Use De Moivres theorem to change the given complex number to the fo.pdf
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Use De Moivre\'s theorem to change the given complex number to the form a + bi, where a and b are real numbers. (- Squareroot 3/2 - 1/2 i)^4 Solution ( -sqrt3/2 -i/2)^4 Polar form = r*e^i*theta r = sqrt[ (-sqrt3/2)^2 + (1/2)^2 )] = 1 theta = tan^-1(-1/2 /-sqrt3/2) = pi +pi/3 = 4pi/3 ( -sqrt3/2 -i/2)^4 = (re^i*theta)^4 = r^4e^i*4theta = 1*e^i*16pi/3 = cos16pi/3 + isin16pi/3 = -0.5 -i*0.866.
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![Use De Moivre's theorem to change the given complex number to the form a + bi, where a and
b are real numbers. (- Squareroot 3/2 - 1/2 i)^4
Solution
( -sqrt3/2 -i/2)^4
Polar form = r*e^i*theta
r = sqrt[ (-sqrt3/2)^2 + (1/2)^2 )] = 1
theta = tan^-1(-1/2 /-sqrt3/2) = pi +pi/3 = 4pi/3
( -sqrt3/2 -i/2)^4 = (re^i*theta)^4 = r^4e^i*4theta
= 1*e^i*16pi/3
= cos16pi/3 + isin16pi/3
= -0.5 -i*0.866](https://image.slidesharecdn.com/usedemoivrestheoremtochangethegivencomplexnumbertothefo-230704085254-8cc2eca2/85/Use-De-Moivres-theorem-to-change-the-given-complex-number-to-the-fo-pdf-1-320.jpg)
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Correct Answer:
Culminating life structure for early adulthood
Explanation:
As per the given theory, settling down takes place during the transition of culminating life
structure for early adulthood.
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Correct Answer:
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Explanation:
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Solution
/* Java Programme for Grades Calculation */
import java.util.Scanner;
public class Student
{
//Private varibles declaration
private String firstName;
private String lastName;
private int sid;
private int[] grades= new int[10];
int[] tempgrades= new int[10];
int noc;
private float gpa=0;
//Constructor method
Scanner in=new Scanner(System.in); //scanner class object for reading input
public static void main(String[] args)
{
Student st=new Student(); //creating object and calling default constructor
st.readInput();
st.calgpa();
}
//Mutator for firstName
public void setfirstName(String firstName)
{
this.firstName = firstName;
}
//Mutator for lastName
public void setlastName(String lastName)
{
this.lastName = lastName;
}
//Mutator for sid
public void setsid(int sid)
{
this.sid = sid;
}
//Mutator for grades
public void setgrades(int[] grades)
{
this.grades = grades;
}
void readInput() //Function to read input data
{
System.out.println(\"Enter First Name\ \"); //Prompt for name
String temp=in.nextLine(); //reading name
setfirstName(temp); //calling mutator for firstName
System.out.println(\"Enter Last Name\ \"); //Prompt for last name
String temp1=in.nextLine(); //read last anme
setlastName(temp1); //calling mutator for lastName
System.out.println(\"Enter Student ID\ \"); //prompt for student id
int temp2=in.nextInt(); //read student id
setsid(temp2); //calling mutator for sid
System.out.println(\"Enter Number of classes ID\ \"); //Prompt for Number of classes
noc=in.nextInt(); //read Number of classes
System.out.println(\"Enter Grades of classes ID\ \"); //Prompt for Grades of classes
for(int i=0;i Gpa : \" + gpa+\"\ \"); //Dispaly output
}
}
Output :
javac Student.java //Compilation of java programme
java Student //running java programme
Sample output :
Enter First Name
Ashok
Enter Last Name
Kumar
Enter Student ID
101
Enter Number of classes ID
4
Enter Grades of classes ID
10
9
8
7
First Name : Ashok Last Name : Kumar Student Id : 101 --> Gpa : 8.5.
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Answer:
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Q6:
Correct Answer:
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Explanation:
As per the given theory, settling down takes place during the transition of culminating life
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Correct Answer:
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Correct Answer:
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A. The statement \"Evolutionists cannot point to any transitional fossils\" is very frustrating to
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Malfunction 2-Replication never starts.
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Malfunction 4-DNA replication has slowed such that it is not finished with the whole genome by
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Solution
Malfunction 1-Two daughter cells result, but one has two copies of the chromosome
concatenated into a big circle.
-TOPO IV Introduces single-stranded break into concatenated chromosomes to release them
from each other. And if it is mutated then concatenated chromosomes are not separated and may
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Malfunction 2-Replication never starts.
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mutated then replication is not initiated I.e. never starts.
Malfunction 3-Cell divides at the poles.
-MinC is known to dimerize and functions as an FtsZ protein inhibitor(It forms a ring at the site
of constriction). And if MinC is mutated then cell divides at poles.
Malfunction 4-DNA replication has slowed such that it is not finished with the whole genome
by the time cell is trying to septate.
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are stored, grade point average (gpa) (float). Write appropriate constructor, mutator and accessor
functions for the class along with the following A function that inputs all values from the user
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Solution
/* Java Programme for Grades Calculation */
import java.util.Scanner;
public class Student
{
//Private varibles declaration
private String firstName;
private String lastName;
private int sid;
private int[] grades= new int[10];
int[] tempgrades= new int[10];
int noc;
private float gpa=0;
//Constructor method
Scanner in=new Scanner(System.in); //scanner class object for reading input
public static void main(String[] args)
{
Student st=new Student(); //creating object and calling default constructor
st.readInput();
st.calgpa();
}
//Mutator for firstName
public void setfirstName(String firstName)
{
this.firstName = firstName;
}
//Mutator for lastName
public void setlastName(String lastName)
{
this.lastName = lastName;
}
//Mutator for sid
public void setsid(int sid)
{
this.sid = sid;
}
//Mutator for grades
public void setgrades(int[] grades)
{
this.grades = grades;
}
void readInput() //Function to read input data
{
System.out.println(\"Enter First Name\ \"); //Prompt for name
String temp=in.nextLine(); //reading name
setfirstName(temp); //calling mutator for firstName
System.out.println(\"Enter Last Name\ \"); //Prompt for last name
String temp1=in.nextLine(); //read last anme
setlastName(temp1); //calling mutator for lastName
System.out.println(\"Enter Student ID\ \"); //prompt for student id
int temp2=in.nextInt(); //read student id
setsid(temp2); //calling mutator for sid
System.out.println(\"Enter Number of classes ID\ \"); //Prompt for Number of classes
noc=in.nextInt(); //read Number of classes
System.out.println(\"Enter Grades of classes ID\ \"); //Prompt for Grades of classes
for(int i=0;i Gpa : \" + gpa+\"\ \"); //Dispaly output
}
}
Output :
javac Student.java //Compilation of java programme
java Student //running java programme
Sample output :
Enter First Name
Ashok
Enter Last Name
Kumar
Enter Student ID
101
Enter Number of classes ID
4
Enter Grades of classes ID
10
9
8
7
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Answer:
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based on pre-industrial, environmental, natality & moratility rates and the population growth
(influence the degree of stochasticity in birth and death rates) and reduction are both explained in
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mortality compared to the growth rate and fertility rate observed in stage IV. Stage IV is
composed of population with life expectancy is lower compared to stage II as a result both birth
rate & death rate are lower. This is leading to proportion of the population less than 15 years of
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Solution
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Different classes of roads are constructed with different geometric design and to meet the
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1. The Arterial Interstate roads/highways are constructed to interconnect states of a country with
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2. Expressways are contructed to provide a function of very hight mobility in a particular
direction or connected some major cities of a country so as to reduce the time of travel. There
access and egress points are limited and therefore the traffic is to and fro from the major end
points only.
3. Similar to major arterial roads, there are minor arterial roads which serve the smaller areas and
lengths. In urban areas their function is to connect major metropolitan cities, while in rural parts,
they connect major cities and towns.
4. The collector roads are constructed to function as the feeder roads for the arterial roads. They
collect the traffic from the local roads and lead them to the arterial roads. There lengths and
amount of traffic puts them into one of the two categories of major or minor collector roads.
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Solution
Answer
Germany\'s talk of a new, more assertive foreign policy that relies more heavily on its military is,
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success on the economic front. The European Union is an economic entity, but economics has
turned from being the binding element to being a centrifugal force. Either something new must
be introduced into the European experiment, or it might come undone.
Berlin believes that holding the European Union together requires adding another dimension that
it heretofore has withheld in its dealing with the bloc: military-political relations. Standing up to
a weakening Russia will appeal to Central European nations, and taking a more active role
overseas would endear Berlin to Paris. Germany\'s allusions that it would expand its international
military operations, particularly in Africa, is a clear nod to France, which has consistently
expressed its desire for a deeper military and political partnership with Germany.
Notably, the drive to bring Germany closer to France in the short term could create tensions
between them in the long term. Last week\'s summit between British Prime Minister David
Cameron and French President Francois Hollande was a reminder that France and the United
Kingdom may have extremely different views regarding the European Union but still see each
other as a military partner and, more important, as a counterweight to Germany.
Of course, Germany is in no position to take military action. It is in a position to posit the
possibility in some vague way, thereby generating political forces that can temporarily hold
things together. Berlin needs to buy time, particularly in Central Europe, where Hungary has
embarked on an independent course and is being watched carefully by others. With the United
States unwilling to become involved, Germany either becomes the counterweight or lives with
the consequences.
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together as it re-evaluates Russia. So far, Germany\'s announcement has been met positively,
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Wealth refers to the monetary value of net assets and income earned and accumulated by an
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In general it is presumed that higher wealth leads to higher welfare. However, that holds true
only an an individual level - individuals raise their standard of living and higher welfare state, as
they gather more wealth. But at an aggregative level, an overall nation\'s wealth doesn\'t define
its welfare state.
That is why, the welfare programs are mostly undertaken by the government to ensure that the
standard of living and overall welfare be raised for those sections of the society who cannot
afford it. Governments spend a considerable portion of their GDP every year for welfare
programs..
Which of the following are ways nutrients can enter the root cells .pdf
Which of the following are ways nutrients can enter the root cells? diffusion and transpiration
fermentation and transpiration none of these choices diffusion and active transport
Solution
10.
The answer is: diffusion and active transport.
Through diffusion and active transport, nutrients can be absorbed by root cells..
What is E coliSolutionE.coli is a bacteria that normally live.pdf
What is E coli?
Solution
E.coli is a bacteria that normally live in the intestine of humans or animals some type of E.coli
can form infection which can cause diarrhea, abdominal pain, and fever if the infection is severe
it can cause bloody diarrhea even kidney failure.
E. coli is most studied free living organism indeed some of the E.coli are beneficial. the major
cause for E.coli infection is the contaminated food.
Molecular biologist love E.coli because they have some features that make them perfect DNA
and protein producing facteries for genetic engineering. They have a short life span of (20 min
betwwen two cell division) can cultivated easily in large amount under very simple condition
they contains circular DNA in addition to Nucleoid. Such extrachromosomal circular DNA
molecule are called plasmids..
Where does mitosis occur in the bodies of vascular plantsSoluti.pdf
Where does mitosis occur in the bodies of vascular plants?
Solution
Ans.
The following cells/ tissue in the vascular plants show mitotic events-
I. Meristem tissues: Apical meristems (primary meristem) of shoot and root is under constant
mitotic divisions. Mitosis in these regions is the primary factor responsible for primary growth
(longitudinal, increase in height). Secondary meristems divide mitotically and responsible for
lateral growth (increase in thickness).
II. Parenchyma cell (ex- xylem and phloem parenchyma) constituting the parenchymatous tissue
show totipotency i.e. under appropriate conditions, these calls can dedifferentiate into embryonic
cells and form the entire plant through mitoses. These cells also may also divide to form new
cells meant for tissue repair..
Which of the following are the three properties of dataA. mean, m.pdf
Which of the following are the three properties of data?
A. mean, median, mode
B. mean, standard deviation, skewness
C. location, dispersion, shape
D. none of the above
What does the standard deviation infer about the mean’s usefulness in portraying it as a
“representative” average value of our data?
A. the larger the standard deviation the more representative the mean is as a “representative”
average
B. the smaller the standard deviation the less representative the mean is as a “representative”
average
C. the smaller the standard deviation the more representative the mean is as a “representative”
average
D. none of the above
If the mean and the median are both larger than the mode, and the mean is greater than the
median, what information about the shape of the distribution can we conclude?
A. it is symmetrical
B. it is bell-shaped
C. it is skewed to the right
D. it is skewed to the left
Solution
1.
It is OPTION B: Mean, standard deviation, skewness
2.
The smaller the standard deviation, the close the data to the mean.
Thus, it is OPTION C: the smaller the standard deviation the more representative the mean is as
a “representative” average
3. It is skewed to the right, as most data points are less than (to the left of the mean).
Hence, SKEWED TO THE RIGHT. [OPTION C].
What is the range of clinical signssymptoms of C. difficile infecti.pdf
What is the range of clinical signs/symptoms of C. difficile infection in humans?
Solution
Clostridium difficile infection (CDI) is a symptomatic infection due to the spore forming
bacterium, clostridium difficile. The CDI is spread by bacterial spores found within faeces. The
main symptoms / signs of clostridium difficile infection includes:
1. Water diarrhoea 10 to 15 times a day
2. Abdominal cramping and pain, which may be severe
3. Rapid heart rate and fever
4. Blood or pus in the stool
5. Nausea, dehydration and loss of appetite.
6. Complications may include pseudomembranous colitis, toxic megacolon, perforation of colon,
and sepsis..
What was Kants ideas regarding retributive punishmentSolution.pdf
What was Kant\'s ideas regarding retributive punishment?
Solution
Kant idea of redistributive justice:
\"Judicial punishment can never be used merely as a means to promote some other good for the
criminal himself or for civil society, but instead it must in all cases be imposed on him only on
the ground that he has committed a crime.\"
Kant regards punishment as a matter of justice, and it must be carried out by the state for the
sake of the law, not for the sake of the criminal or the victim. He argues that if the guilty are not
punished, justice is not done. Further, if justice is not done, then the idea of law itself is
undermined..
What are the major events that occur during each stage in the life cy.pdf
What are the major events that occur during each stage in the life cycle of a cell such as a skin
cell, i.e., interphase, prophase Metaphase Anaphase, Telophase, and Cytokinesis? Assume you
are a court reporter that has to explain these stages to your readers with as little jargon as
possible. How does the dell division differ in prokaryotes and eukaryotes? Clearly these
differences may be difficult for the prokaryotes in the jury to follow unless you how the
similarities and differences.
Solution
Answer:
1. Stages of Mitosis:
a. Interphase: Cell grows and makes a copy of DNA.
b. Prophase: Nuclear envelope breakdown, chromosomes condenses and spindle formation takes
place.
c. Metaphase: Chromosomes are lined up precisely on the metaphase plate, or middle of the cell.
d. Anaphase: Spindle pulls sister chromatids apart.
e. Telophase: Chromatids begin to decondense and become chromatin. Spindle disappears.
f. Cytokinesis: Divide cell and organelles. Actin ring, or cleavage furrow splits cell.
2. Prokaryotic cells
Binary fission:
Prokaryotic cells are simple in structure when compared to eukaryotic cells. They contain non-
membranous organelles, lack a cell nucleus, and have a simplistic genome: only one circular
chromosome of limited size. Therefore, prokaryotic cell division, a process known as binary
fission, is fast. The chromosome is duplicated prior to division. The two copies of the
chromosome attach to opposing sides of the cellular membrane. Cytokinesis, the physical
separation of the cell
Eukaryotic cells
Mitosis: The division of the nucleus, separating the duplicated genome into two sets identical to
the parent\'s.
Cytokinesis: The division of the cytoplasm, separating the organelles and other cellular
components.
Meiosis: The division of the nucleus in sex cells, making one cell into four sex cells identical to
the parent sex cell. Eukaryotic cells divide by separating the duplicated chromosomes, through
movements directed by microtubles. it protect the organ system of our body.
Degradation:
Multicellular organisms replace worn-out cells through cell division. In some animals, however,
cell division eventually halts. In humans this occurs on average, after 52 divisions, known as the
Hayflick limit. The cell is then referred to as senescent. Senescent cells deteriorate and die,
causing the body to age. Cells stop dividing because the telomeres, protective bits of DNA on the
end of a chromosome, become shorter with each division and eventually can no longer protect
the chromosome. Cancer cells, on the other hand, are \"immortal.\" An enzyme called
telomerase, present in large quantites in cancerous cells, rebuilds the telomeres, allowing
division to continue indefinitely.
Daughter chromosomes :
During the metaphase stage of mitosis, chromosomes, which become aligned on the equatorial
plane, take on the shape of an \"X\" as a result of a repelling force between chromosomes. The
lobes of the chromosome in this shape are called \'sister ch.
What challenges do corporations face with regards to social media.pdf
What challenges do corporations face with regards to social media?
Solution
Answer:
With the advent of internet , social media turned the life into new stage. Before internet ,
childrens were busy in playing toys. But today\'s generation busy in playing vedio games online
on the computer. All round change is there. Wherever we look , all are busy on their tabs ,
laptops and mobile phones : chating , vedio calling and making VoIP calls . Social media is good
to bring people together , keep friends in touch all the time. But excess of everything is bad.
Students in colleges and universities , employees in the companies pay more attention on chating
and vedio calling rather than what actually they came for. According to a research conducted
recently Proofpoint found that 8 % companies had terminated employees because of social media
usage. As the employees sharing their sensitive information in the social media that could be
vulnerable to the whole database of the company. Here are few challenges that every should plan
:
1. Integration
2. Governance
3. Culture
4. Human Resources
5. Cyber Bullying Crisis
6. Personal opinions.
There are no organs for excretion and gas exchange in cnidarians. How.pdf
There are no organs for excretion and gas exchange in cnidarians. How do they accomplish
these two functions?
Solution
Humans have lungs which is well developed to breath in and expel out the gases. In fishes, the
gills plays role in respiration. In certain organisms like frogs even skin also act in passive
exchange of gases. Cnidarians like in water but they do not have any gills to respire. They are not
well developed to possess lungs. Diffusion is the process which could be seen in exchange of
gases in cnidarians. The gaseous particles move from higher concentration to lower
concentration through direct diffusion in cnidarians. Cnidarians lack digestive, excretory and
circulatory systems. Gastrovascular cavity plays main role in digestion and distribution of food
particles throughout the body in cnidarians and also in expelling waste materials to out through
its single opening. This gastrovascular cavity has a single opening which both acts as mouth and
anus. This gastrovascular cavity functions in respiration, digestion, and discharge of gametes. All
the food particles/waste particles enter and exit the gastrovascular cavity through its single
opening. Nitrogenous waste materials have been removed in the form of ammonia have been
excreted through body surfaces..
the term with the definition. Duality Of Effects Journal Entry An amo.pdf
the term with the definition. Duality Of Effects Journal Entry An amount that is posted on the
left side of a T-account or ledger B. Assets are initially recorded at the amount paid to acquire
them. C. The mechanism used to record each transaction in the General Posting Journal. Debit D.
A summary of account names and numbers E. The concept that any transaction must have at
least two effects Chart Of Accounts on the accounting equation F. When journal entries are
recorded in the appropriate T-account. G. An amount that is posted on the right side of a T-
account. T-Account A detailed listing of debits and credits for a specific account within the
General Ledger. Credit
Solution
Duality of effects The concept that any transaction must have atleast two effects on the
accounting equation Journal entry The mechanism used to record each transaction in the
General Journal Posting When journal entries are recorded in the appropriate T-account Debit
An amount that is posted on the left side of a T-account or ledger Chart of accounts A summary
of account names and numbers T-account A detailed listing of debits and credits for a sepcific
account within the genral ledger Credit An amount that is posted on the right side of a T-
account.
Two proteins are normally found in the cytosol. They are involved in .pdf
Two proteins are normally found in the cytosol. They are involved in DNA double strand break
repair. When DNA damage is detected, they enter the nucleus and locate on the DNA to fix it.
After the process is complete, they exit the nucleus. Explain how the proteins are activated and
their entry into and exit from the nucleus.
Solution
Different proteins involved in such functions may undergo activation and transport by different
mechanisms. I\'ll describe the most commonly used mechanism. When DNA double strand break
is detected, lets say during the cell-cycle checkpoints, the cyclin-dependent kinases
phosphorylate repairing proteins, causing them to get activated. The phosphorylated, activated
protein may be recognized by transporter proteins that traffick them to the nuclear compartment.
During the repairing process, the phosphorylated proteins would usually lose the phosphate (such
as during trade-offs for the energy required for the repair, or another accessory protein involved
in repair may cause de-phosphorylation). The repair protein, devoid of its phosphate group, may
then be transported out of the nucleus back into the cytosol..
Two components operate in parallel in a device, so the device fails .pdf
Two components operate in parallel in a device, so the device fails when and only when both
components fail. The lifetimes, X1 and X2, of the respective components are independent and
identically distributed with an exponential distribution with theta=2. The costs of operating the
device is Z=2Y1+Y2, where Y1=Min(X1,X2) and Y2=Max(X1, X2). Compute E(Z).
Solution.
That assignment had 2 part. I already complated part 1. and collecte.pdf
That assignment had 2 part. I already complated part 1. and collected the phasor frequency and
magnitude datas. My code is here: I need to write a new code to answer those question. Please
help me i dont know how to do those. Many thanks.
You can find .wav file from this website any .wav file will be
okay.http://www.grsites.com/archive/sounds/category/10/
HERE IS MY PART 1 OF THIS ASSIGNMENT CODE.
clc
clear all
[y, Fs] = audioread(\'chirp.wav\'); % Reading file
Fs = 8192; % Sampling frequency
N=length(y);
T=1/Fs;
Fmax=Fs/2;
T0=N*T; %duration of the signal.
t =[1/Fs:1/Fs:N/Fs] % Time Vector
sound(y,Fs); % Plays .wav file
hold on
subplot(3,1,1);
plot(t,y, \'c\'); % Plots input signal
xlabel(\'Time\');
ylabel(\'Amplitude\');
title(\'Input Signal\')
Xr=fft(y); %use FFT to find DFT
freq1 =[-Fmax:Fs/N:Fmax-(Fs/N)];
%freq1 = linspace(1,Fs,length(y)); %%%frequency vector
subplot(3,1,2);
a=fftshift(Xr);
c=abs(a);
plot(freq1, c, \'r\'); %%%Stem Plot
xlabel(\'Freq\');
ylabel(\'Amplitude\');
title(\'Fourier Transform (Magnitude)\');
Xr_phase=angle(a); %Calculation of %phase
subplot(3,1,3);
plot(freq1,Xr_phase,\'g\') %Use %fftshift to %shift zero frequency component
xlabel(\'Phase\');
ylabel(\'Frequency\');
title(\'Phase Plot\');
%%% saving text file in ASCII foramt%%
% 1st column- frequency%%
% 2nd column- magnitude components%%
% 3rd column- phase (deg) components%%
A(:,1)=freq1;
A(:,2)=c;
A(:,3)=Xr_phase;
save(\'parameters.txt\',\'A\', \'-ASCII\')
THIS PART HAS BEEN COMPLETED NO NEED TO RE DO IT AGAIN. ITS CODE IS IN
THE ABOVE. This picture is only information purposes to understand part 2 of the assignment.
Signal Synthesizer (part-2) 1. Read the analyzed value (magnitude, phase and frequency) from
ASCII files. 2. Re-create sinusoid using the following equation for each set of (magnitude, phase
and frequency) A Cos (2nft G) 3. Combine all sinusoids and plot the signal (this is regenerated
signal). 4. Compare this combined signal with original chirp signal 5. Play the regenerated signal
(It should sound like the original audio signal). 6. Store the synthesized audio signal in a WAV
file (name it chirp synthesized.wav).
Solution
freq2=A(:,1)\'%freq Components;
c2=A(:,2)\';%magnitude components
Xr_phase2=A(:,3)\'; %phase (deg) components%%;
for i=1:1:length(c2)
signalA(i)=c2(i)*cos((2*pi*freq2(i)*t(i))+Xr_phase2(i));
end
figure(2)
plot(t,signalA);
sound(signalA,Fs);.
Book Fundamental of Biostatistics 7th editionProblem pg. 67 #3.1.pdf
Book: Fundamental of Biostatistics 7th edition
Problem: pg. 67 #3.113
Question: Suppose that a baby is born with a birth defect, bu thte baby\'s ancestry is unknown.
Wha tis the posterior probability that the baby will have both parents from population A, both
parents from population B, or mixed ancestry, respectively? Show all work.
Solution.
Stre cell Tran and Homeostasis 5. special proteins in the membrane th.pdf
Stre cell Tran and Homeostasis 5. special proteins in the membrane that aid diffuvion membeane
protein that forms a small hole that anowsions raw through 7, an active transport protein 8
diffusioe with the help of transport proteins 9, the movement of a substance across a membrane
without any help from echer molecules 10. the transport or very large molecules, such as proteins
transport across a membrane in which energy is required active transport b channel protein
facilitated diffusion h. passive transport i, sodium potassium pump transport protein k, vesicle
transport Lesson 3.3: Vocabulary ll Class Date Fil in the blank with the appropriate lennt. I. By
moving substances into and out of cells the process of keeping stable coeditious inside a cel, is
maintained. 2. A protein changes shape as it camiesions or molecules across the membrane, 3,
Exocytosis is the type transport that moves a substance out of the cell. of 4. transport is
movement across the plasma membrane that does not require an input ofenergy. 5. The sodium-
potassium nvolved in the active-transport of ions. 6. Facilitated diffusion needs the help of
proteins refers to the number of particles of a substance per unit of volume. 8. is the type of
vesicle transport that moves a substance into the cell. 9. Energy for active transport is supplied
by molecules of is the diffusion of water. concentration to an area of 1. During active transport, a
substance is moving from an area of concentration.
Solution
answer:1)exocytosis.
2)transport protein.
3)active transport.
4)passive
5)faciliated diffusion
6)channel protein
7)vesicle transport
8)endocytosis
9)diffusion
10)osmosis
11)lower concentration to higher concentration.(sodium potassium pump)..
Let X and Y be two continuous random variables with probability dens.pdf
Let X and Y be two continuous random variables with probability density functions fX (x) and
fY (y).
(a) Determine an expression for the density of their difference, fXY , if X and Y are independent.
(b) Use your expression to compute the density for X Y for X Unif[0, 1] and Y Unif[0,1].
(c) What is the density of the sum X+Y for XUnif [0,1] and Y Unif [0,1]?
(d) How are the answers to parts (b) and (c) similar? What assumptions on X and Y make this
so? Prove your observation is true by manipulating your expression in part (a).
Solution.
Many areas of the country have reservoirs or waterways with locks an.pdf
Many areas of the country have reservoirs or waterways with locks and dams. An interesting
problem is to evalute the potential for hydroelectric generation at a local facility with a spillway.
For example, would a hydroelectric facility be feasible at the columbus (Mississippi) lock and
dam on the Tennessee Tombigbee waterway?
Solution
i think it cant happen because the head is only 27 feet.and the lowest turbine head is kaplan
turbine that itself requires at least 33 feet.
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What are the fuctional need of a roadWhat is road building system.pdf
What are the fuctional need of a road?
What is road building system?
Solution
Prob. 1
Different classes of roads are constructed with different geometric design and to meet the
different funcitonal needs.
1. The Arterial Interstate roads/highways are constructed to interconnect states of a country with
high mobility and access to major cities.
2. Expressways are contructed to provide a function of very hight mobility in a particular
direction or connected some major cities of a country so as to reduce the time of travel. There
access and egress points are limited and therefore the traffic is to and fro from the major end
points only.
3. Similar to major arterial roads, there are minor arterial roads which serve the smaller areas and
lengths. In urban areas their function is to connect major metropolitan cities, while in rural parts,
they connect major cities and towns.
4. The collector roads are constructed to function as the feeder roads for the arterial roads. They
collect the traffic from the local roads and lead them to the arterial roads. There lengths and
amount of traffic puts them into one of the two categories of major or minor collector roads.
5. Then there are the local roads constructed to cater the transportation needs in different
localities. There lengths are limited to their respective locality but if we calculate the total
lengths of the local roads in a country, they form the largest road network by mileage..
Why does Germany have to be assertive in the EUSolutionAnswer.pdf
Why does Germany have to be assertive in the EU?
Solution
Answer
Germany\'s talk of a new, more assertive foreign policy that relies more heavily on its military is,
however, not solely linked to concerns over Russia or the United States. Germany has accepted
that its only option is to rally Europe but as the past six years have shown, it has had limited
success on the economic front. The European Union is an economic entity, but economics has
turned from being the binding element to being a centrifugal force. Either something new must
be introduced into the European experiment, or it might come undone.
Berlin believes that holding the European Union together requires adding another dimension that
it heretofore has withheld in its dealing with the bloc: military-political relations. Standing up to
a weakening Russia will appeal to Central European nations, and taking a more active role
overseas would endear Berlin to Paris. Germany\'s allusions that it would expand its international
military operations, particularly in Africa, is a clear nod to France, which has consistently
expressed its desire for a deeper military and political partnership with Germany.
Notably, the drive to bring Germany closer to France in the short term could create tensions
between them in the long term. Last week\'s summit between British Prime Minister David
Cameron and French President Francois Hollande was a reminder that France and the United
Kingdom may have extremely different views regarding the European Union but still see each
other as a military partner and, more important, as a counterweight to Germany.
Of course, Germany is in no position to take military action. It is in a position to posit the
possibility in some vague way, thereby generating political forces that can temporarily hold
things together. Berlin needs to buy time, particularly in Central Europe, where Hungary has
embarked on an independent course and is being watched carefully by others. With the United
States unwilling to become involved, Germany either becomes the counterweight or lives with
the consequences.
At first, Germany\'s actions seemed confusing and uncharacteristic. But they become more
sensible when you consider that that Berlin is looking for other tools to hold the European Union
together as it re-evaluates Russia. So far, Germany\'s announcement has been met positively,
mainly outside Germany, but the tensions that a stronger and more assertive Berlin exerts on the
European continent and the global stage are sure to come to the fore again. For now, however,
Merkel has no choice..
What is the link between wealth and welfare Do governments have a r.pdf
What is the link between wealth and welfare? Do governments have a responsibility to address
welfare?
Solution
Wealth refers to the monetary value of net assets and income earned and accumulated by an
economic entity. Welfare relates to the increase in overall standard of living, including non-
monetary standards.
In general it is presumed that higher wealth leads to higher welfare. However, that holds true
only an an individual level - individuals raise their standard of living and higher welfare state, as
they gather more wealth. But at an aggregative level, an overall nation\'s wealth doesn\'t define
its welfare state.
That is why, the welfare programs are mostly undertaken by the government to ensure that the
standard of living and overall welfare be raised for those sections of the society who cannot
afford it. Governments spend a considerable portion of their GDP every year for welfare
programs..
Which of the following are ways nutrients can enter the root cells .pdf
Which of the following are ways nutrients can enter the root cells? diffusion and transpiration
fermentation and transpiration none of these choices diffusion and active transport
Solution
10.
The answer is: diffusion and active transport.
Through diffusion and active transport, nutrients can be absorbed by root cells..
What is E coliSolutionE.coli is a bacteria that normally live.pdf
What is E coli?
Solution
E.coli is a bacteria that normally live in the intestine of humans or animals some type of E.coli
can form infection which can cause diarrhea, abdominal pain, and fever if the infection is severe
it can cause bloody diarrhea even kidney failure.
E. coli is most studied free living organism indeed some of the E.coli are beneficial. the major
cause for E.coli infection is the contaminated food.
Molecular biologist love E.coli because they have some features that make them perfect DNA
and protein producing facteries for genetic engineering. They have a short life span of (20 min
betwwen two cell division) can cultivated easily in large amount under very simple condition
they contains circular DNA in addition to Nucleoid. Such extrachromosomal circular DNA
molecule are called plasmids..
Where does mitosis occur in the bodies of vascular plantsSoluti.pdf
Where does mitosis occur in the bodies of vascular plants?
Solution
Ans.
The following cells/ tissue in the vascular plants show mitotic events-
I. Meristem tissues: Apical meristems (primary meristem) of shoot and root is under constant
mitotic divisions. Mitosis in these regions is the primary factor responsible for primary growth
(longitudinal, increase in height). Secondary meristems divide mitotically and responsible for
lateral growth (increase in thickness).
II. Parenchyma cell (ex- xylem and phloem parenchyma) constituting the parenchymatous tissue
show totipotency i.e. under appropriate conditions, these calls can dedifferentiate into embryonic
cells and form the entire plant through mitoses. These cells also may also divide to form new
cells meant for tissue repair..
Which of the following are the three properties of dataA. mean, m.pdf
Which of the following are the three properties of data?
A. mean, median, mode
B. mean, standard deviation, skewness
C. location, dispersion, shape
D. none of the above
What does the standard deviation infer about the mean’s usefulness in portraying it as a
“representative” average value of our data?
A. the larger the standard deviation the more representative the mean is as a “representative”
average
B. the smaller the standard deviation the less representative the mean is as a “representative”
average
C. the smaller the standard deviation the more representative the mean is as a “representative”
average
D. none of the above
If the mean and the median are both larger than the mode, and the mean is greater than the
median, what information about the shape of the distribution can we conclude?
A. it is symmetrical
B. it is bell-shaped
C. it is skewed to the right
D. it is skewed to the left
Solution
1.
It is OPTION B: Mean, standard deviation, skewness
2.
The smaller the standard deviation, the close the data to the mean.
Thus, it is OPTION C: the smaller the standard deviation the more representative the mean is as
a “representative” average
3. It is skewed to the right, as most data points are less than (to the left of the mean).
Hence, SKEWED TO THE RIGHT. [OPTION C].
What is the range of clinical signssymptoms of C. difficile infecti.pdf
What is the range of clinical signs/symptoms of C. difficile infection in humans?
Solution
Clostridium difficile infection (CDI) is a symptomatic infection due to the spore forming
bacterium, clostridium difficile. The CDI is spread by bacterial spores found within faeces. The
main symptoms / signs of clostridium difficile infection includes:
1. Water diarrhoea 10 to 15 times a day
2. Abdominal cramping and pain, which may be severe
3. Rapid heart rate and fever
4. Blood or pus in the stool
5. Nausea, dehydration and loss of appetite.
6. Complications may include pseudomembranous colitis, toxic megacolon, perforation of colon,
and sepsis..
What was Kants ideas regarding retributive punishmentSolution.pdf
What was Kant\'s ideas regarding retributive punishment?
Solution
Kant idea of redistributive justice:
\"Judicial punishment can never be used merely as a means to promote some other good for the
criminal himself or for civil society, but instead it must in all cases be imposed on him only on
the ground that he has committed a crime.\"
Kant regards punishment as a matter of justice, and it must be carried out by the state for the
sake of the law, not for the sake of the criminal or the victim. He argues that if the guilty are not
punished, justice is not done. Further, if justice is not done, then the idea of law itself is
undermined..
What are the major events that occur during each stage in the life cy.pdf
What are the major events that occur during each stage in the life cycle of a cell such as a skin
cell, i.e., interphase, prophase Metaphase Anaphase, Telophase, and Cytokinesis? Assume you
are a court reporter that has to explain these stages to your readers with as little jargon as
possible. How does the dell division differ in prokaryotes and eukaryotes? Clearly these
differences may be difficult for the prokaryotes in the jury to follow unless you how the
similarities and differences.
Solution
Answer:
1. Stages of Mitosis:
a. Interphase: Cell grows and makes a copy of DNA.
b. Prophase: Nuclear envelope breakdown, chromosomes condenses and spindle formation takes
place.
c. Metaphase: Chromosomes are lined up precisely on the metaphase plate, or middle of the cell.
d. Anaphase: Spindle pulls sister chromatids apart.
e. Telophase: Chromatids begin to decondense and become chromatin. Spindle disappears.
f. Cytokinesis: Divide cell and organelles. Actin ring, or cleavage furrow splits cell.
2. Prokaryotic cells
Binary fission:
Prokaryotic cells are simple in structure when compared to eukaryotic cells. They contain non-
membranous organelles, lack a cell nucleus, and have a simplistic genome: only one circular
chromosome of limited size. Therefore, prokaryotic cell division, a process known as binary
fission, is fast. The chromosome is duplicated prior to division. The two copies of the
chromosome attach to opposing sides of the cellular membrane. Cytokinesis, the physical
separation of the cell
Eukaryotic cells
Mitosis: The division of the nucleus, separating the duplicated genome into two sets identical to
the parent\'s.
Cytokinesis: The division of the cytoplasm, separating the organelles and other cellular
components.
Meiosis: The division of the nucleus in sex cells, making one cell into four sex cells identical to
the parent sex cell. Eukaryotic cells divide by separating the duplicated chromosomes, through
movements directed by microtubles. it protect the organ system of our body.
Degradation:
Multicellular organisms replace worn-out cells through cell division. In some animals, however,
cell division eventually halts. In humans this occurs on average, after 52 divisions, known as the
Hayflick limit. The cell is then referred to as senescent. Senescent cells deteriorate and die,
causing the body to age. Cells stop dividing because the telomeres, protective bits of DNA on the
end of a chromosome, become shorter with each division and eventually can no longer protect
the chromosome. Cancer cells, on the other hand, are \"immortal.\" An enzyme called
telomerase, present in large quantites in cancerous cells, rebuilds the telomeres, allowing
division to continue indefinitely.
Daughter chromosomes :
During the metaphase stage of mitosis, chromosomes, which become aligned on the equatorial
plane, take on the shape of an \"X\" as a result of a repelling force between chromosomes. The
lobes of the chromosome in this shape are called \'sister ch.
What challenges do corporations face with regards to social media.pdf
What challenges do corporations face with regards to social media?
Solution
Answer:
With the advent of internet , social media turned the life into new stage. Before internet ,
childrens were busy in playing toys. But today\'s generation busy in playing vedio games online
on the computer. All round change is there. Wherever we look , all are busy on their tabs ,
laptops and mobile phones : chating , vedio calling and making VoIP calls . Social media is good
to bring people together , keep friends in touch all the time. But excess of everything is bad.
Students in colleges and universities , employees in the companies pay more attention on chating
and vedio calling rather than what actually they came for. According to a research conducted
recently Proofpoint found that 8 % companies had terminated employees because of social media
usage. As the employees sharing their sensitive information in the social media that could be
vulnerable to the whole database of the company. Here are few challenges that every should plan
:
1. Integration
2. Governance
3. Culture
4. Human Resources
5. Cyber Bullying Crisis
6. Personal opinions.
There are no organs for excretion and gas exchange in cnidarians. How.pdf
There are no organs for excretion and gas exchange in cnidarians. How do they accomplish
these two functions?
Solution
Humans have lungs which is well developed to breath in and expel out the gases. In fishes, the
gills plays role in respiration. In certain organisms like frogs even skin also act in passive
exchange of gases. Cnidarians like in water but they do not have any gills to respire. They are not
well developed to possess lungs. Diffusion is the process which could be seen in exchange of
gases in cnidarians. The gaseous particles move from higher concentration to lower
concentration through direct diffusion in cnidarians. Cnidarians lack digestive, excretory and
circulatory systems. Gastrovascular cavity plays main role in digestion and distribution of food
particles throughout the body in cnidarians and also in expelling waste materials to out through
its single opening. This gastrovascular cavity has a single opening which both acts as mouth and
anus. This gastrovascular cavity functions in respiration, digestion, and discharge of gametes. All
the food particles/waste particles enter and exit the gastrovascular cavity through its single
opening. Nitrogenous waste materials have been removed in the form of ammonia have been
excreted through body surfaces..
the term with the definition. Duality Of Effects Journal Entry An amo.pdf
the term with the definition. Duality Of Effects Journal Entry An amount that is posted on the
left side of a T-account or ledger B. Assets are initially recorded at the amount paid to acquire
them. C. The mechanism used to record each transaction in the General Posting Journal. Debit D.
A summary of account names and numbers E. The concept that any transaction must have at
least two effects Chart Of Accounts on the accounting equation F. When journal entries are
recorded in the appropriate T-account. G. An amount that is posted on the right side of a T-
account. T-Account A detailed listing of debits and credits for a specific account within the
General Ledger. Credit
Solution
Duality of effects The concept that any transaction must have atleast two effects on the
accounting equation Journal entry The mechanism used to record each transaction in the
General Journal Posting When journal entries are recorded in the appropriate T-account Debit
An amount that is posted on the left side of a T-account or ledger Chart of accounts A summary
of account names and numbers T-account A detailed listing of debits and credits for a sepcific
account within the genral ledger Credit An amount that is posted on the right side of a T-
account.
Two proteins are normally found in the cytosol. They are involved in .pdf
Two proteins are normally found in the cytosol. They are involved in DNA double strand break
repair. When DNA damage is detected, they enter the nucleus and locate on the DNA to fix it.
After the process is complete, they exit the nucleus. Explain how the proteins are activated and
their entry into and exit from the nucleus.
Solution
Different proteins involved in such functions may undergo activation and transport by different
mechanisms. I\'ll describe the most commonly used mechanism. When DNA double strand break
is detected, lets say during the cell-cycle checkpoints, the cyclin-dependent kinases
phosphorylate repairing proteins, causing them to get activated. The phosphorylated, activated
protein may be recognized by transporter proteins that traffick them to the nuclear compartment.
During the repairing process, the phosphorylated proteins would usually lose the phosphate (such
as during trade-offs for the energy required for the repair, or another accessory protein involved
in repair may cause de-phosphorylation). The repair protein, devoid of its phosphate group, may
then be transported out of the nucleus back into the cytosol..
Two components operate in parallel in a device, so the device fails .pdf
Two components operate in parallel in a device, so the device fails when and only when both
components fail. The lifetimes, X1 and X2, of the respective components are independent and
identically distributed with an exponential distribution with theta=2. The costs of operating the
device is Z=2Y1+Y2, where Y1=Min(X1,X2) and Y2=Max(X1, X2). Compute E(Z).
Solution.
That assignment had 2 part. I already complated part 1. and collecte.pdf
That assignment had 2 part. I already complated part 1. and collected the phasor frequency and
magnitude datas. My code is here: I need to write a new code to answer those question. Please
help me i dont know how to do those. Many thanks.
You can find .wav file from this website any .wav file will be
okay.http://www.grsites.com/archive/sounds/category/10/
HERE IS MY PART 1 OF THIS ASSIGNMENT CODE.
clc
clear all
[y, Fs] = audioread(\'chirp.wav\'); % Reading file
Fs = 8192; % Sampling frequency
N=length(y);
T=1/Fs;
Fmax=Fs/2;
T0=N*T; %duration of the signal.
t =[1/Fs:1/Fs:N/Fs] % Time Vector
sound(y,Fs); % Plays .wav file
hold on
subplot(3,1,1);
plot(t,y, \'c\'); % Plots input signal
xlabel(\'Time\');
ylabel(\'Amplitude\');
title(\'Input Signal\')
Xr=fft(y); %use FFT to find DFT
freq1 =[-Fmax:Fs/N:Fmax-(Fs/N)];
%freq1 = linspace(1,Fs,length(y)); %%%frequency vector
subplot(3,1,2);
a=fftshift(Xr);
c=abs(a);
plot(freq1, c, \'r\'); %%%Stem Plot
xlabel(\'Freq\');
ylabel(\'Amplitude\');
title(\'Fourier Transform (Magnitude)\');
Xr_phase=angle(a); %Calculation of %phase
subplot(3,1,3);
plot(freq1,Xr_phase,\'g\') %Use %fftshift to %shift zero frequency component
xlabel(\'Phase\');
ylabel(\'Frequency\');
title(\'Phase Plot\');
%%% saving text file in ASCII foramt%%
% 1st column- frequency%%
% 2nd column- magnitude components%%
% 3rd column- phase (deg) components%%
A(:,1)=freq1;
A(:,2)=c;
A(:,3)=Xr_phase;
save(\'parameters.txt\',\'A\', \'-ASCII\')
THIS PART HAS BEEN COMPLETED NO NEED TO RE DO IT AGAIN. ITS CODE IS IN
THE ABOVE. This picture is only information purposes to understand part 2 of the assignment.
Signal Synthesizer (part-2) 1. Read the analyzed value (magnitude, phase and frequency) from
ASCII files. 2. Re-create sinusoid using the following equation for each set of (magnitude, phase
and frequency) A Cos (2nft G) 3. Combine all sinusoids and plot the signal (this is regenerated
signal). 4. Compare this combined signal with original chirp signal 5. Play the regenerated signal
(It should sound like the original audio signal). 6. Store the synthesized audio signal in a WAV
file (name it chirp synthesized.wav).
Solution
freq2=A(:,1)\'%freq Components;
c2=A(:,2)\';%magnitude components
Xr_phase2=A(:,3)\'; %phase (deg) components%%;
for i=1:1:length(c2)
signalA(i)=c2(i)*cos((2*pi*freq2(i)*t(i))+Xr_phase2(i));
end
figure(2)
plot(t,signalA);
sound(signalA,Fs);.
Book Fundamental of Biostatistics 7th editionProblem pg. 67 #3.1.pdf
Book: Fundamental of Biostatistics 7th edition
Problem: pg. 67 #3.113
Question: Suppose that a baby is born with a birth defect, bu thte baby\'s ancestry is unknown.
Wha tis the posterior probability that the baby will have both parents from population A, both
parents from population B, or mixed ancestry, respectively? Show all work.
Solution.
Stre cell Tran and Homeostasis 5. special proteins in the membrane th.pdf
Stre cell Tran and Homeostasis 5. special proteins in the membrane that aid diffuvion membeane
protein that forms a small hole that anowsions raw through 7, an active transport protein 8
diffusioe with the help of transport proteins 9, the movement of a substance across a membrane
without any help from echer molecules 10. the transport or very large molecules, such as proteins
transport across a membrane in which energy is required active transport b channel protein
facilitated diffusion h. passive transport i, sodium potassium pump transport protein k, vesicle
transport Lesson 3.3: Vocabulary ll Class Date Fil in the blank with the appropriate lennt. I. By
moving substances into and out of cells the process of keeping stable coeditious inside a cel, is
maintained. 2. A protein changes shape as it camiesions or molecules across the membrane, 3,
Exocytosis is the type transport that moves a substance out of the cell. of 4. transport is
movement across the plasma membrane that does not require an input ofenergy. 5. The sodium-
potassium nvolved in the active-transport of ions. 6. Facilitated diffusion needs the help of
proteins refers to the number of particles of a substance per unit of volume. 8. is the type of
vesicle transport that moves a substance into the cell. 9. Energy for active transport is supplied
by molecules of is the diffusion of water. concentration to an area of 1. During active transport, a
substance is moving from an area of concentration.
Solution
answer:1)exocytosis.
2)transport protein.
3)active transport.
4)passive
5)faciliated diffusion
6)channel protein
7)vesicle transport
8)endocytosis
9)diffusion
10)osmosis
11)lower concentration to higher concentration.(sodium potassium pump)..
Let X and Y be two continuous random variables with probability dens.pdf
Let X and Y be two continuous random variables with probability density functions fX (x) and
fY (y).
(a) Determine an expression for the density of their difference, fXY , if X and Y are independent.
(b) Use your expression to compute the density for X Y for X Unif[0, 1] and Y Unif[0,1].
(c) What is the density of the sum X+Y for XUnif [0,1] and Y Unif [0,1]?
(d) How are the answers to parts (b) and (c) similar? What assumptions on X and Y make this
so? Prove your observation is true by manipulating your expression in part (a).
Solution.
Many areas of the country have reservoirs or waterways with locks an.pdf
Many areas of the country have reservoirs or waterways with locks and dams. An interesting
problem is to evalute the potential for hydroelectric generation at a local facility with a spillway.
For example, would a hydroelectric facility be feasible at the columbus (Mississippi) lock and
dam on the Tennessee Tombigbee waterway?
Solution
i think it cant happen because the head is only 27 feet.and the lowest turbine head is kaplan
turbine that itself requires at least 33 feet.
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Chapter 2
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