1.Mitosis2.In meiosis,3.Homologous chromosomes4.Chromosomes ar.pdf
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1.Mitosis 2.In meiosis, 3.Homologous chromosomes 4.Chromosomes are 5.Which of the following is true of mitosis? 6. The phenotype of an individual refers to 7.Chromosomes are contained in 8. In diploid organisms 9.Gametes 10.The genotype of an individual refers toa.results in an egg only. Solution 1.Mitosis :c.results in a daughter cell that has the exact copy of the chromosomes of its parent. Reason: Mitosis results in two identical daughter cells with same number of chromosomes as that of parent cell. 2.Meiosis: d.haploid cells are produced from a single diploid cell. Reason:Form a single dipolid cell, four haploid cells are formed. 3.Homologous chromosomes: c. come in pairs. Reason: Homologous chromosomes means one from father and one from mother.So they come in pairs. 4.Chromosomes are :a.replicated during cell division. Reason: Chromosome replication takes place in mitosis. 5.Which of the following is true of mitosis?a.Mitosis produces identical daughter cells. Reason: Mitosis results in two identical daughter cells with same number of chromosomes as that of parent cell. 6.The phenotype of an individual refers to :d.its visible characteristics. Reason:The observed characteristics of individual is known as phenotype. 7.Chromosomes are contained in :d.the nuclei of eukaryotes. Reason: It is in the nucleus of eukaryotes that chromosomes are contained. 8.In diploid organisms :d.chromosomes occur in homologous pairs. Reason: Homologous chromosomes means one from father and one from mother.So they come in pairs. 9.Gametes : a.are the sex cells, or eggs and sperm. Reason:Gametes are male or female sex cells. 10.The genotype of an individual refers to :c.the number of chromosomes in its body cells. Reason:Genotype refers to genetic constitution of an individual and so refers to the number of chromosomes in its body cells..
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Meiosis pp [compatibility mode]
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Meiosis is a cell division process that results in gamete cells like sperm or egg cells containing half the number of chromosomes as regular body cells. It involves two rounds of cell division and nuclear division, resulting in four daughter cells with only two chromosomes each. This ensures fertilization restores the normal chromosome number.
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Meiosis is the process by which diploid cells undergo two cell divisions to produce four haploid daughter cells. It involves pairing of homologous chromosomes followed by two rounds of cell division. In meiosis I, homologous chromosomes separate into different daughter cells. This reduces the chromosome number by half. Meiosis II then separates sister chromatids, resulting in four haploid gametes or sex cells with only one chromosome of each type. Meiosis ensures genetic variation between offspring by independent assortment of chromosomes and crossing over during prophase I.
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Mitosis and meiosis are two types of cell division. Mitosis produces two daughter cells that are identical to the parent cell and have the same number of chromosomes. Meiosis produces four haploid daughter cells that have half the number of chromosomes of the parent cell and are genetically different from each other. Meiosis is involved in sexual reproduction to create gametes, while mitosis produces somatic cells for growth and repair of organisms.
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Meiosis is the process by which diploid cells in organisms undergo two cell divisions to produce four haploid daughter cells. This process ensures that gametes, or sex cells, have half the normal number of chromosomes. Specifically, meiosis involves one round of DNA replication followed by two cell divisions, reducing the chromosome number from diploid (2N) to haploid (N). The end result is four cells that are genetically different from each other and the original parent cell due to independent assortment and crossing over during meiosis I.
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Meiosis is the process by which diploid cells undergo two cell divisions to produce four haploid daughter cells. It involves pairing of homologous chromosomes followed by two rounds of cell division. In meiosis I, homologous chromosomes separate into different daughter cells. This reduces the chromosome number by half. Meiosis II then separates sister chromatids, resulting in four haploid gametes or sex cells with only one chromosome of each type. Meiosis ensures genetic variation between offspring by independent assortment of chromosomes and crossing over during prophase I.
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Mitosis and meiosis are two types of cell division. Mitosis produces two daughter cells that are identical to the parent cell and have the same number of chromosomes. Meiosis produces four haploid daughter cells that have half the number of chromosomes of the parent cell and are genetically different from each other. Meiosis is involved in sexual reproduction to create gametes, while mitosis produces somatic cells for growth and repair of organisms.
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Meiosis is the process by which diploid cells in organisms undergo two cell divisions to produce four haploid daughter cells. This process ensures that gametes, or sex cells, have half the normal number of chromosomes. Specifically, meiosis involves one round of DNA replication followed by two cell divisions, reducing the chromosome number from diploid (2N) to haploid (N). The end result is four cells that are genetically different from each other and the original parent cell due to independent assortment and crossing over during meiosis I.
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Meiosis Notes
Meiosis is a type of cell division that produces gametes (eggs and sperm) with half the normal number of chromosomes. This allows for fertilization to restore the full chromosome number. Meiosis involves two cell divisions after one round of DNA replication, resulting in four daughter cells each with half the number of chromosomes as the original parent cell. Differences between meiosis and mitosis include that meiosis reduces the chromosome number while mitosis maintains it.
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The document discusses meiosis, the process by which diploid cells undergo two cell divisions to produce four haploid cells. It explains that meiosis is necessary to produce gametes or sex cells that have half the normal number of chromosomes. This allows fertilization to restore the diploid number. The stages of meiosis I and meiosis II are outlined, noting how homologous chromosomes separate in meiosis I and sister chromatids separate in meiosis II, resulting in four unique haploid daughter cells.
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Meiosis
Meiosis is a type of cell division that produces gametes, such as sperm and egg cells, with half the number of chromosomes as regular body cells. This allows offspring to have a full set of chromosomes from each parent. Meiosis involves two cell divisions after one round of DNA replication, resulting in four haploid cells from one original diploid cell. Genetic variation arises in meiosis through independent assortment of chromosomes and recombination during crossover. Nondisjunction is when chromosomes fail to separate properly, resulting in gametes with extra or missing chromosomes and genetic disorders like Down syndrome.
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Meosis and mitosis are both forms of cell division. However, the outcomes of these processes
differ. Consider a diploid organism with two sexes. Select the reasons why meosis typically
produces genetic variation, whereas mitosis does not.
1n gametes produced in mitosis fuse to produce 2n cells in meiosis
Sister chromatids are not genetically identical as a result of crossing over during meiosis
Gametic chromosomes have a different combination of alleles than parental chromosomes as a
result of independent assortment
Meiosis produces four genetically different haploid cells
1n gametes produced in mitosis fuse to produce 2n cells in meiosis
Sister chromatids are not genetically identical as a result of crossing over during meiosis
Gametic chromosomes have a different combination of alleles than parental chromosomes as a
result of independent assortment
Meiosis produces four genetically different haploid cells.
meiosis
This document summarizes the process of meiosis. It begins by explaining that meiosis produces haploid gametes from diploid body cells to allow for sexual reproduction. It then describes the key phases and events of meiosis I and meiosis II, including homologous chromosome pairing, crossing over, and independent assortment. This genetic recombination during meiosis provides variation in offspring. The document concludes by discussing how nondisjunction can occasionally occur, resulting in gametes with an extra or missing chromosome and leading to conditions like Down syndrome.
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Meiosis is the process of cell division that produces gametes with half the normal number of chromosomes. It involves two rounds of division (Meiosis I and Meiosis II) starting with one diploid cell and resulting in four haploid cells. This ensures offspring inherit a mix of traits from both parents and allows for genetic variation between offspring. Meiosis occurs in the testes and ovaries, where homologous chromosomes separate in Meiosis I and sister chromatids separate in Meiosis II.
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This document describes the process of meiosis. It begins by explaining that human somatic cells are diploid, containing 23 pairs of homologous chromosomes, with one chromosome from each parent. Meiosis is then defined as the process that converts diploid cells to haploid cells. The summary proceeds to outline the key stages of meiosis I and meiosis II, including prophase I where homologous chromosomes pair up, metaphase and anaphase where the chromosomes separate, and telophase where the cell divides. The result of meiosis is four haploid cells, with half the number of chromosomes as the original cell.
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Meiosis reduces the number of chromosome sets from diploid to haploid in two cell divisions to produce gametes. It begins with chromosome duplication followed by two cell divisions, meiosis I and meiosis II. In meiosis I, homologous chromosomes separate and move to opposite poles, resulting in haploid daughter cells. Meiosis II then separates the sister chromatids, resulting in four haploid daughter cells each with a random assortment of one chromosome from each homologous pair. This ensures genetic variation in the offspring from sexual reproduction.
Meiosis
Meiosis is a type of cell division that produces gametes, such as sperm and egg cells, with half the number of chromosomes as normal body cells. It involves two nuclear divisions so that the original diploid cell is divided into four haploid cells. This ensures that offspring inherit the correct number of chromosomes from each parent. Important processes in meiosis include homologous chromosomes pairing up and crossing over, which introduces genetic variation. Errors in meiosis can result in chromosomal mutations like nondisjunction and aneuploidy.
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Meiosis is a type of cell division that produces gametes, such as sperm and egg cells, with half the number of chromosomes as regular body cells. This involves two rounds of division called Meiosis I and Meiosis II. In Meiosis I, homologous chromosomes pair up and may exchange genetic material through crossing over. This results in two daughter cells each with half the original number of chromosomes. Meiosis II then separates the sister chromatids, resulting in four haploid gametes total. This ensures fertilization restores the full chromosome number.
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Chromosomes play a key role in inheritance by carrying genes from parents to offspring. During meiosis, chromosome pairs separate and divide, reducing the number of chromosomes in sex cells to half the normal amount. When sperm and egg join during fertilization, they form a new diploid cell with the normal number of chromosomes, half from each parent. Genes located on chromosomes are inherited this way, accounting for both the similarities and variations between parents and their offspring.
Given that tan = 7 in quadrant IV, nd sin 2, cos 2, and tan 2.S.pdf
Give one example of how allele frequencies change from one generation to the next due to
mutation, migration, genetic drift, nonrandom mating, and selection.
Give one example of how allele frequencies change from one generation to the next due to
mutation, migration, genetic drift, nonrandom mating, and selection.
Solution
Answers:
Change of allele frequencies due to the said factors have been discussed below:
Mutation:
Mutations introduces new alleles into a population and can affect allele the allele frequency in
the population.
Migration:
The movement of new individuals into or out of a population results in a change in allele
frequency. Immigration results in addition of new alleles to a population while the reverse results
in a loss.
Genetic drift:
It occurs due to a change in allele frequencies in a population by chance. Genetic drift is a
random process and result in large changes in population within a short interval.
Non random mating:
Unless accompanied by selection, non random mating change the genotypic frequencies but not
the allele frequencies in a population. The most important form of non random mating is
inbreeding.
Selection:
Selection occurs when genotypes have different reproductive successes. It acts as a positive filter
and removes alleles which has negative influence on a population while selects those which have
positive or neutral effects..
How would functionalism perspective view the relationship between me.pdf
How to compute the cost oand ond pertmance bond2 paument bond 1 supply subdivicion band1
in a certan conshuchion proiedt with exomples
Solution
Bid Bond
Contractors who submit bids usually are asked to
provide a bid bond. The bid bond states that the
contractor will enter into a contract when one is
offered and will provide bonding as required.<>Bid
bonds generally are written with a penalty equal to a
percentage of the contract price; usually 5%, 10%,
or 20%. They may also be written with a specific
dollar penalty. If an owner offers a contract to a
selected contractor and the contractor refuses to
enter into the contract, the owner may make a claim
against the bid bond for the difference between the
price of the contract in question and the price of a
substitute contract or the penalty of the bid bond,
whichever is less.
Maintenance bond
Maintenance bonds are used when an owner wants a
warranty period beyond one year. A warranty period
can be extended for an annual fee, but sureties gener-
ally do not go beyond a total of two or three years.
The annual fee for a maintenance bond is a fraction
of the cost of a performance bond.
Performance bond
Performance bonds guarantee to the owner that the
contractor will perform its contractual obligations in
accordance with the plans and specifications. These
bonds can take a variety of forms, ranging from the
very simple to the long and complicated. The cost of a performance bond usually is less than 1%
of the contract price; however, if the contract is under $1 million, the premium may run between
1% and 2%. Bonds may be more costly, depending upon the credit-worthiness of the contractor.
Suppy Bond
Supply bonds guarantee that ordered materials will
be delivered. Such bonds generally are employed if
an item is critical, time-sensitive, hard to find, or
proprietary. Supply bonds may guarantee only a
purchase order, so the terms and conditions of that
order should always be carefully drafted. The cost of
these bonds is usually minimal.
Payment bond
Labor and material payment bonds are companions
to the performance bond. They assure the owner that
the labor, material, and subcontractor costs on the
job will be paid. This assurance is for the use and
benefit of all laborers, material suppliers, and
subcontractors who are eligible by contract or statute
for the protection afforded by the payment bonds.
They can also act as a way to protect the project from
liens.
Subdivision bond
Frequently, general contractors will require from
their subcontractors the same types of bonds required
by the owner. Generals may do this, for example,
when the sub trade is critical to the project, the sub’s
price was much lower than its competitors, the sub is
not well-known to the general, or the general’s surety
requires the bonding of some or all subs as a precondition to bonding the general..
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Meiosis is a type of cell division that produces gametes (eggs and sperm) with half the normal number of chromosomes. This allows for fertilization to restore the full chromosome number. Meiosis involves two cell divisions after one round of DNA replication, resulting in four daughter cells each with half the number of chromosomes as the original parent cell. Differences between meiosis and mitosis include that meiosis reduces the chromosome number while mitosis maintains it.
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The document discusses meiosis, the process by which diploid cells undergo two cell divisions to produce four haploid cells. It explains that meiosis is necessary to produce gametes or sex cells that have half the normal number of chromosomes. This allows fertilization to restore the diploid number. The stages of meiosis I and meiosis II are outlined, noting how homologous chromosomes separate in meiosis I and sister chromatids separate in meiosis II, resulting in four unique haploid daughter cells.
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Meiosis creates sex cells such as eggs and sperm. It results in 4 haploid cells that each have a single set of chromosomes. Meiosis involves two divisions, Meiosis I and Meiosis II, which separates the homologous chromosomes and then the sister chromatids. This leads to genetic variation between the resulting gametes and allows for genetic recombination during crossing over. Nondisjunction during meiosis can result in conditions like Down syndrome if an extra chromosome is present (trisomy 21) or Turner syndrome if a female is missing an X chromosome (monosomy X).
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Meiosis is a type of cell division that produces gametes, such as sperm and egg cells, with half the number of chromosomes as regular body cells. This allows offspring to have a full set of chromosomes from each parent. Meiosis involves two cell divisions after one round of DNA replication, resulting in four haploid cells from one original diploid cell. Genetic variation arises in meiosis through independent assortment of chromosomes and recombination during crossover. Nondisjunction is when chromosomes fail to separate properly, resulting in gametes with extra or missing chromosomes and genetic disorders like Down syndrome.
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produces genetic variation, whereas mitosis does not.
1n gametes produced in mitosis fuse to produce 2n cells in meiosis
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Meiosis is a type of cell division that produces gametes, such as sperm and egg cells, with half the number of chromosomes as normal body cells. It involves two nuclear divisions so that the original diploid cell is divided into four haploid cells. This ensures that offspring inherit the correct number of chromosomes from each parent. Important processes in meiosis include homologous chromosomes pairing up and crossing over, which introduces genetic variation. Errors in meiosis can result in chromosomal mutations like nondisjunction and aneuploidy.
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More from longojasperze84880
Given that tan = 7 in quadrant IV, nd sin 2, cos 2, and tan 2.S.pdf
Give one example of how allele frequencies change from one generation to the next due to
mutation, migration, genetic drift, nonrandom mating, and selection.
Give one example of how allele frequencies change from one generation to the next due to
mutation, migration, genetic drift, nonrandom mating, and selection.
Solution
Answers:
Change of allele frequencies due to the said factors have been discussed below:
Mutation:
Mutations introduces new alleles into a population and can affect allele the allele frequency in
the population.
Migration:
The movement of new individuals into or out of a population results in a change in allele
frequency. Immigration results in addition of new alleles to a population while the reverse results
in a loss.
Genetic drift:
It occurs due to a change in allele frequencies in a population by chance. Genetic drift is a
random process and result in large changes in population within a short interval.
Non random mating:
Unless accompanied by selection, non random mating change the genotypic frequencies but not
the allele frequencies in a population. The most important form of non random mating is
inbreeding.
Selection:
Selection occurs when genotypes have different reproductive successes. It acts as a positive filter
and removes alleles which has negative influence on a population while selects those which have
positive or neutral effects..
How would functionalism perspective view the relationship between me.pdf
How to compute the cost oand ond pertmance bond2 paument bond 1 supply subdivicion band1
in a certan conshuchion proiedt with exomples
Solution
Bid Bond
Contractors who submit bids usually are asked to
provide a bid bond. The bid bond states that the
contractor will enter into a contract when one is
offered and will provide bonding as required.<>Bid
bonds generally are written with a penalty equal to a
percentage of the contract price; usually 5%, 10%,
or 20%. They may also be written with a specific
dollar penalty. If an owner offers a contract to a
selected contractor and the contractor refuses to
enter into the contract, the owner may make a claim
against the bid bond for the difference between the
price of the contract in question and the price of a
substitute contract or the penalty of the bid bond,
whichever is less.
Maintenance bond
Maintenance bonds are used when an owner wants a
warranty period beyond one year. A warranty period
can be extended for an annual fee, but sureties gener-
ally do not go beyond a total of two or three years.
The annual fee for a maintenance bond is a fraction
of the cost of a performance bond.
Performance bond
Performance bonds guarantee to the owner that the
contractor will perform its contractual obligations in
accordance with the plans and specifications. These
bonds can take a variety of forms, ranging from the
very simple to the long and complicated. The cost of a performance bond usually is less than 1%
of the contract price; however, if the contract is under $1 million, the premium may run between
1% and 2%. Bonds may be more costly, depending upon the credit-worthiness of the contractor.
Suppy Bond
Supply bonds guarantee that ordered materials will
be delivered. Such bonds generally are employed if
an item is critical, time-sensitive, hard to find, or
proprietary. Supply bonds may guarantee only a
purchase order, so the terms and conditions of that
order should always be carefully drafted. The cost of
these bonds is usually minimal.
Payment bond
Labor and material payment bonds are companions
to the performance bond. They assure the owner that
the labor, material, and subcontractor costs on the
job will be paid. This assurance is for the use and
benefit of all laborers, material suppliers, and
subcontractors who are eligible by contract or statute
for the protection afforded by the payment bonds.
They can also act as a way to protect the project from
liens.
Subdivision bond
Frequently, general contractors will require from
their subcontractors the same types of bonds required
by the owner. Generals may do this, for example,
when the sub trade is critical to the project, the sub’s
price was much lower than its competitors, the sub is
not well-known to the general, or the general’s surety
requires the bonding of some or all subs as a precondition to bonding the general..
For which column would you create an indexSolutionAnswerIn.pdf
For the following four questions, your answer should be about one paragraph for each, but be
sufficiently detailed. If you wish, you can use numerical calculations involving the Nernst or
Goldman equations as part of your qualitative explanations; however, if you choose to do so, you
must establish and justify your basic assumptions, and you must show all calculations. For the
following scenarios, explain what the changes in resting permeabilities will do to V_m of an
excitable cell. An increase in resting permeability to Na^+ A decrease in resting permeability to
Na^+ An increase in resting permeability to K^+. A decrease in resting permeability to K^+.
Read the accompanying article, Potassium chloride in hospitals a health hazard, and answer the
following three questions. Explain why increasing extracellular K^+ may cause death. For this,
you can assume that resting permeability to K^+ does not change. To properly answer this
question, you must research the problem using Internet, literature, or library sources. Be certain
to provide proper citation. Remember, cutting-and pasting is plagiarism and s forbidden. Explain
why increasing extracellular K^+ caused searing pain to the unfortunate patients. Include
appropriate clinical and/or research references to support your answer. To properly answer this
question, you must research the problem using Internet, literature, or library sources. Be certain
to provide proper citation. Remember, cutting-and pasting is plagiarism and is forbidden. Since
V_m is close to resting E_K, you can approximate V_m as E_K by using the Nernst equation.
Use at least 3 numerical examples of increasing extracellular [K^+], but ones that are
physiologically appropriate extracellular [K^+]. Again, you must research this. Calculate the
approximate changes in V_m; show all calculations, and explain what is happening.
Alternatively, you may employ the Goldman equation, but as mentioned above, you must
establish and justify your basic assumptions, and you must show all calculations.
Solution
2A)An increase in resting permeability to Na+ results in depolarisation of the cell.
2B)An decrease in resting permeability to Na+ makes cell move positive.
2C) An increase in resting permeability to K+ makes cell more positive.
2D)An decrease in resting permeability to K+ makes cell more negative.
Nerst Equation is given by-:
Eion =RT ln (Ion1) /ZF(Ion 2)
R=gas constant =1.987 cal 7mol - deg
T= degree K
Z= ion balance
F=faraday\'s constant 23.062 cal/ V mol
At 37 degree C E in mV = - 61.5/z log10 (Ion 1)/(Ion 2)
According to above equation
When Na+ permeability begins to exceed the membrane potential approaches to ENa +
Decease in Na+ lead to increase in K+ permeability.
Increase in k+permeability makes cell more positive Ek+.
Decease in k+ permeability makes cell negative..
Explain how the Seven Years War had a significant impact on coloni.pdf
Explain for each step in this sample Program? %Select( EMPLID) SELECT EM PUD FROM
ABC SQLExec(\"SELECTCOUNT(*)FROM ABC\", &Count;); MessageBox(0, \"\", 0, 0 \"No
of Rows in ABC Table:\" | & Count); MessageBox(0, \"\", 0, 0, \"Employee ID in State Record:
\" | CNU_AET.EMPLTD.Value); DELETE FROM ABC WHERE EMPLID IN (SELECT TOP
5 EMPLID FROM ABC ORDER BY EMPLID DESC)
Solution
Q10
No of Rows in ABC Table: 10
Employee ID in State Record: AA
No of Rows in ABC Table: 5
Employee ID in State Record: AA.
dw Lesson 17 Question 274 If an IRS examiner believes that a tax prac.pdf
During teh month of August 2014, ADC Co, a CPA service, had the following transactions:
Billed $48,000 in revenues on credit
Received $2,000 from customers\' accounts receivable
Incurred expenses of $6,000 but only paid $5,750 cash for these expenses
Prepaid $1,110 for computer services to be used next month
Incurred wage expense of $100 but to be paid next month
Used office building for the month; depreciation expenseof $4,000 not yet recorded
What was the company\'s accrual basis net income for the month and why?
Solution.
Click on each bony marking term that represents a part of a bone that.pdf
Class and Objects based Problems:
Question 1: Circle -radius: double +Circle() +Circle(r: double) +setRadius(radius: double): void
+getRadius():double +getArea():double +getPerimeter():double
Design a class named Circle with the features shown in the UML Diagram. Write a test class and
create two objects of circle class and ask the user to enter the radius for each object and give the
entered radius as input to circle class to find the area and perimeter. The default value is 1 for
radius. Display the radius, area, and perimeter of circle in this order. Question
2. Design a class named Person, Student and Employee. a. A Person has a Name, Address and
phone Number. b. A Student has an id and major. c. An Employee has a Salary and Employee
Number. Write a constructor, setter and getter for each class. Write display() method in each
class to display all the fields. Write a test program that creates a Person, Student, Employee, and
invokes their display() methods.
Solution
1
#include
using namespace std;
class Circle
{
private:
double radius;
public:
Circle() //default constructor with default radius =1
{
radius = 1;
}
Circle(double r) //parameterized constructor
{
radius = r;
}
void setRadius(double r) //set and get methods
{
radius = r;
}
double getRadius()
{
return radius;
}
double getArea() //calculate area
{
return (3.1412*radius*radius);
}
double getPerimeter() //calculate perimeter
{
return (2*3.1412*radius);
}
};
int main()
{
Circle c1 ; //object 1
double radius;
cout<<\"\ Enter the radius of the circle \";
cin>>radius;
c1.setRadius(radius);
cout<<\"\ Radius= \"<
using namespace std;
class Person
{
private:
string name;
string address;
string phone;
public:
Person(string n,string a,string p) //default constructor
{
name = n;
address =a;
phone =p;
}
void setName(string n) //set and get methods
{
name = n;
}
string getName()
{
return name;
}
void setAddress(string a) //set and get methods
{
address=a;
}
string getAddress()
{
return address;
}
void setPhone(string p) //set and get methods
{
phone = p;
}
string getPhone()
{
return phone;
}
void display()
{
cout<<\"\ Person Name :\"<.
a taxpayer who owns And operates two separate business activities ca.pdf
A study examined waiting times in the X-ray department for a hospital in Jacksonville, Florida.
A clerk recorded the number of patients waiting for service at 9:00 A.M. on 20 consecutive days
and obtained the following results.
Number Waiting
Number of Days Outcome Occurred
0
2
1
5
2
6
3
4
4
3
Total
20
a. Define the experiment the clerk conducted.
b. List the experimental outcomes.
c. Assign probabilities to the experimental outcomes.
d. What method did you use?
Number Waiting
Number of Days Outcome Occurred
0
2
1
5
2
6
3
4
4
3
Total
20
Solution
1) Experiment :
The clerk noted the number of people waiting in the queue at 9.00 AM everyday at the hospital
for 20 days consecutively
2) Experimental outcomes :
The outcomes are X = 0,1,2,3,4 patients
3) Probabilities :
d)
Probabilities are calculated as :
Number of days of occurence / total number of days
Hope this helps. Ask if you have doubtsNumber WaitingNumber of
DaysProbability020.1150.25260.3340.2430.15Total20.
A book of 500 pages contains 750 misprints. What is the probability .pdf
A bird sanctuary has 23 parakeets and is planning a breeding program to increase the number of
birds. The program aims to increase the number of parakeets by 18% each year. The number of
parakeets, N, can be found using the formula N = c(1.18)^t, where c is the initial number of
parakeets and t is the number of years. How many parakeets will there be after 9 years? A 405 B
102 C 244 D 27
Solution
The formula for N, the number of parakeets is N = c( 1.18)t where c is the initial number of
parakeets and t is the number of years. Here, c = 23 and t = 9. Therefore, N = 23(1.18)9 =
23*4.435453859 = 102 ( on rounding off to the nearest whole number.The answer B is correct..
A hypothesis test regarding the population mean is based on The samp.pdf
A golden bar of metal (77 centimeters by 1919 centimeters by 44 centimeters) being transported
by armored car is suspected of being fake, made from a less valuable metal with a thin coating of
pure gold. The bar is found to have a mass of 3.83.8 kilograms. If the specific gravity of gold is
19.319.3, is the bar fake? Justify your answer.
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A.) The bar is fake because the specific gravity of a bar is __________ and it is different from
the specific gravity of gold. (Round your answer to three significant digits.)
or
B. The bar is not fake because its specific gravity is 19.319.3 and it is the same as the specific
gravity of gold.
Solution
The bar is not fake because its specific gravity is 19.319 and it is the same specific gravity of
gold.
What services require auditors to be independent How does an audito.pdf
what is the time complexity for this code: For (int i = 1 to n) {j = i while(j
Solution
The time complexity of an algorithm quantifies the amount of time taken by an algorithm to run
as a function of the length of the string representing the input.
The time complexity of an algorithm is commonly expressed using big O notation, which
excludes coefficients and lower order terms.
When expressed this way, the time complexity is said to be described asymptotically, i.e., as the
input size goes to infinity.
For example, if the time required by an algorithm on all inputs of size n is at most 5n3 + 3n for
any n (bigger than some n0), the asymptotic time complexity is O(n3).
Time complexity is commonly estimated by counting the number of elementary operations
performed by the algorithm, where an elementary operation takes a fixed amount of time to
perform.
Thus the amount of time taken and the number of elementary operations performed by the
algorithm differ by at most a constant factor..
Write 250 words on information on the MacBook, and explanation of th.pdf
Will has $4,355.19 in his checking account. He writes checks for $1,204.90 and $890.99. How
much is left in Will’s account?
A. $1,189.65
B. $2,259.30
C. $2,867.30
D. $2,989.50
Solution
B. To calculate the answer to this question, add up the amount of the checks and subtract that
from the current amount in the checking account.
$4,355.19 – ($1,204.90 + $890.99)
= $2,259.30.
What is special about the skulls of LagomorphsSagittal crestThe.pdf
What is a Relay Log in My SQL? Please reference the question.
Solution
The dictionary definition of relay is to recieve and pass on. It is similar in MySQL. When the
server wants to replicate, it creates many logs that store the binary log events from master to
slave. It contains the set of events that change the data or structure of the database.
There are two parts to it:
> Set of relay log files
> Index file of all relay log files
The step by step process of replication can be described as follows:
The master is connected to IO Thread by reading the binary log events and these are replicated in
the relay log.This way a new copy of the events is created..
What is the difference between an open channel flow and sheet flo.pdf
What is special about the skulls of Lagomorphs?
Sagittal crest
The skull is fenestrated
Deeply roted molars
Incomplete zygomatic arch
Solution
special about the skulls of Lagomorphs is
The skull is fenestrated
Explination= in case of Lagomorphs on various places on skull surface ,the bone is exceedingly
porous or perforated & this screen like structure is called fenestrae or with windows.
BIO cell and molecular bio ANSWER ALL1 which of the fo.pdf
because of a hurricane, your regular laboratory session was cancelled and the gram stain
procedure was performed on cultures incubated for a longer period of time. Examination of the
stain staphylococcus aureus slides revealed a great deal of color variability ranging from intese
blue to shades of pink. Account for this result.
Solution
Staphylococcus aureus is a Gram positive, non motile bacteria. On Gram staining it will appear
purple to blue color since it will retain crystal violet dye. Again, Gram negative bacteria say for
example, E.coli will appear pink in color.
So, here due to longer period of incubation the culture has been contaminated with Gram
negative bacteria. Therefore showing color variability in Gram staining results..
If the neuron membrane becomes more permeable to Na+, Na+ will trans.pdf
If I drive x mph for t hours and then y mph for 2t hours. Then what is expression for the average
speed.
Solution
average speed = (total distance)/(total time)
-----
distance = rate*time
-------------------
Your Problem:
total distance = (xt)+ (2yt) = (x+2y)t miles
----------------------
total time = t + 2t = 3t hours
---------------------------
Therefore, average speed = [(x+2y)t]/(3t) = (x+2y)/3 mph.
The figure below shows the chromosomes in a cell for some species at .pdf
The employees have indicated that 95% of employees in one of the departments are receiving
higher salaries than any other department because their supervisor parties with them each
weekend. Some employees only have an hourly rate which you have to calculate their yearly
salary before completing the calculations. Please calculate the 95% standard deviation for each
department and indicate for me if any department is receiving on average $2k or more on the
high end.
The employees have indicated that 95% of employees in one of the departments are receiving
higher salaries than any other department because their supervisor parties with them each
weekend. Some employees only have an hourly rate which you have to calculate their yearly
salary before completing the calculations. Please calculate the 95% standard deviation for each
department and indicate for me if any department is receiving on average $2k or more on the
high end.
Department 1
Department 2
Department 3
Department 4
Department 5
$22,450
18,550
B
22555
19050
$30,050
27,050
16765
29950
21545
$23,785
26,785
30050
23785
27950
A
25050
29955
B
27550
$27,594
26557
A
27650
25550
A – $9/hr @ 40 hrs week @ 50 weeks a year
B – $11/hr @ 35 hrs wee @ 52 weeks a year
Is any department receiving on average $2k or more on the high end of the standard deviation
using standard deviation at 95%? If so, which department and be sure to explain your answer.
Department 1
Department 2
Department 3
Department 4
Department 5
$22,450
18,550
B
22555
19050
$30,050
27,050
16765
29950
21545
$23,785
26,785
30050
23785
27950
A
25050
29955
B
27550
$27,594
26557
A
27650
25550
Solution.
The Additional Funds Needed Equation (AFN) can be intuitively inte.pdf
Test y = 4x^2/x^2 + 1 for symmetry.
Solution
y =4x2/(x2+1)
for symmetry about y axis:
put -x in place of x
y=4(-x)2/((-x)2+1)
y=4x2/(x2+1) we get same equation. so curve is symmetric about y axis
for symmetry about x axis
put -y in place of y:
-y =4x2/(x2+1)
equation is not same as original one
so not symmetric about x axis
for symmetry about origin
put -y in place of y:, -x in place of x
-y =4(-x)2/((-x)2+1)
-y =4(x)2/(x2+1)
equation is not same as original one
so not symmetric about origin.
Summarise the behavioural differences between subcritical and superc.pdf
Statistics: Random Sample of Packages
Thank you. At a nearby post office, the weight of packages mailed out has a mean of 3.22
pounds and a standard deviation of 1.06 pounds. Consider a random sample of 50 packages
mailed out from this post office in the last year. Find the probability that the average weight of
these packages exceeds 3.4 pounds. Clearly state the distribution of the sample mean and find the
probability. Is your answer exact or approximate?
Solution
****************************************
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 3.4
u = mean = 3.22
n = sample size = 50
s = standard deviation = 1.06
Thus,
z = (x - u) * sqrt(n) / s = 1.200747364
Thus, using a table/technology, the right tailed area of this is
P(z > 1.200747364 ) = 0.114924608 [ANSWER]
************
BY central limit theorem, the distribution of the means will be NORMALLY DISTRIBUTED.
This is an approximate, because we only have a limited sample size..
Similar and Nonsimilar Shapes Name Find the pairs of shapes that are .pdf
Show that, in the process of transferring a tower under the restrictions of the preceding excercise,
we will actually encounter every properly stacked arrangement of n disks on three pegs.
Tower of hanoi problem.
Solution
Let call the three pegs Src (Source), Aux (Auxiliary) and Dst (Destination). To better understand
and appreciate the following solution you should try solving the puzzle for small number of
disks, say, 2,3, and, perhaps, 4. However one solves the problem, sooner or later the bottom disk
will have to be moved from Src to Dst. At this point in time all the remaining disks will have to
be stacked in decreasing size order on Aux. After moving the bottom disk from Src to Dst these
disks will have to be moved from Aux to Dst. Therefore, for a given number N of disks, the
problem appears to be solved if we know how to accomplish the following tasks:.
Please answer the question carefully.Name and briefly describe 4 d.pdf
Plasmid pRIT452 was cut with Pstl. HindlII, and EcoRI. From the data below, determine the
map Pstl HindIll. 6.2 EcoRI PstI HindIII 4.8 2.0 1.7 Pstl EcoRI 5.4 3.8 3.0 EcoRI HindlII 6.2 1.8
1.2 Plasmid pRIT453 was cut with smal, Hindlu, and EcoRI. From the data low, determine the
map. EcoRI 1.6 Hindlll 1.9 Smal 6.6 EcoRI Hindlll 4.5 1.9 1.6 EcoRI Smal 5.7 0.7 Smal HindIII
2.7 2.2 1.9 e the ry on only des o nega
Solution
figure shows pRIT452 map, in the same way ,it can map pRIT453 also. first draw a gel running
like picture then compare fragment with different enzyme and find out exact position
accordingly.
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This slide is special for master students (MIBS & MIFB) in UUM. Also useful for readers who are interested in the topic of contemporary Islamic banking.
DRUGS AND ITS classification slide share
Any substance (other than food) that is used to prevent, diagnose, treat, or relieve symptoms of a
disease or abnormal condition
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
Denis is a dynamic and results-driven Chief Information Officer (CIO) with a distinguished career spanning information systems analysis and technical project management. With a proven track record of spearheading the design and delivery of cutting-edge Information Management solutions, he has consistently elevated business operations, streamlined reporting functions, and maximized process efficiency.
Certified as an ISO/IEC 27001: Information Security Management Systems (ISMS) Lead Implementer, Data Protection Officer, and Cyber Risks Analyst, Denis brings a heightened focus on data security, privacy, and cyber resilience to every endeavor.
His expertise extends across a diverse spectrum of reporting, database, and web development applications, underpinned by an exceptional grasp of data storage and virtualization technologies. His proficiency in application testing, database administration, and data cleansing ensures seamless execution of complex projects.
What sets Denis apart is his comprehensive understanding of Business and Systems Analysis technologies, honed through involvement in all phases of the Software Development Lifecycle (SDLC). From meticulous requirements gathering to precise analysis, innovative design, rigorous development, thorough testing, and successful implementation, he has consistently delivered exceptional results.
Throughout his career, he has taken on multifaceted roles, from leading technical project management teams to owning solutions that drive operational excellence. His conscientious and proactive approach is unwavering, whether he is working independently or collaboratively within a team. His ability to connect with colleagues on a personal level underscores his commitment to fostering a harmonious and productive workplace environment.
Date: May 29, 2024
Tags: Information Security, ISO/IEC 27001, ISO/IEC 42001, Artificial Intelligence, GDPR
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- 1. 1.Mitosis 2.In meiosis, 3.Homologous chromosomes 4.Chromosomes are 5.Which of the following is true of mitosis? 6. The phenotype of an individual refers to 7.Chromosomes are contained in 8. In diploid organisms 9.Gametes 10.The genotype of an individual refers toa.results in an egg only. Solution 1.Mitosis :c.results in a daughter cell that has the exact copy of the chromosomes of its parent. Reason: Mitosis results in two identical daughter cells with same number of chromosomes as that of parent cell. 2.Meiosis: d.haploid cells are produced from a single diploid cell. Reason:Form a single dipolid cell, four haploid cells are formed. 3.Homologous chromosomes: c. come in pairs. Reason: Homologous chromosomes means one from father and one from mother.So they come in pairs. 4.Chromosomes are :a.replicated during cell division. Reason: Chromosome replication takes place in mitosis. 5.Which of the following is true of mitosis?a.Mitosis produces identical daughter cells. Reason: Mitosis results in two identical daughter cells with same number of chromosomes as that of parent cell. 6.The phenotype of an individual refers to :d.its visible characteristics. Reason:The observed characteristics of individual is known as phenotype. 7.Chromosomes are contained in :d.the nuclei of eukaryotes. Reason: It is in the nucleus of eukaryotes that chromosomes are contained. 8.In diploid organisms :d.chromosomes occur in homologous pairs. Reason: Homologous chromosomes means one from father and one from mother.So they come in pairs. 9.Gametes : a.are the sex cells, or eggs and sperm. Reason:Gametes are male or female sex cells.
- 2. 10.The genotype of an individual refers to :c.the number of chromosomes in its body cells. Reason:Genotype refers to genetic constitution of an individual and so refers to the number of chromosomes in its body cells.