
digitalWrite(led1,LOW);
digitalWrite(led2,LOW);
digitalWrite(led3,LOW);
break;
}
case 1: {
digitalWrite(led1,HIGH);
digitalWrite(led2,LOW);
digitalWrite(led3,LOW);
break;
}
case 2: {
digitalWrite(led1,LOW);
digitalWrite(led2,HIGH);
digitalWrite(led3,LOW);
break;
}
case 3: {
digitalWrite(led1,HIGH);
digitalWrite(led2,HIGH);
digitalWrite(led3,LOW);
break;
}
case 4: {
digitalWrite(led1,LOW);
digitalWrite(led2,LOW);
digitalWrite(led3,HIGH);
break;
}
case 5: {
digitalWrite(led1,HIGH);
digitalWrite(led2,LOW);
digitalWrite(led3,HIGH);
break;
}
case 6: {
digitalWrite(led1,LOW);
digitalWrite(led2,HIGH);
digitalWrite(led3,HIGH);
break;
}
case 7: {
digitalWrite(led1,HIGH);
digitalWrite(led2,HIGH);
digitalWrite(led3,HIGH);
break;
}
}
Serial.println(sensorValue); // print out the value you read:
Serial.println(dataConv);
delay(1000); // delay in between reads for stability
}
const int led1 = 2;
const int led2 = 3;
const int led3 = 4;
void setup() {
pinMode(led1, OUTPUT);
pinMode(led2, OUTPUT);
pinMode(led3, OUTPUT);
// initialize serial communication at 9600 bits per second:
Serial.begin(9600);
}
// the loop routine runs over and over again forever:
void loop() {
// read the input on analog pin 0:
int sensorValue = analogRead(A0);
int dataConv = sensorValue*(8.0/1024);
//write analog equvivalant data on led pins
switch(dataConv)
{
case 0: {
digitalWrite(led1,LOW);
digitalWrite(led2,LOW);
digitalWrite(led3,LOW);
break;
}
case 1: {
digitalWrite(led1,HIGH);
digitalWrite(led2,LOW);
digitalWrite(led3,LOW);
break;
}
case 2: {
digitalWrite(led1,LOW);
digitalWrite(led2,HIGH);
digitalWrite(led3,LOW);
break;
}
case 3: {
digitalWrite(led1,HIGH);
digitalWrite(led2,HIGH);
digitalWrite(led3,LOW);
break;
}
case 4: {
digitalWrite(led1,LOW);
digitalWrite(led2,LOW);
digitalWrite(led3,HIGH);
break;
}
case 5: {
digitalWrite(led1,HIGH);
digitalWrite(led2,LOW);
digitalWrite(led3,HIGH);
break;
}
case 6: {
digitalWrite(led1,LOW);
digitalWrite(led2,HIGH);
digitalWrite(led3,HIGH);
break;
}
case 7: {
digitalWrite(led1,HIGH);
digitalWrite(led2,HIGH);
digitalWrite(led3,HIGH);
break;
}
}
Serial.println(sensorValue); // print out the value you read:
Serial.println(dataConv);
delay(1000); // delay in between reads for stability
}
Solution
// Random LED Dots - from noise source
// Ed Nisley - KE4ANU - September 2015
//----------
// Pin assignments
const byte PIN_HEARTBEAT = 8; // DO - heartbeat LED
const byte PIN_SYNC = A3; // DO.
What is the need for a shadow page table (One or two brief sentences.pdf
What is the need for a shadow page table? (One or two brief sentences please)
Solution
Virtualization allows multiple operating system instances to run concurrently on a single
computer. Virtualization is the creation of a virtual rather than actual version of something, such
as an operating system, a server, a storage device or network resources.
Virtualization is where the framework divides the resource into one or more execution
environments Even something as simple as partitioning a hard drive is considered virtualization
because you take one drive and partition it to create two separate hard drives.
Types of virtualization:
Storage virtualization: is the pooling of physical storage from multiple network storage devices
into what appears to be a single storage device
Data virtualization: is abstracting the traditional technical details of data and data management,
such as location, performance or format..
Network virtualization: is a method of combining the available resources in a network by
splitting up the available bandwidth into channels, each of which is independent from the others.
Server virtualization: the partitioning a physical server into smaller virtual servers.
Shadow Tables are objects used to improve the way machines, networks and programs handle
information.
Shadow page tables are used by the hypervisor to keep track of the state in which the guest
thinks its page tables should be.
The guest can\'t be allowed access to the hardware page tables because then it would essentially
have control of the machine. So, the hypervisor keeps the \"real\" mappings guest virtual -> host
physical in the hardware when the relevant guest is executing, and keeps a representation of the
page table..
What step of protein synthesis is shown in the figure elongation st.pdf
What step of protein synthesis is shown in the figure? elongation stage of transcription
termination stage of translation mRNA processing elongation stage of translation initiation
stage of transcription
Solution
elongation stage of transcription
Explanation: newly synthesized mRNA separates from the DNA template as the RNA plymerase
continues elongation. multiple RNA plymerases transcribe simultaneously..
Water is forced through a contraction causing low pressure. The wat.pdf
Water is forced through a contraction causing low pressure. The water is observed to boil at a
pressure of -13.01 psi. If atmospheric pressure is 14.70 psi, what is the temperature, in °F, of the
water?
Solution
the water is start boiling at 212 °F
it means the temperature of water at 14.70psi is 212 °F +239.538 °F =451.53 °F.
Using a PIC 18 Write a time overflow interrupt routine that detects .pdf
Using a PIC 18 Write a time overflow interrupt routine that detects the interrupt source if it
overflows, increments the timer overflow counter, and finally clears the interrupt flag
Solution
ORG 00H
GOTO MAIN
ORG 0X08H
MOVLW 01H
ADDWF TMR0L
BCF INTCON,TMR0IF
RETIF
MAIN
ORG 100H
BCF INTCON,TMR0IF
BSF T0CON,TMR0ON
BSF INTCON,TMR0IE
BSF INTC0N,GIE
HERE GOTO HERE
END.
The three types of mixtures are and The weak bond forming a bridge .pdf
The three types of mixtures are and The weak bond forming a bridge between a hydrogen atom
and nitrogen or oxygen is a bond. A is an inorganic compound which has a bitter taste, feels
slipper and is a proton acceptor.
Solution
10. Three types of mixtures are suspensions, solutions and colloids
11.answer is hydrogen bond.
A hydrogen bond is the electrostatic attraction between two polar groups that occurs when a
hydrogen (H) atom covalently bound to a highly electronegative atom such as nitrogen (N),
oxygen (O), or fluorine (F) experiences the electrostatic field of another highly electronegative
atom nearby.
12.answer is Base..
Some commercial airplanes recirculate approximately 50 of the cabin .pdf
Some commercial airplanes recirculate approximately 50% of the cabin air in order to increase
fuel efficiency. The researchers studied 1096 airline passengers, among which some traveled on
airplanes that recirculated air and others traveled on planes that did not recirculate air. Of the 515
passengers who flew on planes that did not recirculate air, 106 reported post-flight respiratory
symptoms, while 112 of the 581 passengers on planes that did recirculate air reported such
symptoms. Is there sufficient evidence to conclude that the proportion of passengers with post-
flight respiratory symptoms differs for planes that do and do not recirculate air? Test the
appropriate hypotheses using a = 0.05. You may assume that it is reasonable to regard these two
samples as being independently selected and as representative of the two populations of interest.
(Use a statistical computer package to calculate the P-value. Use Pdo not recirculate - Pdo
recirculate- Round your test statistic to two decimal places and your P-value to four decimal
places.)
Solution
Null , There Is No Significance between them Ho: p1 = p2
Alternate , There Is Significance between them H1: p1 != p2
Test Statistic
Sample 1 : X1 =106, n1 =515, P1= X1/n1=0.206
Sample 2 : X2 =112, n2 =581, P2= X2/n2=0.193
Finding a P^ value For Proportion P^=(X1 + X2 ) / (n1+n2)
P^=0.199
Q^ Value For Proportion= 1-P^=0.801
we use Test Statistic (Z) = (P1-P2)/(P^Q^(1/n1+1/n2))
Zo =(0.206-0.193)/Sqrt((0.199*0.801(1/515+1/581))
Zo =0.54
| Zo | =0.54
Critical Value
The Value of |Z | at LOS 0.05% is 1.96
We got |Zo| =0.54 & | Z | =1.96
Make Decision
Hence Value of |Zo | < | Z | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) -Ha : ( P != 0.5403 ) = 0.589
Hence Value of P0.05 < 0.589,Here We Do not Reject Ho
No sufficient evidence to condude that the proportion of passengers
with post-flight respratory symptoms differs for planes that do and do not recirculate air
ANS:
Z = 0.54
P = 0.589.
Solve for 02 sin 2+sin 5=0. Separate your answers with commas.pdf
Solve for 0??<2?: sin 2?+sin 5?=0. Separate your answers with commas.
?= Solve for 0 lessthanorequalto theta
Solution
Use formula : sinx +siny = 2sin(x+y)/2cos(x-y)/2
So, sin 2+sin 5=0.
2sin(7)/2cos(3/2) =0
sin(7)/2 =0
= 0, 2pi/7 , 4pi/7, 6pi/7 , 8pi/7 , 10pi/7 , 12pi/7 ,
cos(3)/2 =0
= pi/3 , pi , 5pi/3.
Required 1. Briefly describe different computer controls available .pdf
Required: 1. Briefly describe different computer controls available to an organization.
a. Identify how these controls help to maintain system security and reduce risk.
b. Do companies need a formal internal policy that addresses security violations? Why/why not?
c. Consider what type of actions a company should take against employees that circumvent
computer
Solution
Security Controls:
Administrative Controls, Technical or Logical Controls, Physical Controls
The constituents of Security:
Vulnerability, Threat, Risk, Exposure, Countermeasure or Safeguard
The data vendor grants access needs or may choose to give this function to business unit
managers. And it is the data owner who will contract with security violations pertaining to the
data he is liable for defending.
Subjects that should be measured when developing a security policy:
Classification Controls:
Qualitative Risk Analysis Techniques:.
a. What is Snort [1]b. According to “Snort Users Manual” (downloa.pdf
a. What is Snort? [1]
b. According to “Snort Users Manual” (downloadable under “Docs” menu), what are the three
modes in which Snort operates and how it behaves under each mode? Show an example
command to put Snort in each mode. [6]
c. Explain the difference between the rule headers and rule options [2]
d. List 8 different types of actions Snort may take upon a rule match. Describe what each action
type does. [2]
Solution
a. Snort: Snort is an open source network capable of performing real-time traffic study and
packet-logging on IP networks. It can transmit the protocol analysis, content searching &
matching and is used to detect a range of attacks and investigates, such as buffer run over,
cautious port scans, OS fingerprinting efforts.
d. Types of snort actions:.
Proponents of decentralized forms of socialism usually supportA co.pdf
Proponents of decentralized forms of socialism usually support
A command economy
Control of the economy by a vanguard party of intellectuals
Control of the economy by various local groups of workers
State control of the economy
The Phoenix Suns
A command economy
Control of the economy by a vanguard party of intellectuals
Control of the economy by various local groups of workers
State control of the economy
The Phoenix Suns
Solution
Proponents of decentralized forms of socialism usually support the following things such a
1.A command economy.
2.Control of the economy by a vanguard party of intellectuals.
3.Control of the economy by various local groups of workers.
4.State control of the economy.
5.The phoenix suns..
Please convert the following C code to assembly Y86int i,j; ......pdf
Please convert the following C code to assembly Y86
int i,j; .....
if (i >j) {
i= i+5;
} else {
i=0;
j++;
}
Solution
Answer:
MIPS Assembly Language Code :
.zero 1
main:
push rbp
mov rbp, rsp
mov eax, DWORD PTR [rbp-4]
cmp eax, DWORD PTR [rbp-8]
jle .L2
add DWORD PTR [rbp-4], 5
jmp .L3
.L2:
mov DWORD PTR [rbp-4], 0
add DWORD PTR [rbp-8], 1
.L3:
mov eax, 0
pop rbp
ret
__static_initialization_and_destruction_0(int, int):
push rbp
mov rbp, rsp
sub rsp, 16
mov DWORD PTR [rbp-4], edi
mov DWORD PTR [rbp-8], esi
cmp DWORD PTR [rbp-4], 1
jne .L7
cmp DWORD PTR [rbp-8], 65535
jne .L7
mov edi, OFFSET FLAT:std::__ioinit
call std::ios_base::Init::Init()
mov edx, OFFSET FLAT:__dso_handle
mov esi, OFFSET FLAT:std::__ioinit
mov edi, OFFSET FLAT:std::ios_base::Init::~Init()
call __cxa_atexit
.L7:
nop
leave
ret
push rbp
mov rbp, rsp
mov esi, 65535
mov edi, 1
call __static_initialization_and_destruction_0(int, int)
pop rbp
ret.
Outline the four steps in the data collection process.Explain why .pdf
Outline the four steps in the data collection process.
Explain why it is so important to be systematic in collecting data.
Solution
Step 1: Identify issues and/or opportunities for collecting dataordata collection goals
what problem are you trying to solve by collecting this data?
Step 2:Develop operational definitions and procedures.
Here we need to be very clear as to what we are measuring, how it is to be measured, and who is
to measure it. Often times we will employ sampling in which case we need to define a sampling
plan.
. Who will the data be collected about?
. Who will the group of interest be compared to?
. What locations or geographical areas will the data be gathered from?
. What categories will be used to identify the group of interest and comparator group?
Step 3 : How should data be collected and What sources of data should be used to collect
information?
There are several methods of collecting data: online-surveys, phone interviews, focus groups and
yes, even dreaded handwritten surveys. Each type possesses its own advantages and
disadvantages. Whichever method you use, there are bound to be obstacles to overcome along
the way.
What sources of data should be used to collect information?
Pre-existing or official data
Survey data
Interviews and focus groups
Observed data
. Two types of data are used in any field of research: qualitative and quantitative.
Qualitative data:
Typically, data is called “qualitative” if it is in the form of words, but may also include any
information that is not numerical in form, such as photographs, videos and sound recordings.
Qualitative methods are aimed at describing a specific context, event, people or relationship in a
broad contextual way, by trying to understand the underlying reasons for behaviour, thoughts
and feelings.
Common qualitative research methods include observation, one-on-one interviews, focus groups
and intensive case studies
Quantitative data:
Typically, data is called “quantitative” if it is in the form of numbers.
A quantitative approach can be used to count events or the number of people who represent a
particular background.
Common quantitative tools include surveys, questionnaires and statistical data (such as Statistics
Canada census information).
It is important to note that all quantitative data is based on qualitative judgment. In other words,
numbers cannot be interpreted by themselves, without understanding the assumptions that
underlie them.
A good research effort involves the use of both types. Both approaches, while distinct, can
overlap and rely on the other to produce meaningful data, analysis and results.
Step 4 : Analyze and interpret data and Act on results
Explaining the technical steps involved in analyzing and interpreting data is beyond the scope of
this guide. An organization will have to determine whether it has the internal capacity and
expertise to analyze and interpret data itself, or whether it will need the help of an external
consultant.
. A summa.
Name the author of 1847 review article that conclusively demonstrated.pdf
Name the author of 1847 review article that conclusively demonstrated the validity of the
\"energy concept.\" Please state the \"energy concept\" that developed in Europe during the first
half of the 19^th century. Who was the first person to state the First Law in modern
mathematical terms AND in what year was his work published?
Solution
Hermann von Hermholtz was the author of validity of energy concept in 1847. The energy
concept in the first half of 19th century, i.e., the first law is the version of law of conservation of
energy, which was adopted for thermodynamic systems. It states that total energy of isolated
system is constant, energy can neither be created nor destroyed. It is often stated as the change in
the internal energy of closed system is equal to the amount of heat supplied to the system, minus
the amount of work done by the system on its surroundings and further the impossibility of
perpetual motion machines.
Rudolf clausius was the first person to state first law in modern mathematical terms and his well
esteemed work was published in 1850..
Many of our National Parks are inhabited by large numbers of black b.pdf
Many of our National Parks are inhabited by large numbers of black bears (Ursus americanus),
and their use of the Park can be a problem for humans. These creatures\' travels are governed by
the need for food, as well as a high degree of natural curiosity. As it happens, some humans
inadvertently or purposefully feed these bears, leading them to harass the human population.
When bears become troublesome, one strategy is to transplant them
Solution
Let us prepare chi square table as follows
H0: Transplant and gender are not associated
Ha: There is associaton between them
?2 Calculations
Two-way Contingency Table
?2 = 10.432, df = 2, ?2/df = 5.22 , P(?2 > 10.432) = 0.0054
As p value is less than alpha
reject null hypothesis.
It is statistically evident that transplant outcome was related to genderBlack Bears:
Transplantation Fate by Gender
MaleFemaleSuccessful32
26.85
( 0.99)17
22.15
( 1.20)49Returned34
43.29
( 1.99)45
35.71
( 2.42)79Nuisance14
9.86
( 1.74)4
8.14
( 2.10)188066146.
Implement the ADT stack by using an array stack to contain its entri.pdf
Implement the ADT stack by using an array stack to contain its entries. Expand the array
dynamically, as necessary. Maintain the stack’s bottom entry in stack[stack.length – 1].
Your Stack class must implement StackInterface (provided). You may use ArrayStack.java
(provided) as the starting point for your implementation.
You must implement a comprehensive set of unit tests using the main() method (and private
utility methods) in ArrayStack.java.
ArrayStack.java
public class ArrayStack implements StackInterface
{
private T[] stack; // Array of stack entries
private int topIndex; // Index of top entry
private boolean initialized = false;
private static final int DEFAULT_CAPACITY = 50;
private static final int MAX_CAPACITY = 10000;
public ArrayStack()
{
this(DEFAULT_CAPACITY);
} // end default constructor
public ArrayStack(int initialCapacity)
{
checkCapacity(initialCapacity);
// The cast is safe because the new array contains null entries
@SuppressWarnings(\"unchecked\")
T[] tempStack = (T[])new Object[initialCapacity];
stack = tempStack;
topIndex = -1;
initialized = true;
} // end constructor
// < Implementations of the stack operations go here. >
// < Implementations of the private methods go here; checkCapacity and
// checkInitialization are analogous to those in Chapter 2. >
// . . .
} // end ArrayStack
StackInterface.java
public interface StackInterface
{
/** Adds a new entry to the top of this stack.
@param newEntry An object to be added to the stack. */
public void push(T newEntry);
/** Removes and returns this stack\'s top entry.
@return The object at the top of the stack.
@throws EmptyStackException if the stack is empty before the operation. */
public T pop();
/** Retrieves this stack\'s top entry.
@return The object at the top of the stack.
@throws EmptyStackException if the stack is empty. */
public T peek();
/** Detects whether this stack is empty.
@return True if the stack is empty. */
public boolean isEmpty();
/** Removes all entries from this stack. */
public void clear();
} // end StackInterface
Solution
StackTester.java
public class StackTester
{
public static void main(String[] args)
{
ArrayStack aS = new ArrayStack();
aS.push(3);
aS.push(5);
aS.push(10);
aS.push(11);
System.out.println(\"\"+ aS.peek() + \", \" + aS.pop() + \", \" + aS.peek2());
aS.remove(1);
}
}
ArrayStack.java
import java.util.Arrays;
public class ArrayStack implements StackInterface
{
private T[] stack;
private int topIndex;
private boolean initialized = false;
private static final int DEFAULT_CAPACITY = 50;
private static final int MAX_CAPACITY = 10000;
public ArrayStack()
{
this(DEFAULT_CAPACITY);
}
public ArrayStack(int initialCapacity)
{
checkCapacity(initialCapacity);
@SuppressWarnings(\"unchecked\")
T[] tempStack = (T[])new Object[initialCapacity];
stack = tempStack;
topIndex = -1;
initialized = true;
}
public boolean isEmpty()
{
return topIndex < 0;
}
private void ensureCapacity()
{
if(topIndex == stack.length - 1)
{
int newLength = 2*stack.length;
.
How is the energy transferred from the source to the load in any elec.pdf
How is the energy transferred from the source to the load in any electromagnetic circuit? Does
the magnitude of the reflection coefficient increase, decrease, or stay the same as you move away
from the load on a lossless transmission line? Why?
Solution
Wave’ is a common term for a number of different ways in which energy is transferred:
Waves transfer energy but not mass
When we watch surf waves coming into shore, it’s easy to think that individual water particles
are moving towards us, but that’s not actually the case. The particles involved in waves move
back and forth perpendicularly to the way the wave is going, but don’t move significantly in the
direction of the wave. The particles ‘take part’ in the wave by bumping into one another and
transferring energy. This is why energy can be transferred, even though the average position of
the particles doesn’t change.
How does this work? It can help to think of a buoy bobbing in the ocean. The buoy is moved up
and down by the waves that pass by it, but doesn’t move directionally across the water.
You could also think about a Mexican wave at a sports match. The wave moves around the
arena, but the audience members don’t move around with it – they only stand up and sit down (a
perpendicular movement to the wave direction).
Particles in a water wave exchange kinetic energy for potential energy
When particles in water become part of a wave, they start to move up or down. This means that
kinetic energy (energy of movement) has been transferred to them. As the particles move further
away from their normal position (up towards the wave crest or down towards the trough), they
slow down. This means that some of theirkinetic energy has been converted into potential energy
– the energy of particles in a wave oscillates between kinetic and potential energy.
Thinking about potential energy can help us understand why tsunamis can be so damaging.
When a tsunami approaches the shore, it shoals (becomes much higher), so the water particles
are displaced further from equilibrium. They acquire a lot of potential energy, and this is released
when the wave interacts with land.
--------------------------------------------------------------------------------------------------------------------
-------------------------------------------------
The magnitude of reflection coefficeint decreases
that the magnitude of the reflection coefficient does not depend on the length of the line, only the
load impedance and the impedance of the transmission line. Also, note that if ZL=Z0, then the
line is \"matched\". In this case, there is no mismatch loss and all power is transferred to the load.
At this point, you should begin to understand the importance of impedance matching: grossly
mismatched impedances will lead to most of the power reflected away from the load.
Note that the reflection coefficient can be a real or a complex number..
Given Rhombus RHOB Prove HSR HSO Solution We know that .pdf
Given: Rhombus RHOB Prove: HSR HSO
Solution
We know that in traingles hsr and hso , the diagonal is equally split by other diagonal
So,
RS = SO.........1
HS is common side..........2
And, all sides of rhombus are equal, So, HR=HO.............3
Hence, since the 3 sides are equal, the triangles are congruent..
Find the real part, the imaginary part, and the absolute value of sin.pdf
Find the real part, the imaginary part, and the absolute value of sin (x - iy).
Solution
sin(x-iy)=sinx.cosiy-cosx.siniy
Since siniy=i sinhy ,cosiy=coshy
Sin(x-iy)=sinx.coshy- cosx.i sinhy
= sinx.1/2 (e^y +e^-y) -cosx.i.1/2(e^x-e^-x)
Real part=1/2 sinx(e^y+e^-y)
Imaginary part=-1/2 cosx(e^x-e^-x)
Absolute value=1/2{sinx(e^y+e^-y)+cosx(e^x-e^-x)}.
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Unit 8 - Information and Communication Technology (Paper I).pdf
Find the probability mass function (PMF) of L A source of informatio.pdf
- 1. Find the probability mass function (PMF) of L A source of information randomly generates symbols from a five letter alphabet {a, b, c. d, e}. The probability of each symbol is as follows: P(a) = 1/2; P(b) = 1/4; P(c) = 1/8; P(d) = 1/16; and P(e) = 1/16. These symbols are now encoded into binary codes using the scheme shown below. Let the random variable L denote the length of the binary code (for example, the length of the binary code for the symbol c is 3 bits). Find the probability mass function (PMF) of L. Solution PMF of L: L can take value 1, 2, 3, 4. For L=0, P(L=0)=1/2 For L=1, P(L=1)=1/4 For L=2, P(L=2)=1/8 For L=4, P(L=4)=(1/16)+(1/16)=1/8