I need an inttroduction for my paper on Conducting and Analyzing Sta.pdf
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I need an inttroduction for my paper on Conducting and Analyzing Statistcal Tests Solution Statistical tests are of various types, depending upon the nature of the study. Statistical tests provide a method for making quantitative decisions about a particular sample. Statistical tests mainly test the hypothesis that is made about the significance of an observed sample. Statistics.
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If Mendel had chosen factors to follow in his pea plants poorly,.pdf
If Mendel had chosen \'factors\' to follow in his pea plants poorly, and had unwittingly chosen to
observe gene traits on the same chromosome:
a. which one of his laws would he not have been able to figure out? why?
Solution
Mendel performed classical experiments on pea plant considering following points:
Based upon these assumption, Mendel performed some experiments and resulted in following
conclusions:
Till that time, it was not known that these \'factors\' were actually genes which are very active at
molecular levels. These genes can indeed show segregation and independent assortment but for
that to take place, a minimum prescribed distance should be present between two genes. If two
genes are located very close to each other, it is very highly probabale that these gene-pair will be
inherited as a single unit in the successive generations. Thus, the law of segregation and
independent assortment fails here.
Tthus, if Mendel was able to figure out the molecular nature of genes/factors, the laws of
segregation and resultantly, the law of independent assortmuent would have failed..
If the significance level of a test is increased (i.e. increased fr.pdf
If the significance level of a test is increased (i.e. increased from 0.05 to 0.1), will the Type II
error increase, decrease, or stay the same? Explain.
Solution
The two type of error rates moves in opposite of other. Thus if we increase the type I error (that
is level of significance ) then type II error will decrease..
Identify the contributions to Microbiology made by Redi, Tyndall, an.pdf
Identify the contributions to Microbiology made by Redi, Tyndall, and Pasteur.
Solution
Francesco Redi (1626-1697) was an Italian physician, who first refused the theory of
spontaneous generation. He refused the idea of spontaneous generation of maggots (larva of
flies) from the meat.
John Tyndall (1820-1893) is a physicist of 19th century. He studied diamagnetism and also the
made discoveries regarding the physical properties of air and the infrared radiation. He studied
the effect of radiant energy on the air.
Louis Pasteur was a French chemist during the 19th century. He disproved the theory of
spontaneous generation and won the debate on this topic and Academy’s prize. He filtered the air
through a gun-cotton filter and observed the presence of microbes on the filter under the
microscope..
I need help with two things. First Id like someone to check over m.pdf
I need help with two things. First I\'d like someone to check over my code for the 3 //TODO
statements below to verify that they\'re correct
public static HashSet doHashSetInsertMedian(int numItems) {
System.out.print(\"doHashSetInsertMedian: \");
HashSet set = new HashSet<>();
// TODO Write code that adds integers 0 through (numItems - 1)
// to set, inside a loop.
for(int i=0; i doLinkedHashSetRemoveMedian(int numItems) {
System.out.print(\"doLinkedHashSetRemoveMedian: \");
LinkedHashSet set = new LinkedHashSet<>();
// TODO Write code that adds integers 0 through (numitems - 1)
// to set.
long startTime = getTimestamp();
// TODO Write code that removes the median element
// from set, repeatedly until the set is empty.
long endTime = getTimestamp();
long totalTime = endTime - startTime;
System.out.println(totalTime);
return set;
}
Solution
PART 1
Hi friend, first part you have done correctly except a small mistake. In first TODO you missed
\'=\' sign:
public static HashSet doHashSetInsertMedian(int numItems) {
System.out.print(\"doHashSetInsertMedian: \");
HashSet set = new HashSet<>();
// TODO Write code that adds integers 0 through (numItems - 1)
// to set, inside a loop.
for(int i=0; i<=numItems-1; i++){
set.add(i);
}
// TODO Write code that removes integer (numItems / 2)
// from set.
int median = numItems / 2;
set.remove(median);
long startTime = getTimestamp();
// TODO Write code that adds integer (numItems / 2)
// to set.
set.add(median);
long endTime = getTimestamp();
long totalTime = endTime - startTime;
System.out.println(totalTime);
return set;
}
PART 2:
public static LinkedHashSet doLinkedHashSetRemoveMedian(int numItems) {
System.out.print(\"doLinkedHashSetRemoveMedian: \");
LinkedHashSet set = new LinkedHashSet<>();
// TODO Write code that adds integers 0 through (numitems - 1)
// to set.
for(int i=0; i<=numitems-1; i++)
set.add(i);
long startTime = getTimestamp();
// TODO Write code that removes the median element
// from set, repeatedly until the set is empty.
int median;
int current_size = set.size();
while(set.size() > 0){
current_size = set.size(); //getting current size of set
median = current_size/2; // finding current median
set.remove(median); // removing median
}
long endTime = getTimestamp();
long totalTime = endTime - startTime;
System.out.println(totalTime);
return set;
}.
Explain Internal Controls and Corporate Governance.SolutionInt.pdf
Explain Internal Controls and Corporate Governance.
Solution
Internal Control is generally defined as \" the process designed , implemented and maintained by
those charged with governance ,management to provide reasonable assurance with regard to the
financial reporting ,effectiveness and efficiency of operations ,protection of assets and
compliance with prevalent laws and regulations.
Objectives of Internal Control :
(i) To ensure that the financial transactions are preformed within the general or specific
authorisation limits set by the management of the company
(ii) The accounting transactions are properly recorded with regards to correct amounts, as per
applicable financial reporting framework, in the right accounting period and per accounting
policies and principles
(iii) Assets are safeguarded against unauthorised use, disposition and physically protected against
damage
Corporate Governance :
Corporate governance is defined as \" the system by which an enterprise is directed and
controlled to achieve the objective of increasing the shareholders value by enhancing economic
performance.\" Corporate governance typically involves relationship among the board of
directors, controlling shareholders and other stakeholders of the company. Issues that corporate
governance ideally deals with issues which include roles of the CEO and board of directors,
controls assurance and risk management. Corporate governance is a key aspect to enhance
shareholders value, drive the enterprise to do best business and achieve the company\'s
objectives. Corporate governance also serves as an mechanism to oversee the company\'s
performance in financial and operational terms. The Sarbanes Oxley Act 2002 provides various
sections and clauses for an effective working of the Corporate Governance in US companies..
Environmental Laws and Acts Project Research all of the following law.pdf
Environmental Laws and Acts Project Research all of the following laws, acts, and/or treaties
and answer the questions below. Forest Reserve Act (1891) a. Amendment Year: b. International
or National?: Description of Function/Environmental Issues Affected: d. Agency/Group
Responsible for Regulation and Enforcement: 2. Lacey Act Draft Year: b. Amendment Year:
International or National?: Description of Function/Environmental Issues Affected d. e.
Agency/Group Responsible for Regulation and Enforcement: 3. Antiquities Act (1906) ent Year:
nternational or National?: c. Description of Function/Environmental Issues Affected: d.
Responsible for Regulation and Enforcement: 4, Public Health Service Act Amendment Year: b.
International or National? c, Description of Function/Environmental ssues Affected: d.
Agency/Group Responsible for Regulation and Enforcement: 5. National Park Service Act
(1916) a. Amendment Year: b. International or National? c. Description of
Function/Environmental lssues Affected: d. Agency/Group Responsible for Regulation and
Enforcement: 6. Migratory Bird Act (1918) a. Amendment Year: b. International or National?: c.
Description of Function/Environmentallssues Affected: d. Agency/Group Responsible for
Regulation and Enforcement: 7. Taylor Grazing Act (1934) a. Amendment Year: b. nternational
or National? c. Description of Function/Environmental lssues Affected: d. Agency/Group
Responsible for Regulation and Enforcement: 8, Soil Conservation Act (1935) a. Amendment
Year: b, International or National? c. Description of Function/E Issues Affected nvironmental d.
Agency/Group Responsible for Regulation and Enforcement: 9. Federal Insecticide, Fungicide,
and Rodenticide Act (1947) a. Amendment Year: b, International or National? c. Description of
Function/Environmental lssues Affected:
Solution
Please find the answers below as tabulated:
Answer 2:
Lacey act
Answer 3
:Antiquities act(1906)
Answer 4:
Public Health Service Act (1912)
Governed by Service Commission Corps and United States Department of Health and Human
Services and enacted by Office of the Assistant Secretary for Preparedness and Response along
with Office of Global AffairsActDrafted yearAmendment
yearNatiaonl/internationalDescriptionagency/groupAnswer 1: Forest reserve
act1891NoneNational (for U.S.)This law gave the President the power to keep the forest area
and residential area separate. This helped in keeping the forest reserves intact while managing
the residential areas.Formulated and executed by the United States congress
Answer 2:
Lacey act1990Amended in 2008 and entitled as Food, Conservation, and Energy ActNationalA
law of conservation of illegal trade of fauna and flora and parts thereof such as fisheries, plant
products etc.The United States Federal Environmental Legislation
Answer 3
:Antiquities act(1906)1906Amended in 1950 to incorporate some domains of
antiquitiesNationalThis law was formulated in order to create monuments from the land of the
country in.
discuss the drawbacks of programmed and interrupt driven io and des.pdf
discuss the drawbacks of programmed and interrupt driven i/o and describe in general the
functionality of the DNA
Solution
Programmed I/O
Programmed I/O (PIO) refers to data transfers initiated by a CPU under driver software control
to access registers or memory on a device.
The CPU issues a command then waits for I/O operations to be complete. As the CPU is faster
than the I/O module, the problem with programmed I/O is that the CPU has to wait a long time
for the I/O module of concern to be ready for either reception or transmission of data. The CPU,
while waiting, must repeatedly check the status of the I/O module, and this process is known as
Polling. As a result, the level of the performance of the entire system is severely degraded.
Programmed I/O basically works in these ways:
Interrupt
The CPU issues commands to the I/O module then proceeds with its normal work until
interrupted by I/O device on completion of its work.
For input, the device interrupts the CPU when new data has arrived and is ready to be retrieved
by the system processor. The actual actions to perform depend on whether the device uses I/O
ports, memory mapping.
For output, the device delivers an interrupt either when it is ready to accept new data or to
acknowledge a successful data transfer. Memory-mapped and DMA-capable devices usually
generate interrupts to tell the system they are done with the buffer.
Although Interrupt relieves the CPU of having to wait for the devices, but it is still inefficient in
data transfer of large amount because the CPU has to transfer the data word by word between I/O
module and memory.
The main limitation of programmed I/O and interrupt driven I/O is given below:
Programmed I/O
Each instructions selects one I/O device (by number) and transfers a single character (byte)
Example: microprocessor controlled video terminal.
Four registers: input status and character, output status and character.
Interrupt-driven I/O
Primary disadvantage of programmed I/O is that CPU spends most of its time in a tight loop
waiting for the device to become ready. This is called busy waiting.
With interrupt-driven I/O, the CPU starts the device and tells it to generate an interrupt when it is
finished.
Done by setting interrupt-enable bit in status register.
Still requires an interrupt for every character read or written.
Interrupting a running process is an expensive business (requires saving context).
Requires extra hardware (DMA controller chip).
All these limitation can be overcome by the Introduction of DMA (Direct Memory Access)
To write block of 32 bytes from memory address 100 to device 4
1. CPU writes 32, 100, 4 into the first three DMA registers (memory address, count, device
number)
2. CPU puts code for WRITE (say 1) into fourth (direction) DMA register, which signals DMA
controller to begin operation
3. Controller reads (via bus request as CPU would) byte 100 from memory
4. Controller makes I/O request to write to device 4
5. Controller increments m.
do it on excel and please kindly upload the excel file. or explain m.pdf
do it on excel and please kindly upload the excel file. or explain me how to do this question on
excel. I really need help.
Solution
In numerical analysis, Newton\'s method (also known as the Newton–Raphson method), named
after IsaacNewton and Joseph Raphson, is a method for finding successively better
approximations to the roots (or zeroes) of a real-valued function. x : f ( x ) = 0 . {\\displaystyle
x:f(x)=0\\,.}.
Recommended
If Mendel had chosen factors to follow in his pea plants poorly,.pdf
If Mendel had chosen \'factors\' to follow in his pea plants poorly, and had unwittingly chosen to
observe gene traits on the same chromosome:
a. which one of his laws would he not have been able to figure out? why?
Solution
Mendel performed classical experiments on pea plant considering following points:
Based upon these assumption, Mendel performed some experiments and resulted in following
conclusions:
Till that time, it was not known that these \'factors\' were actually genes which are very active at
molecular levels. These genes can indeed show segregation and independent assortment but for
that to take place, a minimum prescribed distance should be present between two genes. If two
genes are located very close to each other, it is very highly probabale that these gene-pair will be
inherited as a single unit in the successive generations. Thus, the law of segregation and
independent assortment fails here.
Tthus, if Mendel was able to figure out the molecular nature of genes/factors, the laws of
segregation and resultantly, the law of independent assortmuent would have failed..
If the significance level of a test is increased (i.e. increased fr.pdf
If the significance level of a test is increased (i.e. increased from 0.05 to 0.1), will the Type II
error increase, decrease, or stay the same? Explain.
Solution
The two type of error rates moves in opposite of other. Thus if we increase the type I error (that
is level of significance ) then type II error will decrease..
Identify the contributions to Microbiology made by Redi, Tyndall, an.pdf
Identify the contributions to Microbiology made by Redi, Tyndall, and Pasteur.
Solution
Francesco Redi (1626-1697) was an Italian physician, who first refused the theory of
spontaneous generation. He refused the idea of spontaneous generation of maggots (larva of
flies) from the meat.
John Tyndall (1820-1893) is a physicist of 19th century. He studied diamagnetism and also the
made discoveries regarding the physical properties of air and the infrared radiation. He studied
the effect of radiant energy on the air.
Louis Pasteur was a French chemist during the 19th century. He disproved the theory of
spontaneous generation and won the debate on this topic and Academy’s prize. He filtered the air
through a gun-cotton filter and observed the presence of microbes on the filter under the
microscope..
I need help with two things. First Id like someone to check over m.pdf
I need help with two things. First I\'d like someone to check over my code for the 3 //TODO
statements below to verify that they\'re correct
public static HashSet doHashSetInsertMedian(int numItems) {
System.out.print(\"doHashSetInsertMedian: \");
HashSet set = new HashSet<>();
// TODO Write code that adds integers 0 through (numItems - 1)
// to set, inside a loop.
for(int i=0; i doLinkedHashSetRemoveMedian(int numItems) {
System.out.print(\"doLinkedHashSetRemoveMedian: \");
LinkedHashSet set = new LinkedHashSet<>();
// TODO Write code that adds integers 0 through (numitems - 1)
// to set.
long startTime = getTimestamp();
// TODO Write code that removes the median element
// from set, repeatedly until the set is empty.
long endTime = getTimestamp();
long totalTime = endTime - startTime;
System.out.println(totalTime);
return set;
}
Solution
PART 1
Hi friend, first part you have done correctly except a small mistake. In first TODO you missed
\'=\' sign:
public static HashSet doHashSetInsertMedian(int numItems) {
System.out.print(\"doHashSetInsertMedian: \");
HashSet set = new HashSet<>();
// TODO Write code that adds integers 0 through (numItems - 1)
// to set, inside a loop.
for(int i=0; i<=numItems-1; i++){
set.add(i);
}
// TODO Write code that removes integer (numItems / 2)
// from set.
int median = numItems / 2;
set.remove(median);
long startTime = getTimestamp();
// TODO Write code that adds integer (numItems / 2)
// to set.
set.add(median);
long endTime = getTimestamp();
long totalTime = endTime - startTime;
System.out.println(totalTime);
return set;
}
PART 2:
public static LinkedHashSet doLinkedHashSetRemoveMedian(int numItems) {
System.out.print(\"doLinkedHashSetRemoveMedian: \");
LinkedHashSet set = new LinkedHashSet<>();
// TODO Write code that adds integers 0 through (numitems - 1)
// to set.
for(int i=0; i<=numitems-1; i++)
set.add(i);
long startTime = getTimestamp();
// TODO Write code that removes the median element
// from set, repeatedly until the set is empty.
int median;
int current_size = set.size();
while(set.size() > 0){
current_size = set.size(); //getting current size of set
median = current_size/2; // finding current median
set.remove(median); // removing median
}
long endTime = getTimestamp();
long totalTime = endTime - startTime;
System.out.println(totalTime);
return set;
}.
Explain Internal Controls and Corporate Governance.SolutionInt.pdf
Explain Internal Controls and Corporate Governance.
Solution
Internal Control is generally defined as \" the process designed , implemented and maintained by
those charged with governance ,management to provide reasonable assurance with regard to the
financial reporting ,effectiveness and efficiency of operations ,protection of assets and
compliance with prevalent laws and regulations.
Objectives of Internal Control :
(i) To ensure that the financial transactions are preformed within the general or specific
authorisation limits set by the management of the company
(ii) The accounting transactions are properly recorded with regards to correct amounts, as per
applicable financial reporting framework, in the right accounting period and per accounting
policies and principles
(iii) Assets are safeguarded against unauthorised use, disposition and physically protected against
damage
Corporate Governance :
Corporate governance is defined as \" the system by which an enterprise is directed and
controlled to achieve the objective of increasing the shareholders value by enhancing economic
performance.\" Corporate governance typically involves relationship among the board of
directors, controlling shareholders and other stakeholders of the company. Issues that corporate
governance ideally deals with issues which include roles of the CEO and board of directors,
controls assurance and risk management. Corporate governance is a key aspect to enhance
shareholders value, drive the enterprise to do best business and achieve the company\'s
objectives. Corporate governance also serves as an mechanism to oversee the company\'s
performance in financial and operational terms. The Sarbanes Oxley Act 2002 provides various
sections and clauses for an effective working of the Corporate Governance in US companies..
Environmental Laws and Acts Project Research all of the following law.pdf
Environmental Laws and Acts Project Research all of the following laws, acts, and/or treaties
and answer the questions below. Forest Reserve Act (1891) a. Amendment Year: b. International
or National?: Description of Function/Environmental Issues Affected: d. Agency/Group
Responsible for Regulation and Enforcement: 2. Lacey Act Draft Year: b. Amendment Year:
International or National?: Description of Function/Environmental Issues Affected d. e.
Agency/Group Responsible for Regulation and Enforcement: 3. Antiquities Act (1906) ent Year:
nternational or National?: c. Description of Function/Environmental Issues Affected: d.
Responsible for Regulation and Enforcement: 4, Public Health Service Act Amendment Year: b.
International or National? c, Description of Function/Environmental ssues Affected: d.
Agency/Group Responsible for Regulation and Enforcement: 5. National Park Service Act
(1916) a. Amendment Year: b. International or National? c. Description of
Function/Environmental lssues Affected: d. Agency/Group Responsible for Regulation and
Enforcement: 6. Migratory Bird Act (1918) a. Amendment Year: b. International or National?: c.
Description of Function/Environmentallssues Affected: d. Agency/Group Responsible for
Regulation and Enforcement: 7. Taylor Grazing Act (1934) a. Amendment Year: b. nternational
or National? c. Description of Function/Environmental lssues Affected: d. Agency/Group
Responsible for Regulation and Enforcement: 8, Soil Conservation Act (1935) a. Amendment
Year: b, International or National? c. Description of Function/E Issues Affected nvironmental d.
Agency/Group Responsible for Regulation and Enforcement: 9. Federal Insecticide, Fungicide,
and Rodenticide Act (1947) a. Amendment Year: b, International or National? c. Description of
Function/Environmental lssues Affected:
Solution
Please find the answers below as tabulated:
Answer 2:
Lacey act
Answer 3
:Antiquities act(1906)
Answer 4:
Public Health Service Act (1912)
Governed by Service Commission Corps and United States Department of Health and Human
Services and enacted by Office of the Assistant Secretary for Preparedness and Response along
with Office of Global AffairsActDrafted yearAmendment
yearNatiaonl/internationalDescriptionagency/groupAnswer 1: Forest reserve
act1891NoneNational (for U.S.)This law gave the President the power to keep the forest area
and residential area separate. This helped in keeping the forest reserves intact while managing
the residential areas.Formulated and executed by the United States congress
Answer 2:
Lacey act1990Amended in 2008 and entitled as Food, Conservation, and Energy ActNationalA
law of conservation of illegal trade of fauna and flora and parts thereof such as fisheries, plant
products etc.The United States Federal Environmental Legislation
Answer 3
:Antiquities act(1906)1906Amended in 1950 to incorporate some domains of
antiquitiesNationalThis law was formulated in order to create monuments from the land of the
country in.
discuss the drawbacks of programmed and interrupt driven io and des.pdf
discuss the drawbacks of programmed and interrupt driven i/o and describe in general the
functionality of the DNA
Solution
Programmed I/O
Programmed I/O (PIO) refers to data transfers initiated by a CPU under driver software control
to access registers or memory on a device.
The CPU issues a command then waits for I/O operations to be complete. As the CPU is faster
than the I/O module, the problem with programmed I/O is that the CPU has to wait a long time
for the I/O module of concern to be ready for either reception or transmission of data. The CPU,
while waiting, must repeatedly check the status of the I/O module, and this process is known as
Polling. As a result, the level of the performance of the entire system is severely degraded.
Programmed I/O basically works in these ways:
Interrupt
The CPU issues commands to the I/O module then proceeds with its normal work until
interrupted by I/O device on completion of its work.
For input, the device interrupts the CPU when new data has arrived and is ready to be retrieved
by the system processor. The actual actions to perform depend on whether the device uses I/O
ports, memory mapping.
For output, the device delivers an interrupt either when it is ready to accept new data or to
acknowledge a successful data transfer. Memory-mapped and DMA-capable devices usually
generate interrupts to tell the system they are done with the buffer.
Although Interrupt relieves the CPU of having to wait for the devices, but it is still inefficient in
data transfer of large amount because the CPU has to transfer the data word by word between I/O
module and memory.
The main limitation of programmed I/O and interrupt driven I/O is given below:
Programmed I/O
Each instructions selects one I/O device (by number) and transfers a single character (byte)
Example: microprocessor controlled video terminal.
Four registers: input status and character, output status and character.
Interrupt-driven I/O
Primary disadvantage of programmed I/O is that CPU spends most of its time in a tight loop
waiting for the device to become ready. This is called busy waiting.
With interrupt-driven I/O, the CPU starts the device and tells it to generate an interrupt when it is
finished.
Done by setting interrupt-enable bit in status register.
Still requires an interrupt for every character read or written.
Interrupting a running process is an expensive business (requires saving context).
Requires extra hardware (DMA controller chip).
All these limitation can be overcome by the Introduction of DMA (Direct Memory Access)
To write block of 32 bytes from memory address 100 to device 4
1. CPU writes 32, 100, 4 into the first three DMA registers (memory address, count, device
number)
2. CPU puts code for WRITE (say 1) into fourth (direction) DMA register, which signals DMA
controller to begin operation
3. Controller reads (via bus request as CPU would) byte 100 from memory
4. Controller makes I/O request to write to device 4
5. Controller increments m.
do it on excel and please kindly upload the excel file. or explain m.pdf
do it on excel and please kindly upload the excel file. or explain me how to do this question on
excel. I really need help.
Solution
In numerical analysis, Newton\'s method (also known as the Newton–Raphson method), named
after IsaacNewton and Joseph Raphson, is a method for finding successively better
approximations to the roots (or zeroes) of a real-valued function. x : f ( x ) = 0 . {\\displaystyle
x:f(x)=0\\,.}.
Describe the structure and location of the parathyroid glands.Desc.pdf
Describe the structure and location of the parathyroid glands.
Describe the structure and location of the adrenal glands. Include the organization of the
following parts and their major role: Cortex and Medulla
Explain how the adrenal medulla and the sympathetic nervous system interact.
Describe the circadian cycle of melatonin.
Describe how the HPA axis controls sex-hormone release.
Describe what happens in the body in diabetes mellitus and explain the difference between Type
1 and Type 2 diabetes.
Solution
1.Structure and location of parathyroid glands:
Structure :
Loaction :These glands are situated behind the left and right lobes of the thyroid.
2.Structure and location of adrenal glands:
Structure :
Cortex :It is the outermost layer and is made up of three zones:
Zona glomerulosa :The outermost layer of the adrenal cortex is the zona glomerulosa and is
made up of oval cells that separated by thin strands of connective tissue.They secrete aldosterone
that function to regulate blood pressure.
Zona fasciculata:Lies next to zona glomerulosa.It is the largest layer made up of columnar
cells.They secrete hormone cortisol that is essential in maintaining homeostasis.
Zona reticularis:Lies next to zona fasciculata and near to medulla.Cells are in form of irregular
cords and they secrete androgen hormones for male reproductive functions.
Medulla :Present in the centre of the adrenal gland.Presence of chromaffin cells that secrete
hormones catecholamines adrenaline and noradrenaline. in response to fright and fear.
Location of adrenal gland:The adrenal glands are located on both sides of the body in the
retroperitoneum above the kidneys medially.
3.Adrenal and sympathetic nervous sytem interaction : In response to fear and fright , the
sympathetic nervous sysyem is activated that in turn activates the adrenal medulla to secrete
norepinephrine.Nor epinephrine prepares the body to respond to the situation immediately by
increasing the blood flow to our skeletal muscles to act accordingly.
4.Circadian cycle of melatonin: Melatonin is secreted by the pineal gland controls our sleep and
wake up cycle in a day.The internal clock of our body known as circadian rhythm regulates the
amount of melatonin produced by pineal gland.Melatonin levels rise in the dark after the sunset
and are at elevated levels at the night tht makes us asleep.In the morning,with sunrise the levels
drop making us awake.
5.HPA axis controls sex hormones: Sex difference is the result of interactions between the
gonadal hormone system and the HPA axis.In adulthood,androgen levels increase HPA activity
while and estrogen level decreases HPA activity.Monoamine neurotransmitters norepinephrine
and serotonin are released from HPA axis that determine differences in sex.Signals that originate
from the brain and periphery are integrated together to influence the level of HPA axis activity,
and these are sex-specific.
6.Diabetes Mellitus : In diabetes mellitus, the blood sugar level increase.
Describe how the composition of MALT tissue bypasses the need for.pdf
Describe how the composition of MALT tissue bypasses the need for dendritic cells to travel to
the lymph node to present antigen to B and T cells. What benefit does this provide, given that gut
mucosal membranes are a major avenue for pathogen entry? Is this benefit provided during the
first or subsequent encounters with a specific pathogen? Why?
Solution
Ans Problem 3. Problem Type: Comprehension:
MALT (mucosa-associated lymphoid tissue) is scattered along all mucosal linings of body and is
responsible for initiating immune responses to antigens that are specifically encountered along
these mucosal linings.
Mucosal epithelium contains special flat epithelial cells called M (which are membrane-like)
cells or FAE (also known as follicle associated epithelial cells). These special cells take small
amounts of gut antigens that enter the lumen of gut, and pass them to antigen producing cells and
lymphocytes of the MALT. M Cells also produce B-lymphocytes in gut react to micro-organisms
that tend to cross mucosa, and make IgA type of antibodies, which are then secreted directly onto
the gut lumen. These antibodies help in prevention of micro-organisms entering the gut by
sticking to the epithelium of gut, and can neutralise toxins and viruses both first and
subsequently also due to presence of memory B cells and T cells that then move to mucosal
effector sites..
C. wye D. delta E. not defined Question 6 What is the configuration o.pdf
C. wye D. delta E. not defined Question 6 What is the configuration of R3, R4 and R6? O A.
wye B. not defined O C. delta D. parallel E. Series Question 7 What is the configuration of R5,
R7 and R8? R2 LR3
Solution
its a delta connection
Standard 3-phase circuits or networks take on two major forms with names that represent the
way in which the resistances are connected, a Star connected network which has the symbol of
the letter, (wye) and a Delta connected network which has the symbol of a triangle, (delta).
A bullet that lodges in the heart would a. be located in the ventra.pdf
A bullet that lodges in the heart would: a. be located in the ventral body cavity. b. penetrate the
visceral peritoneum c. be located in the vertebral cavity. d. penetrate the parietal pleura. e. be
located laterally to a bullet that lodges in the lung.
Solution
3. d. A bullet that lodges in the heart would pentrate the parietal pluera to reach the heart..
Year d) interactions with other species for space or resources-classi.pdf
Year d) interactions with other species for space or resources-classic experiment with Paramecia
ook in at pop: Grown separately One tends \"Or O\' hod P to have a Place or o P kaudatum higer
tenders aurelia than the owner to hide They can Grown in mixed culture lire together o o a. 2 4 6
8 10 12 14 16 18 Days Other considerations: a) interactions from competition Paramecia exa c)
age ofreproduction Guvenile to senescence) reproductive value d) seasonal effects a weather
(regular and catastrophic) thrips example e) effects of habitat degradation fKocarrying capacity)
is effectively unknown for natural pops of all species
Solution
d) Interaction with other species for space or resources-classical example of paramecia
1st diagram: Here both the species of paramecium P. aurelia and P. caudatumare are gwown
separately. In this case there is no competition between both species. Therefore these are
growing in number and number is increasing with time (days).
2nd diagram: Here both the species of paramecium P. aurelia and P. caudatumare gwown are
grown together. Here you can see that one species population is increasing (P. aurelia) and other
is decreasing (P. caudatumare). Here ‘Principle of Competitive Exclusion’ works. When two
species are too similar, they cannot co-exsits because one of them will prove superior (dominant)
competitor. In other words we can say that two species cannot co-exist when they are actively
dependent on same resources.
d) Seasonal effect of weather (regular or catastrophic) strips example
Change in weather also effects the growth of a particular species. One species grow well in
winters whereas other in summer or autumn. As the favourable season changes the regular
growth is inhibited and whole population is destroyed (catastrophe).
Example: Agricultural crops.
Write a program that creates a linked list of families. Each family .pdf
Write a program that creates a linked list of families. Each family contains a husband, his wife
and a set of children. Each family is linked to another by a pointer called nextFamily. Husband is
linked to his wife by a pointer called myWife. The children are linked to their mother through a
pointer called myChildren. Children are linked by a pointer called mySibling.
Therefore, there are three types of nodes. The Husband node (class) contains the following
information:
SSN of type long
firstName of type string
lastName of type string
nextFamily of Husband*
myWife of type Wife*
The Wife class contains the following information:
SSN of type long
firstName of type string
lastName of type string
myChildren of Child*
The Child class contains the following information:
SSN of type long
firstName of type string
lastName of type string
mySibling of Child*
Your program reads a set of commands from the transaction file and acts accordingly. The
commands in the transaction file include
AddHusband
RemoveHusband
AddWife
RemoveWife
AddAChild
RemoveAChild
PrintAFamily
PrintAllFamilies
AddHusbad comes with three more information: his SSN, his first name and his last name. When
you get this command, you need to create a node of type Husband and insert it to the top of the
Link list of the families
RemveHusband comes with one more information: the SSN. This node should be deleted. In
order to delete this node, first you need to remove the children (if any), then remove the wife (if
he is married) and then remove the node itself
AddWife comes with 4 more information, the SSN of the wife, her first name, her last name and
finally the SSN of her husband. Based on the SSN of her husband, her husband must be found in
the linked list and the wife node should be linked to it.
RemoveWife, comes with the SSN of her husband. In this case, the husband should be searched
and his wife node should be deleted. If the wife node is attached to the children, first remove all
the children and then remove the wide node
AddAchild comes with 4 more information, the SSN of the child, child first name, child last
name and finally the SSN of the father. Based on the SSN of the father, the father must be found
in the linked list and the child should be attached to the mother.
RemoveAchild, comes with the SSN of the father and SSN of the child. In this case, the father
should be searched and his wife node should be followed to find the child that should be deleted.
If you find the child, you need to remove it from the family
PrintAFamily comes with the SSN of the father. You need to search that family in the linked list
and print the information of the father. The mother (if any) and the children (if any) on the screen
PrintAllFamilies, goes through the entire linked list of the family and prints the information of
each family one by one.
Test your program with the following transaction file
AddHusband 100100100 Jim Smith
AddHusband 200200200 Joe Brown
AddHusband 300300300 Kevin Tarr
A.
Why do Radicals maintain capitalism will always tend to enter into c.pdf
Why do Radicals maintain capitalism will always tend to enter into crisis?
Solution
Keynes challenged Say’s law that supply will create its own demand, and argues it is
incompetent to tackle the problems of unemployment and of the trade cycle. The radicals believe
that capitalism has reduced the need for work, blurred the edges between work and free time and
loosened the relationship between work and wages. Information is corroding the market’s ability
to form prices correctly because markets are based on scarcity while information is abundant.
They feel the system’s defence mechanism is to form monopolies. Goods and services are
appearing no longer responsive to the dictates of the market and the managerial hierarchy.
Parallel currencies, time banks, and cooperatives have been proliferated and the whole swaths of
economic life are beginning to move to a different rhythm.
*****.
Write a C program that converts an NFA file to a DFA, where the lang.pdf
Write a C program that converts an NFA file to a DFA, where the language of the NFA consists
of lowercase letters of the alphabet. Then have it be able to determine whether strings in a string
file are in the language or not. The program must take the two files as command line arguments
(do not prompt the user).
Solution
#include
#include
#define STATES 256
#define SYMBOLS 20
int N_symbols;
int NFA_states;
char *NFAtab[STATES][SYMBOLS];
int DFA_states; /* number of DFA states */
int DFAtab[STATES][SYMBOLS];
/*Print state-transition table.*/
void put_dfa_table(
int tab[][SYMBOLS], /* DFA table */
int nstates, /* number of states */
int nsymbols) /* number of input symbols */
{
int i, j;
puts(\"STATE TRANSITION TABLE\");
/* input symbols: \'0\', \'1\', ... */
printf(\" | \");
for (i = 0; i < nsymbols; i++) printf(\" %c \", \'0\'+i);
printf(\"\ -----+--\");
for (i = 0; i < nsymbols; i++) printf(\"-----\");
printf(\"\ \");
for (i = 0; i < nstates; i++) {
printf(\" %c | \", \'A\'+i); /* state */
for (j = 0; j < nsymbols; j++)
printf(\" %c \", \'A\'+tab[i][j]);
printf(\"\ \");
}
}
/*Initialize NFA table.*/
void init_NFA_table()
{
NFAtab[0][0] = \"12\";
NFAtab[0][1] = \"13\";
NFAtab[1][0] = \"12\";
NFAtab[1][1] = \"13\";
NFAtab[2][0] = \"4\";
NFAtab[2][1] = \"\";
NFAtab[3][0] = \"\";
NFAtab[3][1] = \"4\";
NFAtab[4][0] = \"4\";
NFAtab[4][1] = \"4\";
NFA_states = 5;
DFA_states = 0;
N_symbols = 2;
}
/*String \'t\' is merged into \'s\' in an alphabetical order.*/
void string_merge(char *s, char *t)
{
char temp[STATES], *r=temp, *p=s;
while (*p && *t) {
if (*p == *t) {
*r++ = *p++; t++;
} else if (*p < *t) {
*r++ = *p++;
} else
*r++ = *t++;
}
*r = \'\\0\';
if (*p) strcat(r, p);
else if (*t) strcat(r, t);
strcpy(s, temp);
}
/*Get next-state string for current-state string.*/
void get_next_state(char *nextstates, char *cur_states,
char *nfa[STATES][SYMBOLS], int n_nfa, int symbol)
{
int i;
char temp[STATES];
temp[0] = \'\\0\';
for (i = 0; i < strlen(cur_states); i++)
string_merge(temp, nfa[cur_states[i]-\'0\'][symbol]);
strcpy(nextstates, temp);
}
int state_index(char *state, char statename[][STATES], int *pn)
{
int i;
if (!*state) return -1; /* no next state */
for (i = 0; i < *pn; i++)
if (!strcmp(state, statename[i])) return i;
strcpy(statename[i], state); /* new state-name */
return (*pn)++;
}
/*
Convert NFA table to DFA table.
Return value: number of DFA states.
*/
int nfa_to_dfa(char *nfa[STATES][SYMBOLS], int n_nfa,
int n_sym, int dfa[][SYMBOLS])
{
char statename[STATES][STATES];
int i = 0; /* current index of DFA */
int n = 1; /* number of DFA states */
char nextstate[STATES];
int j;
strcpy(statename[0], \"0\"); /* start state */
for (i = 0; i < n; i++) { /* for each DFA state */
for (j = 0; j < n_sym; j++) { /* for each input symbol */
get_next_state(nextstate, statename[i], nfa, n_nfa, j);
dfa[i][j] = state_index(nextstate, statename, &n);
}
}
return n; /* number of DFA states */
}
void main()
{
init_NFA_table();
DFA_states = nfa_to_dfa(NFAtab, N.
Which of these hypotheses cannot be tested by an experiment Biologi.pdf
Which of these hypotheses cannot be tested by an experiment? Biological interactions between
barnacle species are responsible for their distribution patterns. Night-feeding fishes rely on their
sense of smell to locate prey. The growth of salt marsh gross is limited by nitrogen availability.
Certain humpback whale behaviors are important for attracting a mate. Goldfish prefer to eat
brightly-colored foods. TRUE/FALSE The oceans cover less than half of the earth\'s surface.
Ocean productivity is the amount of food produced by marine organisms and the number of
organisms this food supports. Aristotle developed one of the first taxonomical classification
schemes. Today it is generally believed that animals cannot live below the lit regions of the
ocean. Ecology is: a type of life style. the study of the interaction of organisms and their
environment. a place where organisms live. what happens when a predator consumes a prey
item. the study of the physical environment in an ecosystem. Organisms that live under similar
environmental conditions and location, and interact directly or indirectly are part of the same:
ecosystem. biosystem. abiotic habitat. biotic habitat. population. An example of abiotic
factors influencing marine organisms is: predation. competition for food. parasitism. salinity.
bacterial disease. Habitat refers to where an organism: lives. feeds. reproduces. forages. All
of the above.
Solution
Answers
15. False
Oceans cover about 70%of the surface of the earth
16. True
17. True
Aristotle developed the first classification system which divided all known organisms into two
groups plants and animals
18. False
Multiple choice
19. b study of interaction of organism\'s their environment
Ecology is the scientific analysis and study of interactions among organisms and their
environment . It is an interdisciplinary field that includes biology, geography and science .
20. a ecosystem
Interactions describes the different ways organisms obtain their food and energy . Some
organisms can make their own food and other organisms have to their food by eating other
organism
22. a lives
A habitat is an ecological or environmental area that is inhabited by a particular species of
animals plants or other type of organism . It is the natural environment in which an organism
lives ..
Which is the mode of inheritance that is most likely [That is the mo.pdf
Which is the mode of inheritance that is most likely? [That is the mode of inheritance that would
mean that the pattern would HAVE to be as described, not just that is could have come out this
way.] sex-limited, autosomal dominant Y-linked X-linked recessive autosomal recessive X-
linked dominant autosomal dominant autosomal sex-influenced
Solution
Non-autosomal mode of inheritance is the one related to sex-chromosomes. The alleles
associated with sex-chromosomes represent a differential kind of inheritance pattern and display
a differential phenotype response in offsprings based upon their sex. Majority of the human
diseases related to sex-chromosomes are X-linked in nature such as color blindness, hemophilia,
Duchenne muscular dystrophy, Vitamin D resistant rickets or X-linked hypophosphatemia, Rett
syndrome, Incontinentia pigmention and Goltz syndrome.
Since both parents, a male and a female carry an X chromosome, they both are potential donors
of X-linked allele of a disease. Since the offsprings can acquire a differential X-chromosome
profile either from mother or father, their genetic make up will be dependent upon that of their
parents and thus, their disease condition (carrier, affected, unaffected).
Thus, X-linked (dominant or recessive) mode of sex-chromosome related diseases need to be
explained theoratically and observationally, and not only be described as general procedures of
genetic segregation. Since X-linked recessive diseases are more likely to cause ambiguity among
inheritance pattern, a more detailed explanation would be requried for X-linked recessive alleles..
Which of the following is a good example of homoplasy (convergent ev.pdf
Which of the following is a good example of homoplasy (convergent evolution)?
a. Dolphins and humans
b. Apes and monkeys
c. Dolphins and fish
d. Dolphins and whales
e. Cats and bats
Wings of bats are:
a. Plesiomorphic (ancestral) feature for mammals
b. Plesiomorphic feature for bats and birds
c. Synapomorphic (shared-derived) feature for vertebrates
d. Synapomorphic feature for bats
e. Synapomorphic feature for bats and birds
Solution
1). c. Dolphins and fish
Analogous structures (convergent evolution) are not found in the common ancestors but the same
structure or similar structure with the same function can be found in one or more groups. In such
cases, it is not clear to tract the ancestry but the homologous structures are helpful to track the
ancestry. For example, both the dolphins and sharks share the fins but the dolphins are mammals,
while the shark are fish..
What happens at the mid-blastula transition in Xenopus Which part o.pdf
What happens at the mid-blastula transition in Xenopus? Which part of the Xenopus blastula
becomes the mesoderm? Where does the dorsal lip tissue end up when, Xenopus gastrulation is
finished? Where does the vegetal pole end up when gastrulation is finished? How is the central
nervous system formed? Where does the neural crest come from and what does it become?
What are the main tissue types formed from the mesoderm? What happens during the cortical
rotation? What is the principal determinant for endoderm formation in Xenopus? What are the
main mesoderm-inducing factors in Xenopus?
Solution
2. Midblastula transition refers to a number of dramatic changes that occur at this stage in
Xenopus. The important event that happens during this stage is activation of zygotic genes. The
Cell cycle is lengthened, the asynchronous cycle of different cells. Development of cellular
motility.
3. The blastula is composed of three different regions out of which the equatorial or marginal
zone, which compose the walls of the blastocoel differentiate primarily into mesodermal tissue.
4. The lip of the blastopore nearer to the animal pole is termed the dorsal lip of the blastopore.
The dorsal lip of the blastopore forms due to the contraction of bottle cells. It becomes the head
and the dorsal midline structures all along the body. It forms notochord, flanking parts of
somites, floor plate of neural tube.
5. The vegetal plate undergoes primary invagination to produce the archenteron or primitive gut.
Primary invagination is thought to result from changes in the shape of cells in the vegetal plate.
6. During the development, the dorsal ectoderm rolls up into a tube, which becomes the brain at
the anterior end and the spinal cord at the posterior end.
7. The neural crest is a population of migratory cells, derived from the neural folds. In the head,
it forms much of the skeleton. It forms dorsal root ganglia, enteric ganglia and pigment cells.
8. Mesoderm develops into notochord, somites, kidneys, gonads, limbs, lateral plate, blood
islands, heart and blood vessels during the Xenopus development.
9. On the surface of the dorsal egg cortex egg toward the animal pole and internally components
associated with cortex move from a vegetal to a dorsal position..
What is an example of Mendel’s first hypothesis regarding patterns o.pdf
What is an example of Mendel’s first hypothesis regarding patterns of inheritance? What
important vocabulary word do we learn in this hypothesis?
Solution
Gregor Johann Mendel studied the pea plant in his garden. He selected characters that were
easily identifiable. He grew the plants he selected to make sure they were pure – i.e. they were
pure line. A pure line is a plant that does not show any variation for the particular character
under study. Based on his studies, and the careful records he maintained, Mendel proposed his
first hypothesis as;
Vocabulary
Trait
Factors
Genes
Alleles
Homozygous
Heterozygous
Dominant
Recessive
F1.
What are three concepts that characterize the components of a system.pdf
What are three concepts that characterize the components of a system? Which component do you
think is most important? Why?
Solution
There are three concepts
organization considered as Central processing unit
the input devices are resources and employees
output devices are production the goods and services
I like the CPU, because cpu works properly all input and output work niecly, apart from that Cpu
integrating of input and output. Same like organization also integrate the employee as well as
resources and production of goods and services ..
True or False Explain your answer. If 2 , then cos 2 0..pdf
True or False? Explain your answer. If /2 < < , then cos /2 < 0.
Solution
If /2 < < , then cos /2 < 0.
This statement is FALSE.
Since, range of is:
/2 < <
Then, range of / is:
/4 < /2 < /2
which means /2 lies in the first quadrant where all the functions are positive.
Hence, it cannot be negative or less than zero or < 0..
C++ Help, I cant find whats wrong in my program. The sample outp.pdf
C++ Help, I can\'t find what\'s wrong in my program. The sample output is below and the
program I wrote is below that. I\'m having issues with the \"12 hour mode is turned on\", and
now I\'m getting all sorts of errors. Anything helps. Thank you.
-In this project, the “current time” should be stored in 24-hour format. Only when the user
actually interacts with the program will the output and input be sometimes in 12-hour format.
Implement a main menu that will allow the user to (1) enter the current time, (2) add a forward
delta to the “current time”, (3) add a backwards delta to the “current time”, (4) display the
current time, (5) toggle the interface mode between 24-hour and 12-hour (should impact both
getTime and printTime). The program should identify invalid input for the main menu. When
entering the current time, only allow hours between 0 and 23 and minutes between 0 and 59.
Accept only ‘a’ and ‘p’ when entering the time in 12-hour mode. When entering time deltas, only
allow positive values or 0 (this is true for both forward and backwards time deltas).
Sample Output: (user input in italics)
1 - Enter a time
2 - Add delta to time
3 - Subtract delta from time
4 - Display current time
5 - Toggle 24-hour mode
6 - Exit
> 1
Enter the time with whitespace separating the hours and minutes: 5 50
1 - Enter a time
2 - Add delta to time
3 - Subtract delta from time
4 - Display current time
5 - Toggle 24-hour mode
6 - Exit
> 4
The current time is 05:50
1 - Enter a time
2 - Add delta to time
3 - Subtract delta from time
4 - Display current time
5 - Toggle 24-hour mode
6 - Exit
> 2
Enter an increment of hours and minutes (separated by a space): 0 120
The new time would be 07:50
1 - Enter a time
2 - Add delta to time
3 - Subtract delta from time
4 - Display current time
5 - Toggle 24-hour mode
6 - Exit
> 2
Enter an increment of hours and minutes (separated by a space): 2 20
The new time would be 08:10
1 - Enter a time
2 - Add delta to time
3 - Subtract delta from time
4 - Display current time
5 - Toggle 24-hour mode
6 - Exit
> 5
12-hour mode turned on
1 - Enter a time
2 - Add delta to time
3 - Subtract delta from time
4 - Display current time
5 - Toggle 24-hour mode
6 - Exit
> 1
Enter the time with whitespace separating the hours, minutes, and either \'a\' for am or \'p\' for
pm: 0 15 a
1 - Enter a time
2 - Add delta to time
3 - Subtract delta from time
4 - Display current time
5 - Toggle 24-hour mode
6 - Exit
> 4
The current time is 12:15 am
1 - Enter a time
2 - Add delta to time
3 - Subtract delta from time
4 - Display current time
5 - Toggle 24-hour mode
6 - Exit
> 3
Enter a decrement of hours and minutes (separated by a space): 1 15 The new time would be
11:00 pm
1 - Enter a time
2 - Add delta to time
3 - Subtract delta from time
4 - Display current time
5 - Toggle 24-hour mode
6 - Exit
> 3
Enter a decrement of hours and minutes (separated by a space): 0 90 The new time would be
10:45 pm
*****This is the program I wrote:
#include
using namespace.
Based on the sections Multicellularity in the Volvocine Algae and.pdf
Based on the sections \"Multicellularity in the Volvocine Algae\" and \"Mutational/Functional
Genetics\", what important aspect of multicellularity does the glsA gene regulate in Volvox? cell
communication cellular differentiation bulk transport cell adhesion
Solution
The glsA gene is involved in asymmetric cell division and germ cell specification during
embryogenesis – these are aspects of cell differentiation. Hence the correct answer is Cellular
differentiation..
The parallel axis theorum... can only be used to find the moment of .pdf
The parallel axis theorum... can only be used to find the moment of inertia about an axis
through the centroid can only be used to find the moment of inertia about horizontal axes can be
used to find the moment of inertia about any axis can only be used to find the moment of inertia
about perpendicular axes
Solution
can be used to find the moment of interia about any axis.
The many organisms designated as protists are closely related to bac.pdf
The many organisms designated as protists are closely related to bacteria a clad. all unicellular
all microscopic highly diverse and not all closely related to one another
Solution
Ans. Correct option E.
All other kingdoms have specific cell wall composition, for example peptidoglycan in Bacteria,
cellulose in Plantae, chitin in Fungi and no cell wall in Animalia. Similarly, all eukaryotic
kingdoms, except Protista, exhibit specific mode of nutrition, for example, photoautotrophs in
Plantae, Saprophytic in Fungi, and heterotrophic in Animalia.
The members of this taxa show great variations in their cell wall composition, mode of nutrition,
mode of reproduction, etc. Moreover, they exhibit great diversity in their physiology, too. In
spite of these diversifications, the member of this taxa share two common features-
I. these are eukaryotes, and
II. are Unicellular.
So, the taxon was framed to include all unicellular eukaryotes, that are even not closely related to
each other, and which could not fit in any other eukaryotic kingdoms.
Option A: Incorrect: Protists are eukaryotes.
Option B. Incorrect: It the taxon consisting of members that may show no immediate
evolutionary relationship (a clade is a group of organisms sharing common ancestry).
Option C. and D. Incorrect: Bacteria are also unicellular but not placed in this group. Moreover,
many members of the group predominantly exist in filamentous forms, colony forms. For
example, kelps can be serval meters long, visible by naked eyes..
The following table gives the average monthly exchange rate between t.pdf
The following table gives the average monthly exchange rate between the U.S. dollar and the
euro for 20096. It shows that 1 euro was equivalent to 1.289 U.S dollars in January 2009.
Develop a trend line that could be used to predict the exchange rate for 2010. Use this model to
predict the exchange rate for January 2010 and February 2010.
Solution.
Overview on Edible Vaccine: Pros & Cons with Mechanism
This ppt include the description of the edible vaccine i.e. a new concept over the traditional vaccine administered by injection.
TESDA TM1 REVIEWER FOR NATIONAL ASSESSMENT WRITTEN AND ORAL QUESTIONS WITH A...
TESDA TM1 REVIEWER FOR NATIONAL ASSESSMENT WRITTEN AND ORAL QUESTIONS WITH ANSWERS.
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Describe the structure and location of the parathyroid glands.Desc.pdf
Describe the structure and location of the parathyroid glands.
Describe the structure and location of the adrenal glands. Include the organization of the
following parts and their major role: Cortex and Medulla
Explain how the adrenal medulla and the sympathetic nervous system interact.
Describe the circadian cycle of melatonin.
Describe how the HPA axis controls sex-hormone release.
Describe what happens in the body in diabetes mellitus and explain the difference between Type
1 and Type 2 diabetes.
Solution
1.Structure and location of parathyroid glands:
Structure :
Loaction :These glands are situated behind the left and right lobes of the thyroid.
2.Structure and location of adrenal glands:
Structure :
Cortex :It is the outermost layer and is made up of three zones:
Zona glomerulosa :The outermost layer of the adrenal cortex is the zona glomerulosa and is
made up of oval cells that separated by thin strands of connective tissue.They secrete aldosterone
that function to regulate blood pressure.
Zona fasciculata:Lies next to zona glomerulosa.It is the largest layer made up of columnar
cells.They secrete hormone cortisol that is essential in maintaining homeostasis.
Zona reticularis:Lies next to zona fasciculata and near to medulla.Cells are in form of irregular
cords and they secrete androgen hormones for male reproductive functions.
Medulla :Present in the centre of the adrenal gland.Presence of chromaffin cells that secrete
hormones catecholamines adrenaline and noradrenaline. in response to fright and fear.
Location of adrenal gland:The adrenal glands are located on both sides of the body in the
retroperitoneum above the kidneys medially.
3.Adrenal and sympathetic nervous sytem interaction : In response to fear and fright , the
sympathetic nervous sysyem is activated that in turn activates the adrenal medulla to secrete
norepinephrine.Nor epinephrine prepares the body to respond to the situation immediately by
increasing the blood flow to our skeletal muscles to act accordingly.
4.Circadian cycle of melatonin: Melatonin is secreted by the pineal gland controls our sleep and
wake up cycle in a day.The internal clock of our body known as circadian rhythm regulates the
amount of melatonin produced by pineal gland.Melatonin levels rise in the dark after the sunset
and are at elevated levels at the night tht makes us asleep.In the morning,with sunrise the levels
drop making us awake.
5.HPA axis controls sex hormones: Sex difference is the result of interactions between the
gonadal hormone system and the HPA axis.In adulthood,androgen levels increase HPA activity
while and estrogen level decreases HPA activity.Monoamine neurotransmitters norepinephrine
and serotonin are released from HPA axis that determine differences in sex.Signals that originate
from the brain and periphery are integrated together to influence the level of HPA axis activity,
and these are sex-specific.
6.Diabetes Mellitus : In diabetes mellitus, the blood sugar level increase.
Describe how the composition of MALT tissue bypasses the need for.pdf
Describe how the composition of MALT tissue bypasses the need for dendritic cells to travel to
the lymph node to present antigen to B and T cells. What benefit does this provide, given that gut
mucosal membranes are a major avenue for pathogen entry? Is this benefit provided during the
first or subsequent encounters with a specific pathogen? Why?
Solution
Ans Problem 3. Problem Type: Comprehension:
MALT (mucosa-associated lymphoid tissue) is scattered along all mucosal linings of body and is
responsible for initiating immune responses to antigens that are specifically encountered along
these mucosal linings.
Mucosal epithelium contains special flat epithelial cells called M (which are membrane-like)
cells or FAE (also known as follicle associated epithelial cells). These special cells take small
amounts of gut antigens that enter the lumen of gut, and pass them to antigen producing cells and
lymphocytes of the MALT. M Cells also produce B-lymphocytes in gut react to micro-organisms
that tend to cross mucosa, and make IgA type of antibodies, which are then secreted directly onto
the gut lumen. These antibodies help in prevention of micro-organisms entering the gut by
sticking to the epithelium of gut, and can neutralise toxins and viruses both first and
subsequently also due to presence of memory B cells and T cells that then move to mucosal
effector sites..
C. wye D. delta E. not defined Question 6 What is the configuration o.pdf
C. wye D. delta E. not defined Question 6 What is the configuration of R3, R4 and R6? O A.
wye B. not defined O C. delta D. parallel E. Series Question 7 What is the configuration of R5,
R7 and R8? R2 LR3
Solution
its a delta connection
Standard 3-phase circuits or networks take on two major forms with names that represent the
way in which the resistances are connected, a Star connected network which has the symbol of
the letter, (wye) and a Delta connected network which has the symbol of a triangle, (delta).
A bullet that lodges in the heart would a. be located in the ventra.pdf
A bullet that lodges in the heart would: a. be located in the ventral body cavity. b. penetrate the
visceral peritoneum c. be located in the vertebral cavity. d. penetrate the parietal pleura. e. be
located laterally to a bullet that lodges in the lung.
Solution
3. d. A bullet that lodges in the heart would pentrate the parietal pluera to reach the heart..
Year d) interactions with other species for space or resources-classi.pdf
Year d) interactions with other species for space or resources-classic experiment with Paramecia
ook in at pop: Grown separately One tends \"Or O\' hod P to have a Place or o P kaudatum higer
tenders aurelia than the owner to hide They can Grown in mixed culture lire together o o a. 2 4 6
8 10 12 14 16 18 Days Other considerations: a) interactions from competition Paramecia exa c)
age ofreproduction Guvenile to senescence) reproductive value d) seasonal effects a weather
(regular and catastrophic) thrips example e) effects of habitat degradation fKocarrying capacity)
is effectively unknown for natural pops of all species
Solution
d) Interaction with other species for space or resources-classical example of paramecia
1st diagram: Here both the species of paramecium P. aurelia and P. caudatumare are gwown
separately. In this case there is no competition between both species. Therefore these are
growing in number and number is increasing with time (days).
2nd diagram: Here both the species of paramecium P. aurelia and P. caudatumare gwown are
grown together. Here you can see that one species population is increasing (P. aurelia) and other
is decreasing (P. caudatumare). Here ‘Principle of Competitive Exclusion’ works. When two
species are too similar, they cannot co-exsits because one of them will prove superior (dominant)
competitor. In other words we can say that two species cannot co-exist when they are actively
dependent on same resources.
d) Seasonal effect of weather (regular or catastrophic) strips example
Change in weather also effects the growth of a particular species. One species grow well in
winters whereas other in summer or autumn. As the favourable season changes the regular
growth is inhibited and whole population is destroyed (catastrophe).
Example: Agricultural crops.
Write a program that creates a linked list of families. Each family .pdf
Write a program that creates a linked list of families. Each family contains a husband, his wife
and a set of children. Each family is linked to another by a pointer called nextFamily. Husband is
linked to his wife by a pointer called myWife. The children are linked to their mother through a
pointer called myChildren. Children are linked by a pointer called mySibling.
Therefore, there are three types of nodes. The Husband node (class) contains the following
information:
SSN of type long
firstName of type string
lastName of type string
nextFamily of Husband*
myWife of type Wife*
The Wife class contains the following information:
SSN of type long
firstName of type string
lastName of type string
myChildren of Child*
The Child class contains the following information:
SSN of type long
firstName of type string
lastName of type string
mySibling of Child*
Your program reads a set of commands from the transaction file and acts accordingly. The
commands in the transaction file include
AddHusband
RemoveHusband
AddWife
RemoveWife
AddAChild
RemoveAChild
PrintAFamily
PrintAllFamilies
AddHusbad comes with three more information: his SSN, his first name and his last name. When
you get this command, you need to create a node of type Husband and insert it to the top of the
Link list of the families
RemveHusband comes with one more information: the SSN. This node should be deleted. In
order to delete this node, first you need to remove the children (if any), then remove the wife (if
he is married) and then remove the node itself
AddWife comes with 4 more information, the SSN of the wife, her first name, her last name and
finally the SSN of her husband. Based on the SSN of her husband, her husband must be found in
the linked list and the wife node should be linked to it.
RemoveWife, comes with the SSN of her husband. In this case, the husband should be searched
and his wife node should be deleted. If the wife node is attached to the children, first remove all
the children and then remove the wide node
AddAchild comes with 4 more information, the SSN of the child, child first name, child last
name and finally the SSN of the father. Based on the SSN of the father, the father must be found
in the linked list and the child should be attached to the mother.
RemoveAchild, comes with the SSN of the father and SSN of the child. In this case, the father
should be searched and his wife node should be followed to find the child that should be deleted.
If you find the child, you need to remove it from the family
PrintAFamily comes with the SSN of the father. You need to search that family in the linked list
and print the information of the father. The mother (if any) and the children (if any) on the screen
PrintAllFamilies, goes through the entire linked list of the family and prints the information of
each family one by one.
Test your program with the following transaction file
AddHusband 100100100 Jim Smith
AddHusband 200200200 Joe Brown
AddHusband 300300300 Kevin Tarr
A.
Why do Radicals maintain capitalism will always tend to enter into c.pdf
Why do Radicals maintain capitalism will always tend to enter into crisis?
Solution
Keynes challenged Say’s law that supply will create its own demand, and argues it is
incompetent to tackle the problems of unemployment and of the trade cycle. The radicals believe
that capitalism has reduced the need for work, blurred the edges between work and free time and
loosened the relationship between work and wages. Information is corroding the market’s ability
to form prices correctly because markets are based on scarcity while information is abundant.
They feel the system’s defence mechanism is to form monopolies. Goods and services are
appearing no longer responsive to the dictates of the market and the managerial hierarchy.
Parallel currencies, time banks, and cooperatives have been proliferated and the whole swaths of
economic life are beginning to move to a different rhythm.
*****.
Write a C program that converts an NFA file to a DFA, where the lang.pdf
Write a C program that converts an NFA file to a DFA, where the language of the NFA consists
of lowercase letters of the alphabet. Then have it be able to determine whether strings in a string
file are in the language or not. The program must take the two files as command line arguments
(do not prompt the user).
Solution
#include
#include
#define STATES 256
#define SYMBOLS 20
int N_symbols;
int NFA_states;
char *NFAtab[STATES][SYMBOLS];
int DFA_states; /* number of DFA states */
int DFAtab[STATES][SYMBOLS];
/*Print state-transition table.*/
void put_dfa_table(
int tab[][SYMBOLS], /* DFA table */
int nstates, /* number of states */
int nsymbols) /* number of input symbols */
{
int i, j;
puts(\"STATE TRANSITION TABLE\");
/* input symbols: \'0\', \'1\', ... */
printf(\" | \");
for (i = 0; i < nsymbols; i++) printf(\" %c \", \'0\'+i);
printf(\"\ -----+--\");
for (i = 0; i < nsymbols; i++) printf(\"-----\");
printf(\"\ \");
for (i = 0; i < nstates; i++) {
printf(\" %c | \", \'A\'+i); /* state */
for (j = 0; j < nsymbols; j++)
printf(\" %c \", \'A\'+tab[i][j]);
printf(\"\ \");
}
}
/*Initialize NFA table.*/
void init_NFA_table()
{
NFAtab[0][0] = \"12\";
NFAtab[0][1] = \"13\";
NFAtab[1][0] = \"12\";
NFAtab[1][1] = \"13\";
NFAtab[2][0] = \"4\";
NFAtab[2][1] = \"\";
NFAtab[3][0] = \"\";
NFAtab[3][1] = \"4\";
NFAtab[4][0] = \"4\";
NFAtab[4][1] = \"4\";
NFA_states = 5;
DFA_states = 0;
N_symbols = 2;
}
/*String \'t\' is merged into \'s\' in an alphabetical order.*/
void string_merge(char *s, char *t)
{
char temp[STATES], *r=temp, *p=s;
while (*p && *t) {
if (*p == *t) {
*r++ = *p++; t++;
} else if (*p < *t) {
*r++ = *p++;
} else
*r++ = *t++;
}
*r = \'\\0\';
if (*p) strcat(r, p);
else if (*t) strcat(r, t);
strcpy(s, temp);
}
/*Get next-state string for current-state string.*/
void get_next_state(char *nextstates, char *cur_states,
char *nfa[STATES][SYMBOLS], int n_nfa, int symbol)
{
int i;
char temp[STATES];
temp[0] = \'\\0\';
for (i = 0; i < strlen(cur_states); i++)
string_merge(temp, nfa[cur_states[i]-\'0\'][symbol]);
strcpy(nextstates, temp);
}
int state_index(char *state, char statename[][STATES], int *pn)
{
int i;
if (!*state) return -1; /* no next state */
for (i = 0; i < *pn; i++)
if (!strcmp(state, statename[i])) return i;
strcpy(statename[i], state); /* new state-name */
return (*pn)++;
}
/*
Convert NFA table to DFA table.
Return value: number of DFA states.
*/
int nfa_to_dfa(char *nfa[STATES][SYMBOLS], int n_nfa,
int n_sym, int dfa[][SYMBOLS])
{
char statename[STATES][STATES];
int i = 0; /* current index of DFA */
int n = 1; /* number of DFA states */
char nextstate[STATES];
int j;
strcpy(statename[0], \"0\"); /* start state */
for (i = 0; i < n; i++) { /* for each DFA state */
for (j = 0; j < n_sym; j++) { /* for each input symbol */
get_next_state(nextstate, statename[i], nfa, n_nfa, j);
dfa[i][j] = state_index(nextstate, statename, &n);
}
}
return n; /* number of DFA states */
}
void main()
{
init_NFA_table();
DFA_states = nfa_to_dfa(NFAtab, N.
Which of these hypotheses cannot be tested by an experiment Biologi.pdf
Which of these hypotheses cannot be tested by an experiment? Biological interactions between
barnacle species are responsible for their distribution patterns. Night-feeding fishes rely on their
sense of smell to locate prey. The growth of salt marsh gross is limited by nitrogen availability.
Certain humpback whale behaviors are important for attracting a mate. Goldfish prefer to eat
brightly-colored foods. TRUE/FALSE The oceans cover less than half of the earth\'s surface.
Ocean productivity is the amount of food produced by marine organisms and the number of
organisms this food supports. Aristotle developed one of the first taxonomical classification
schemes. Today it is generally believed that animals cannot live below the lit regions of the
ocean. Ecology is: a type of life style. the study of the interaction of organisms and their
environment. a place where organisms live. what happens when a predator consumes a prey
item. the study of the physical environment in an ecosystem. Organisms that live under similar
environmental conditions and location, and interact directly or indirectly are part of the same:
ecosystem. biosystem. abiotic habitat. biotic habitat. population. An example of abiotic
factors influencing marine organisms is: predation. competition for food. parasitism. salinity.
bacterial disease. Habitat refers to where an organism: lives. feeds. reproduces. forages. All
of the above.
Solution
Answers
15. False
Oceans cover about 70%of the surface of the earth
16. True
17. True
Aristotle developed the first classification system which divided all known organisms into two
groups plants and animals
18. False
Multiple choice
19. b study of interaction of organism\'s their environment
Ecology is the scientific analysis and study of interactions among organisms and their
environment . It is an interdisciplinary field that includes biology, geography and science .
20. a ecosystem
Interactions describes the different ways organisms obtain their food and energy . Some
organisms can make their own food and other organisms have to their food by eating other
organism
22. a lives
A habitat is an ecological or environmental area that is inhabited by a particular species of
animals plants or other type of organism . It is the natural environment in which an organism
lives ..
Which is the mode of inheritance that is most likely [That is the mo.pdf
Which is the mode of inheritance that is most likely? [That is the mode of inheritance that would
mean that the pattern would HAVE to be as described, not just that is could have come out this
way.] sex-limited, autosomal dominant Y-linked X-linked recessive autosomal recessive X-
linked dominant autosomal dominant autosomal sex-influenced
Solution
Non-autosomal mode of inheritance is the one related to sex-chromosomes. The alleles
associated with sex-chromosomes represent a differential kind of inheritance pattern and display
a differential phenotype response in offsprings based upon their sex. Majority of the human
diseases related to sex-chromosomes are X-linked in nature such as color blindness, hemophilia,
Duchenne muscular dystrophy, Vitamin D resistant rickets or X-linked hypophosphatemia, Rett
syndrome, Incontinentia pigmention and Goltz syndrome.
Since both parents, a male and a female carry an X chromosome, they both are potential donors
of X-linked allele of a disease. Since the offsprings can acquire a differential X-chromosome
profile either from mother or father, their genetic make up will be dependent upon that of their
parents and thus, their disease condition (carrier, affected, unaffected).
Thus, X-linked (dominant or recessive) mode of sex-chromosome related diseases need to be
explained theoratically and observationally, and not only be described as general procedures of
genetic segregation. Since X-linked recessive diseases are more likely to cause ambiguity among
inheritance pattern, a more detailed explanation would be requried for X-linked recessive alleles..
Which of the following is a good example of homoplasy (convergent ev.pdf
Which of the following is a good example of homoplasy (convergent evolution)?
a. Dolphins and humans
b. Apes and monkeys
c. Dolphins and fish
d. Dolphins and whales
e. Cats and bats
Wings of bats are:
a. Plesiomorphic (ancestral) feature for mammals
b. Plesiomorphic feature for bats and birds
c. Synapomorphic (shared-derived) feature for vertebrates
d. Synapomorphic feature for bats
e. Synapomorphic feature for bats and birds
Solution
1). c. Dolphins and fish
Analogous structures (convergent evolution) are not found in the common ancestors but the same
structure or similar structure with the same function can be found in one or more groups. In such
cases, it is not clear to tract the ancestry but the homologous structures are helpful to track the
ancestry. For example, both the dolphins and sharks share the fins but the dolphins are mammals,
while the shark are fish..
What happens at the mid-blastula transition in Xenopus Which part o.pdf
What happens at the mid-blastula transition in Xenopus? Which part of the Xenopus blastula
becomes the mesoderm? Where does the dorsal lip tissue end up when, Xenopus gastrulation is
finished? Where does the vegetal pole end up when gastrulation is finished? How is the central
nervous system formed? Where does the neural crest come from and what does it become?
What are the main tissue types formed from the mesoderm? What happens during the cortical
rotation? What is the principal determinant for endoderm formation in Xenopus? What are the
main mesoderm-inducing factors in Xenopus?
Solution
2. Midblastula transition refers to a number of dramatic changes that occur at this stage in
Xenopus. The important event that happens during this stage is activation of zygotic genes. The
Cell cycle is lengthened, the asynchronous cycle of different cells. Development of cellular
motility.
3. The blastula is composed of three different regions out of which the equatorial or marginal
zone, which compose the walls of the blastocoel differentiate primarily into mesodermal tissue.
4. The lip of the blastopore nearer to the animal pole is termed the dorsal lip of the blastopore.
The dorsal lip of the blastopore forms due to the contraction of bottle cells. It becomes the head
and the dorsal midline structures all along the body. It forms notochord, flanking parts of
somites, floor plate of neural tube.
5. The vegetal plate undergoes primary invagination to produce the archenteron or primitive gut.
Primary invagination is thought to result from changes in the shape of cells in the vegetal plate.
6. During the development, the dorsal ectoderm rolls up into a tube, which becomes the brain at
the anterior end and the spinal cord at the posterior end.
7. The neural crest is a population of migratory cells, derived from the neural folds. In the head,
it forms much of the skeleton. It forms dorsal root ganglia, enteric ganglia and pigment cells.
8. Mesoderm develops into notochord, somites, kidneys, gonads, limbs, lateral plate, blood
islands, heart and blood vessels during the Xenopus development.
9. On the surface of the dorsal egg cortex egg toward the animal pole and internally components
associated with cortex move from a vegetal to a dorsal position..
What is an example of Mendel’s first hypothesis regarding patterns o.pdf
What is an example of Mendel’s first hypothesis regarding patterns of inheritance? What
important vocabulary word do we learn in this hypothesis?
Solution
Gregor Johann Mendel studied the pea plant in his garden. He selected characters that were
easily identifiable. He grew the plants he selected to make sure they were pure – i.e. they were
pure line. A pure line is a plant that does not show any variation for the particular character
under study. Based on his studies, and the careful records he maintained, Mendel proposed his
first hypothesis as;
Vocabulary
Trait
Factors
Genes
Alleles
Homozygous
Heterozygous
Dominant
Recessive
F1.
What are three concepts that characterize the components of a system.pdf
What are three concepts that characterize the components of a system? Which component do you
think is most important? Why?
Solution
There are three concepts
organization considered as Central processing unit
the input devices are resources and employees
output devices are production the goods and services
I like the CPU, because cpu works properly all input and output work niecly, apart from that Cpu
integrating of input and output. Same like organization also integrate the employee as well as
resources and production of goods and services ..
True or False Explain your answer. If 2 , then cos 2 0..pdf
True or False? Explain your answer. If /2 < < , then cos /2 < 0.
Solution
If /2 < < , then cos /2 < 0.
This statement is FALSE.
Since, range of is:
/2 < <
Then, range of / is:
/4 < /2 < /2
which means /2 lies in the first quadrant where all the functions are positive.
Hence, it cannot be negative or less than zero or < 0..
C++ Help, I cant find whats wrong in my program. The sample outp.pdf
C++ Help, I can\'t find what\'s wrong in my program. The sample output is below and the
program I wrote is below that. I\'m having issues with the \"12 hour mode is turned on\", and
now I\'m getting all sorts of errors. Anything helps. Thank you.
-In this project, the “current time” should be stored in 24-hour format. Only when the user
actually interacts with the program will the output and input be sometimes in 12-hour format.
Implement a main menu that will allow the user to (1) enter the current time, (2) add a forward
delta to the “current time”, (3) add a backwards delta to the “current time”, (4) display the
current time, (5) toggle the interface mode between 24-hour and 12-hour (should impact both
getTime and printTime). The program should identify invalid input for the main menu. When
entering the current time, only allow hours between 0 and 23 and minutes between 0 and 59.
Accept only ‘a’ and ‘p’ when entering the time in 12-hour mode. When entering time deltas, only
allow positive values or 0 (this is true for both forward and backwards time deltas).
Sample Output: (user input in italics)
1 - Enter a time
2 - Add delta to time
3 - Subtract delta from time
4 - Display current time
5 - Toggle 24-hour mode
6 - Exit
> 1
Enter the time with whitespace separating the hours and minutes: 5 50
1 - Enter a time
2 - Add delta to time
3 - Subtract delta from time
4 - Display current time
5 - Toggle 24-hour mode
6 - Exit
> 4
The current time is 05:50
1 - Enter a time
2 - Add delta to time
3 - Subtract delta from time
4 - Display current time
5 - Toggle 24-hour mode
6 - Exit
> 2
Enter an increment of hours and minutes (separated by a space): 0 120
The new time would be 07:50
1 - Enter a time
2 - Add delta to time
3 - Subtract delta from time
4 - Display current time
5 - Toggle 24-hour mode
6 - Exit
> 2
Enter an increment of hours and minutes (separated by a space): 2 20
The new time would be 08:10
1 - Enter a time
2 - Add delta to time
3 - Subtract delta from time
4 - Display current time
5 - Toggle 24-hour mode
6 - Exit
> 5
12-hour mode turned on
1 - Enter a time
2 - Add delta to time
3 - Subtract delta from time
4 - Display current time
5 - Toggle 24-hour mode
6 - Exit
> 1
Enter the time with whitespace separating the hours, minutes, and either \'a\' for am or \'p\' for
pm: 0 15 a
1 - Enter a time
2 - Add delta to time
3 - Subtract delta from time
4 - Display current time
5 - Toggle 24-hour mode
6 - Exit
> 4
The current time is 12:15 am
1 - Enter a time
2 - Add delta to time
3 - Subtract delta from time
4 - Display current time
5 - Toggle 24-hour mode
6 - Exit
> 3
Enter a decrement of hours and minutes (separated by a space): 1 15 The new time would be
11:00 pm
1 - Enter a time
2 - Add delta to time
3 - Subtract delta from time
4 - Display current time
5 - Toggle 24-hour mode
6 - Exit
> 3
Enter a decrement of hours and minutes (separated by a space): 0 90 The new time would be
10:45 pm
*****This is the program I wrote:
#include
using namespace.
Based on the sections Multicellularity in the Volvocine Algae and.pdf
Based on the sections \"Multicellularity in the Volvocine Algae\" and \"Mutational/Functional
Genetics\", what important aspect of multicellularity does the glsA gene regulate in Volvox? cell
communication cellular differentiation bulk transport cell adhesion
Solution
The glsA gene is involved in asymmetric cell division and germ cell specification during
embryogenesis – these are aspects of cell differentiation. Hence the correct answer is Cellular
differentiation..
The parallel axis theorum... can only be used to find the moment of .pdf
The parallel axis theorum... can only be used to find the moment of inertia about an axis
through the centroid can only be used to find the moment of inertia about horizontal axes can be
used to find the moment of inertia about any axis can only be used to find the moment of inertia
about perpendicular axes
Solution
can be used to find the moment of interia about any axis.
The many organisms designated as protists are closely related to bac.pdf
The many organisms designated as protists are closely related to bacteria a clad. all unicellular
all microscopic highly diverse and not all closely related to one another
Solution
Ans. Correct option E.
All other kingdoms have specific cell wall composition, for example peptidoglycan in Bacteria,
cellulose in Plantae, chitin in Fungi and no cell wall in Animalia. Similarly, all eukaryotic
kingdoms, except Protista, exhibit specific mode of nutrition, for example, photoautotrophs in
Plantae, Saprophytic in Fungi, and heterotrophic in Animalia.
The members of this taxa show great variations in their cell wall composition, mode of nutrition,
mode of reproduction, etc. Moreover, they exhibit great diversity in their physiology, too. In
spite of these diversifications, the member of this taxa share two common features-
I. these are eukaryotes, and
II. are Unicellular.
So, the taxon was framed to include all unicellular eukaryotes, that are even not closely related to
each other, and which could not fit in any other eukaryotic kingdoms.
Option A: Incorrect: Protists are eukaryotes.
Option B. Incorrect: It the taxon consisting of members that may show no immediate
evolutionary relationship (a clade is a group of organisms sharing common ancestry).
Option C. and D. Incorrect: Bacteria are also unicellular but not placed in this group. Moreover,
many members of the group predominantly exist in filamentous forms, colony forms. For
example, kelps can be serval meters long, visible by naked eyes..
The following table gives the average monthly exchange rate between t.pdf
The following table gives the average monthly exchange rate between the U.S. dollar and the
euro for 20096. It shows that 1 euro was equivalent to 1.289 U.S dollars in January 2009.
Develop a trend line that could be used to predict the exchange rate for 2010. Use this model to
predict the exchange rate for January 2010 and February 2010.
Solution.
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This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
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Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
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Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
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