STEP 1 Null Hypothesis H0 P1 = P2 Alternate Hypothesis Ha P1.pdf
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STEP 1: Null Hypothesis H0: P1 = P2 Alternate Hypothesis Ha: P1 ? P2 STEP 2: Analysis Plan for significance level: 0.05 STEP 3: Analyze Sample Data n1: Group 1 sample size = 400 n2: Group 2 sample size = 400 Proportion p1: = 0.0675 Proportion p2: = 0.1 Pooled Sample Proportion p = (p1 * n1 + p2 * n2) / (n1 + n2) = [(0.0675 * 400) + (0.1 * 400)] / (400 + 400) = 67 / 800 = 0.0838 Where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 2, and n2 is the size of sample 2. Standard Error: SE = sqrt{p * (1 - p ) * [(1/n1) + (1/n2)]} = sqrt{0.0838 * (1 - 0.0838 ) * [(1/400) + (1/400)]} = sqrt{0.000384} = 0.0196 STEP 4: Test Statistic z-score: z = (p1 - p2) / SE = (0.0675 - 0.1)/0.0196 = -1.659 For two-tailed test, the p-value is the probability that the z-score is less than -1.659 and more than 1.659 Use the Normal Distribution Table to find P(z < -1.659) = 0.049, and P(z >1.659) = 0.049 The Table for Standard Normal Distribution is organized as a cummulative \'area\' from the LEFT corresponding to the STANDARDIZED VARIABLE z. The Standard Normal Distribution is also symmetric (called a \'Bell Curve\') which means its an interpretive procedure to Look-Up the \'area\' from the Table. For STANDARDIZED VARIABLE z = -1.659 the corresponding LEFT \'area\' = 0.049 And due to Table\'s cummulative nature, the corresponding RIGHT \'area\' = 0.049 Solution STEP 1: Null Hypothesis H0: P1 = P2 Alternate Hypothesis Ha: P1 ? P2 STEP 2: Analysis Plan for significance level: 0.05 STEP 3: Analyze Sample Data n1: Group 1 sample size = 400 n2: Group 2 sample size = 400 Proportion p1: = 0.0675 Proportion p2: = 0.1 Pooled Sample Proportion p = (p1 * n1 + p2 * n2) / (n1 + n2) = [(0.0675 * 400) + (0.1 * 400)] / (400 + 400) = 67 / 800 = 0.0838 Where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 2, and n2 is the size of sample 2. Standard Error: SE = sqrt{p * (1 - p ) * [(1/n1) + (1/n2)]} = sqrt{0.0838 * (1 - 0.0838 ) * [(1/400) + (1/400)]} = sqrt{0.000384} = 0.0196 STEP 4: Test Statistic z-score: z = (p1 - p2) / SE = (0.0675 - 0.1)/0.0196 = -1.659 For two-tailed test, the p-value is the probability that the z-score is less than -1.659 and more than 1.659 Use the Normal Distribution Table to find P(z < -1.659) = 0.049, and P(z >1.659) = 0.049 The Table for Standard Normal Distribution is organized as a cummulative \'area\' from the LEFT corresponding to the STANDARDIZED VARIABLE z. The Standard Normal Distribution is also symmetric (called a \'Bell Curve\') which means its an interpretive procedure to Look-Up the \'area\' from the Table. For STANDARDIZED VARIABLE z = -1.659 the corresponding LEFT \'area\' = 0.049 And due to Table\'s cummulative nature, the corresponding RIGHT \'area\' = 0.049.
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![STEP 1:
Null Hypothesis H0: P1 = P2
Alternate Hypothesis Ha: P1 ? P2
STEP 2:
Analysis Plan for significance level: 0.05
STEP 3:
Analyze Sample Data
n1: Group 1 sample size = 400
n2: Group 2 sample size = 400
Proportion p1: = 0.0675
Proportion p2: = 0.1
Pooled Sample Proportion
p = (p1 * n1 + p2 * n2) / (n1 + n2)
= [(0.0675 * 400) + (0.1 * 400)] / (400 + 400)
= 67 / 800
= 0.0838
Where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2,
n1 is the size of sample 2, and n2 is the size of sample 2.
Standard Error:
SE = sqrt{p * (1 - p ) * [(1/n1) + (1/n2)]}
= sqrt{0.0838 * (1 - 0.0838 ) * [(1/400) + (1/400)]}
= sqrt{0.000384}
= 0.0196
STEP 4:](https://image.slidesharecdn.com/step1nullhypothesish0p1p2alternatehypothesishap1-230409115512-974f9b61/85/STEP-1-Null-Hypothesis-H0-P1-P2-Alternate-Hypothesis-Ha-P1-pdf-1-320.jpg)
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int x,y;
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domain, so its not onto
onto=false;
}
///store the results of your findints
// 0th index isfunction?
// 1st index isOneone?
//2ind index isonto?
analysis[0]=function;
analysis[1]=oneone;
analysis[2]=onto;
return analysis;
}
public static void main(String[] args) {
int a,b,c,m,n;
HashMap fdomain=new HashMap();
HashMap> fcodomain=new HashMap>();
HashMap gdomain=new HashMap();
HashMap> gcodomain=new HashMap>(.
Embedded systemIt is a computer system with a dedicated function .pdf
Embedded system:
It is a computer system with a dedicated function within a larger mechanical or electrical system,
often with real-time computing constraints. It is embedded as part of a complete device often
including hardware and mechanical parts.They can control many devices in common use today.
Solution
Embedded system:
It is a computer system with a dedicated function within a larger mechanical or electrical system,
often with real-time computing constraints. It is embedded as part of a complete device often
including hardware and mechanical parts.They can control many devices in common use today..
Factors.javaimport java.io.; import java.util.Scanner; class .pdf
Factors.java
import java.io.*;
import java.util.Scanner;
class Factors
{
public static void main(String args[])
{
int a,i;
Scanner in = new Scanner(System.in);
System.out.println(\"Enter an integer\");
a = in.nextInt();
System.out.println(\"You entered integer \"+a);
System.out.print(\"\ \");
System.out.print(\"The factors are : \");
for(i=1;i<=a/2;i++)
{
if(a%i==0)
System.out.print(i+\",\");
}
System.out.print(a);
}
}
Output:
Enter an integer
44
You entered integer 44
The factors are : 1,2,4,11,22,44
Solution
Factors.java
import java.io.*;
import java.util.Scanner;
class Factors
{
public static void main(String args[])
{
int a,i;
Scanner in = new Scanner(System.in);
System.out.println(\"Enter an integer\");
a = in.nextInt();
System.out.println(\"You entered integer \"+a);
System.out.print(\"\ \");
System.out.print(\"The factors are : \");
for(i=1;i<=a/2;i++)
{
if(a%i==0)
System.out.print(i+\",\");
}
System.out.print(a);
}
}
Output:
Enter an integer
44
You entered integer 44
The factors are : 1,2,4,11,22,44.
Dividend recieved deduction causes a temporary difference between ta.pdf
Dividend recieved deduction causes a temporary difference between taxable and pretax
accounting income. Thus answer will be dividend recieved deduction.
Solution
Dividend recieved deduction causes a temporary difference between taxable and pretax
accounting income. Thus answer will be dividend recieved deduction..
d)y = (x+7.1)^2 x-4This is how you should represent the Fortra.pdf
d)
y = (x+7.1)^2 / x-4
This is how you should represent the Fortran expression..
Solution
d)
y = (x+7.1)^2 / x-4
This is how you should represent the Fortran expression...
Answer 7. Danielle will produce eggs with allele i as her blood grou.pdf
Answer 7. Danielle will produce eggs with allele i as her blood group is O i.e. no antigen is
present in her blood.
Michael can produce sperm with either A or B allele as his blood group is AB allele i.e. both A
and B alleles are present.
Ia
Ib
Ii
Ia Ii
Ib Ii
The possible blood groups of the offspring is Ia Ii (A type) and Ib Ii (B type)
As Michael can contribute either A or B allele, offspring will definitely have A or B allele.
Hence, offspring with O allele (i.e. without A or B allele) is not possible.
Ia
Ib
Ii
Ia Ii
Ib Ii
Solution
Answer 7. Danielle will produce eggs with allele i as her blood group is O i.e. no antigen is
present in her blood.
Michael can produce sperm with either A or B allele as his blood group is AB allele i.e. both A
and B alleles are present.
Ia
Ib
Ii
Ia Ii
Ib Ii
The possible blood groups of the offspring is Ia Ii (A type) and Ib Ii (B type)
As Michael can contribute either A or B allele, offspring will definitely have A or B allele.
Hence, offspring with O allele (i.e. without A or B allele) is not possible.
Ia
Ib
Ii
Ia Ii
Ib Ii.
Answersb. multipath (when two or more signals arrive at the recev.pdf
Answers:
b. multipath (when two or more signals arrive at the recevier it is termes as Multipath)
Solution
Answers:
b. multipath (when two or more signals arrive at the recevier it is termes as Multipath).
A population is a group of individuals that belong to the same speci.pdf
A population is a group of individuals that belong to the same species and confined to a
geographic area where they potentially interbreed. The genetic information carried by individuals
of a population is known as gene pool. At a geographical location, the population may get well
adapted to the environment and are highly homozygous with favorable alleles at high frequency.
But the evidences available reveal that the populations exhibit high degree of heterzygosity. The
diversity of the population may not be detected phenotypically because genetic diversity is often
concealed.
The genetic variation can be detected by artificial selection by which the phenotypic
characteristics are affected. The variations at the genetic levels are mostly estimated by
comparing the nucleotide sequences of the individuals. The amount of genetic variation can be
determined by detectable variation at the protein level and at the nucleotide level. Sometimes the
nucleotide substitutions result in the silent variations in the gene.
Mutation is a sudden heritable change that can occur naturally or it can be induced. The gene
shuffling causes inheritable variations during the production of gametes, by the process of
crossing over during sexual reproduction. Thus sexual reproduction can produce different
phenotypes with various combinations of genes and does not alter the allele frequencies of
particular alleles responsible for specific traits.
The amount of genetic variation can be determined by detectable variation at the protein level
and at the nucleotide level. Sometimes the nucleotide substitutions result in the silent variations
in the gene. Studies on various organisms and on different genes resulted in the enormous
nucleotide diversity. The extensive genetic variability in most populations exhibited at the DNA
level. Almost all genes exhibit the diversity from individual to individual. Especially the alleles
represent the variations that are widely distributed among the individuals of the population.
Genetic diversity in small populations is reduced because of genetic drift, high degree of
inbreeding and reduction in gene flow. These effects are exhibited adversely on a species
survival. For example, inbreeding in a population causes the loss of heterzygosity and promotes
homozygosity. The recessive alleles received by offspring may be rare and lethal and affect the
survival of the offspring. Thus, the low level of genetic diversity in a species extends harmful
effects on the survival of the species and populations cannot get rid of \"bad or lethal genes\" due
to homozygosity.
Solution
A population is a group of individuals that belong to the same species and confined to a
geographic area where they potentially interbreed. The genetic information carried by individuals
of a population is known as gene pool. At a geographical location, the population may get well
adapted to the environment and are highly homozygous with favorable alleles at high frequency.
But .
2. Write the equilibrium constant expression forSolution2. Wr.pdf
2. Write the equilibrium constant expression for:
Solution
2. Write the equilibrium constant expression for:.
Best Digital Marketing Institute In NOIDA
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More from deepakangel
Halfway titration Half of the base is neutralise.pdf
Halfway titration: Half of the base is neutralised with HCl and becomes water plus
its conjugate acid. This weak base, plus its conjugate acid forms a buffer solution with
equimolar of both species. They obey the formula where pH = pKa +log(base/acid). Since they
are equimolar, base/acid =1 and log 1 = 0 pH = pKa Ka = Kw/Kb = 2.27 x 10^-11 pH = -
log(2.27 x 10^-11) = 10.6 (A)
Solution
Halfway titration: Half of the base is neutralised with HCl and becomes water plus
its conjugate acid. This weak base, plus its conjugate acid forms a buffer solution with
equimolar of both species. They obey the formula where pH = pKa +log(base/acid). Since they
are equimolar, base/acid =1 and log 1 = 0 pH = pKa Ka = Kw/Kb = 2.27 x 10^-11 pH = -
log(2.27 x 10^-11) = 10.6 (A).
The HTML canvas element is used to draw graphics on a web page.T.pdf
The HTML element is used to draw graphics on a web page.
The graphic to the left is created with . It shows four elements: a red rectangle, a gradient
rectangle, a multicolor rectangle, and a multicolor text.
In HTML, JavaScript statements are \"instructions\" to be \"executed\" by the web browser.
This statement tells the browser to write \"Hello Dolly.\" inside an HTML element with
id=\"demo\":
Dropdown
Link 1Link 2Link 3
Use any element to open the dropdown menu, e.g. a , or
element.
Use a container element (like
) to create the dropdown menu and add the dropdown links inside it.
Wrap a
element around the button and the
to position the dropdown menu correctly with CSS.
house size color
clouds colorSendClick Button To Popup Form Using JavascriptPopup
The tag defines a clickable button.
Inside a element you can put content, like text or images. This is the difference between this
element and buttons created with the element.
Tip: Always specify the type attribute for a element. Different browsers use different default
types for the element.
// Validating Empty Fieldfunctioncheck_empty()
{
if(document.getElementById(\'name\').value==\"\"||document.getElementById(\'email\').value==
\"\"||document.getElementById(\'msg\').value==\"\")
{
alert(\"Fill All Fields !\");
}
else
{
document.getElementById(\'form\').submit();alert(\"Form Submitted Successfully...\");
}
}
//Function To Display Popupfunctiondiv_show()
{
document.getElementById(\'abc\').style.display=\"block\";
}
//Function to Hide Popupfunctiondiv_hide()
{
document.getElementById(\'abc\').style.display=\"none\";
}
Solution
The HTML element is used to draw graphics on a web page.
The graphic to the left is created with . It shows four elements: a red rectangle, a gradient
rectangle, a multicolor rectangle, and a multicolor text.
In HTML, JavaScript statements are \"instructions\" to be \"executed\" by the web browser.
This statement tells the browser to write \"Hello Dolly.\" inside an HTML element with
id=\"demo\":
Dropdown
Link 1Link 2Link 3
Use any element to open the dropdown menu, e.g. a , or
element.
Use a container element (like
) to create the dropdown menu and add the dropdown links inside it.
Wrap a
element around the button and the
to position the dropdown menu correctly with CSS.
house size color
clouds colorSendClick Button To Popup Form Using JavascriptPopup
The tag defines a clickable button.
Inside a element you can put content, like text or images. This is the difference between this
element and buttons created with the element.
Tip: Always specify the type attribute for a element. Different browsers use different default
types for the element.
// Validating Empty Fieldfunctioncheck_empty()
{
if(document.getElementById(\'name\').value==\"\"||document.getElementById(\'email\').value==
\"\"||document.getElementById(\'msg\').value==\"\")
{
alert(\"Fill All Fields !\");
}
else
{
document.getElementById(\'form\').submit();alert(\"Form Submitted Successfully...\");
}
}
//Functio.
Fluorescein has a OH group and a COOH group. If a.pdf
Fluorescein has a OH group and a COOH group. If a solvent containing NaOH is
added it can deprotonate those group creating a more soluble salt. ex. O-Na+ and COO-Na+
These salts are soluble.
Solution
Fluorescein has a OH group and a COOH group. If a solvent containing NaOH is
added it can deprotonate those group creating a more soluble salt. ex. O-Na+ and COO-Na+
These salts are soluble..
Solution 3Capital = Total Assets – Current Liabilities .pdf
Solution
3
Capital = Total Assets – Current Liabilities
= $159,000 - $48,000
= $111,000
Income before interest and after taxes = EBIT x (1- Tax rate)
= 26000 x (1-.4167)
= $15,165.80
Return on capital(2008) = Income before interest and after taxes/ Capital
= $15,165.80 / $111,000
= 13.66%
Return on capital (2005) = (10000 x (1-.4444)/ (93000 -21000)
= 5555.55/72000
= 7.71%
Improvement = 13.66%-7.71%
= 5.94%.
Q1ROMThe boot loader basically put Hex code into already there .pdf
Q1:
ROM
The boot loader basically put Hex code into already there ROM inside controller, whenever
microcontroller been powered up code from ROM get up into RAM and get executed and hence
job got performed.
Q2:
INCA command is executed only 1 tine
Solution
Q1:
ROM
The boot loader basically put Hex code into already there ROM inside controller, whenever
microcontroller been powered up code from ROM get up into RAM and get executed and hence
job got performed.
Q2:
INCA command is executed only 1 tine.
Raising the temperature changes the solubility of a substance in wat.pdf
Raising the temperature changes the solubility of a substance in water. Solubility increases with
increasing temperature. The rate of solution in water also increases.
Crushing or grinding the solid does not change the solubility but increases rate of solution.
Adding more of the solid also does not change the solubility of that substance but increases rate
of solution
Agitating the solid/solvent mixture does not change the solubility of that substance but increases
rate of solution
Solution
Raising the temperature changes the solubility of a substance in water. Solubility increases with
increasing temperature. The rate of solution in water also increases.
Crushing or grinding the solid does not change the solubility but increases rate of solution.
Adding more of the solid also does not change the solubility of that substance but increases rate
of solution
Agitating the solid/solvent mixture does not change the solubility of that substance but increases
rate of solution.
MAN pages.These pages describes commands functions and there prope.pdf
MAN pages.
These pages describes commands functions and there proper execution.
Pages are arranged in the 9 sections :
Section 1 : Covers the commands that is normally provided by the user in the terminal.
Section 2 to 5 : Programmers interface to the UNIX system is documented.
Section 6: Covers the amusements and games included in UNIX system.
Section 7 : Covers the device drivers of the system.
Section 8: Covers the commands use by administrator to manage the system.
Section 9 : UNIX kernels functions, which programmers uses while writing device drivers.
Solution
MAN pages.
These pages describes commands functions and there proper execution.
Pages are arranged in the 9 sections :
Section 1 : Covers the commands that is normally provided by the user in the terminal.
Section 2 to 5 : Programmers interface to the UNIX system is documented.
Section 6: Covers the amusements and games included in UNIX system.
Section 7 : Covers the device drivers of the system.
Section 8: Covers the commands use by administrator to manage the system.
Section 9 : UNIX kernels functions, which programmers uses while writing device drivers..
If a person has inability to present antigens to Tc cells (cytotoxic.pdf
If a person has inability to present antigens to Tc cells (cytotoxic T cells), then there will be
deficiency of molecules like granzyme, perforin and granulysin. These molecules are required to
kill abnormal cells or cells infected with viruses.
If a person has inability to present antigens to Th1 cells, then there will be deficiency of
molecules like tumour necrosis factor (TNF)-beta, interferon-gamma and interleukin (IL)-2.
These molecules help in phagocytosis of foreign bodies (like bacteria and protozoa) and promote
cell mediated immunity.
If a person has inability to present antigens to Th2 cells, then there will be deficiency of
molecules like IL-4, IL-5, IL-10, and IL-13. These molecules aid in eosinophil activation and
induce antibody production against extracellular parasites.
As is clear from above, the critical role of these molecules shows that a person with these T Cell
deficiencies can not survive properly in normal environment and will be prone to recurrent
infections or illnesses. However, if the person is treated with antiobiotics and other therapeutic
regimes then chances of survival will increase depending on the severity of T Cell deficiency.
Solution
If a person has inability to present antigens to Tc cells (cytotoxic T cells), then there will be
deficiency of molecules like granzyme, perforin and granulysin. These molecules are required to
kill abnormal cells or cells infected with viruses.
If a person has inability to present antigens to Th1 cells, then there will be deficiency of
molecules like tumour necrosis factor (TNF)-beta, interferon-gamma and interleukin (IL)-2.
These molecules help in phagocytosis of foreign bodies (like bacteria and protozoa) and promote
cell mediated immunity.
If a person has inability to present antigens to Th2 cells, then there will be deficiency of
molecules like IL-4, IL-5, IL-10, and IL-13. These molecules aid in eosinophil activation and
induce antibody production against extracellular parasites.
As is clear from above, the critical role of these molecules shows that a person with these T Cell
deficiencies can not survive properly in normal environment and will be prone to recurrent
infections or illnesses. However, if the person is treated with antiobiotics and other therapeutic
regimes then chances of survival will increase depending on the severity of T Cell deficiency..
i didnt get your question its unclear do you mean conjunctive .pdf
i didnt get your question its unclear do you mean
conjunctive normal form (CNF)
Solution
i didnt get your question its unclear do you mean
conjunctive normal form (CNF).
Here is the code with comments for the question. Sample output is al.pdf
Here is the code with comments for the question. Sample output is also attached for differnet
inputs. If it helped, please dont forget to rate the answer.Thank you very much.
import java.io.File;
import java.io.FileNotFoundException;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Scanner;
//Class to read an input file an then process the first line as a b c m n
// f is a relation on set from 1...a to 1...b
// g is a relation on set 1...b to 1..c
// m rows of relation f mappings
// n rows of relation g mappings
// outputs if f is a function, g is funcion, f is onto and g is one-one. Also if f and g are functions,
then print gof() mappings
public class Functions {
//function to load number of rows specified from the file where the domain is 1....domainSize
and codomain is 1, 2, .... codomainsize
// returns a boolean array of size 3. value in 0 th index says where the the relation is a function
or not. return[0] = true means the relation is a function, not a function if its false
//return[1] indicates wheter the relation is one-one . so return[1] is true means the function is
one one, if false not one-one
//return[2] indicates wheter the relation is onto. os return[2] is true means the loaded function
is onto otherwise not onto if false
private static boolean[] analyzeRelation(Scanner scanner,int domainSize,int codomainSize,int
rows,HashMapdomain,HashMap> codomain) throws Exception
{
boolean function=true,onto=true,oneone=true;
boolean analysis[]=new boolean[3]; //stores the return values , isfunction? ,isoneone?
isonto?
//since a singel element in co-domain can be mapped to multiple elements in domain, we
maintain an arraylist to tell which elements in domain it is mapped to .If the number of elements
is 1 for
//each elemeent in codomain, then its one-one
for(int i=1;i<=codomainSize;i++)
codomain.put(i, new ArrayList()); //initialize the arraylist for each of the co-domain
elements
int count=0;
String l,line,nums[];
int idx1,idx2;
int x,y;
while(scanner.hasNext() && count1) //if the co-domain element is mapped to more than 1,
its not oneoone, save the findings
oneone=false;
count++;
}
for(int i=1;i<=domainSize;i++) //now check that every element in the domain has a value in
the map, otherwise its not a function
{
if(domain.get(i)==null) //an element in domain is not mapped. so not function
function=false;
}
//now check the codomain elements, if every element is mapped , then its onto, other its not
for(int i=1;i<=codomainSize;i++)
{
if(codomain.get(i).size()<1) //some element in codomain is not mapped to any element in
domain, so its not onto
onto=false;
}
///store the results of your findints
// 0th index isfunction?
// 1st index isOneone?
//2ind index isonto?
analysis[0]=function;
analysis[1]=oneone;
analysis[2]=onto;
return analysis;
}
public static void main(String[] args) {
int a,b,c,m,n;
HashMap fdomain=new HashMap();
HashMap> fcodomain=new HashMap>();
HashMap gdomain=new HashMap();
HashMap> gcodomain=new HashMap>(.
Embedded systemIt is a computer system with a dedicated function .pdf
Embedded system:
It is a computer system with a dedicated function within a larger mechanical or electrical system,
often with real-time computing constraints. It is embedded as part of a complete device often
including hardware and mechanical parts.They can control many devices in common use today.
Solution
Embedded system:
It is a computer system with a dedicated function within a larger mechanical or electrical system,
often with real-time computing constraints. It is embedded as part of a complete device often
including hardware and mechanical parts.They can control many devices in common use today..
Factors.javaimport java.io.; import java.util.Scanner; class .pdf
Factors.java
import java.io.*;
import java.util.Scanner;
class Factors
{
public static void main(String args[])
{
int a,i;
Scanner in = new Scanner(System.in);
System.out.println(\"Enter an integer\");
a = in.nextInt();
System.out.println(\"You entered integer \"+a);
System.out.print(\"\ \");
System.out.print(\"The factors are : \");
for(i=1;i<=a/2;i++)
{
if(a%i==0)
System.out.print(i+\",\");
}
System.out.print(a);
}
}
Output:
Enter an integer
44
You entered integer 44
The factors are : 1,2,4,11,22,44
Solution
Factors.java
import java.io.*;
import java.util.Scanner;
class Factors
{
public static void main(String args[])
{
int a,i;
Scanner in = new Scanner(System.in);
System.out.println(\"Enter an integer\");
a = in.nextInt();
System.out.println(\"You entered integer \"+a);
System.out.print(\"\ \");
System.out.print(\"The factors are : \");
for(i=1;i<=a/2;i++)
{
if(a%i==0)
System.out.print(i+\",\");
}
System.out.print(a);
}
}
Output:
Enter an integer
44
You entered integer 44
The factors are : 1,2,4,11,22,44.
Dividend recieved deduction causes a temporary difference between ta.pdf
Dividend recieved deduction causes a temporary difference between taxable and pretax
accounting income. Thus answer will be dividend recieved deduction.
Solution
Dividend recieved deduction causes a temporary difference between taxable and pretax
accounting income. Thus answer will be dividend recieved deduction..
d)y = (x+7.1)^2 x-4This is how you should represent the Fortra.pdf
d)
y = (x+7.1)^2 / x-4
This is how you should represent the Fortran expression..
Solution
d)
y = (x+7.1)^2 / x-4
This is how you should represent the Fortran expression...
Answer 7. Danielle will produce eggs with allele i as her blood grou.pdf
Answer 7. Danielle will produce eggs with allele i as her blood group is O i.e. no antigen is
present in her blood.
Michael can produce sperm with either A or B allele as his blood group is AB allele i.e. both A
and B alleles are present.
Ia
Ib
Ii
Ia Ii
Ib Ii
The possible blood groups of the offspring is Ia Ii (A type) and Ib Ii (B type)
As Michael can contribute either A or B allele, offspring will definitely have A or B allele.
Hence, offspring with O allele (i.e. without A or B allele) is not possible.
Ia
Ib
Ii
Ia Ii
Ib Ii
Solution
Answer 7. Danielle will produce eggs with allele i as her blood group is O i.e. no antigen is
present in her blood.
Michael can produce sperm with either A or B allele as his blood group is AB allele i.e. both A
and B alleles are present.
Ia
Ib
Ii
Ia Ii
Ib Ii
The possible blood groups of the offspring is Ia Ii (A type) and Ib Ii (B type)
As Michael can contribute either A or B allele, offspring will definitely have A or B allele.
Hence, offspring with O allele (i.e. without A or B allele) is not possible.
Ia
Ib
Ii
Ia Ii
Ib Ii.
Answersb. multipath (when two or more signals arrive at the recev.pdf
Answers:
b. multipath (when two or more signals arrive at the recevier it is termes as Multipath)
Solution
Answers:
b. multipath (when two or more signals arrive at the recevier it is termes as Multipath).
A population is a group of individuals that belong to the same speci.pdf
A population is a group of individuals that belong to the same species and confined to a
geographic area where they potentially interbreed. The genetic information carried by individuals
of a population is known as gene pool. At a geographical location, the population may get well
adapted to the environment and are highly homozygous with favorable alleles at high frequency.
But the evidences available reveal that the populations exhibit high degree of heterzygosity. The
diversity of the population may not be detected phenotypically because genetic diversity is often
concealed.
The genetic variation can be detected by artificial selection by which the phenotypic
characteristics are affected. The variations at the genetic levels are mostly estimated by
comparing the nucleotide sequences of the individuals. The amount of genetic variation can be
determined by detectable variation at the protein level and at the nucleotide level. Sometimes the
nucleotide substitutions result in the silent variations in the gene.
Mutation is a sudden heritable change that can occur naturally or it can be induced. The gene
shuffling causes inheritable variations during the production of gametes, by the process of
crossing over during sexual reproduction. Thus sexual reproduction can produce different
phenotypes with various combinations of genes and does not alter the allele frequencies of
particular alleles responsible for specific traits.
The amount of genetic variation can be determined by detectable variation at the protein level
and at the nucleotide level. Sometimes the nucleotide substitutions result in the silent variations
in the gene. Studies on various organisms and on different genes resulted in the enormous
nucleotide diversity. The extensive genetic variability in most populations exhibited at the DNA
level. Almost all genes exhibit the diversity from individual to individual. Especially the alleles
represent the variations that are widely distributed among the individuals of the population.
Genetic diversity in small populations is reduced because of genetic drift, high degree of
inbreeding and reduction in gene flow. These effects are exhibited adversely on a species
survival. For example, inbreeding in a population causes the loss of heterzygosity and promotes
homozygosity. The recessive alleles received by offspring may be rare and lethal and affect the
survival of the offspring. Thus, the low level of genetic diversity in a species extends harmful
effects on the survival of the species and populations cannot get rid of \"bad or lethal genes\" due
to homozygosity.
Solution
A population is a group of individuals that belong to the same species and confined to a
geographic area where they potentially interbreed. The genetic information carried by individuals
of a population is known as gene pool. At a geographical location, the population may get well
adapted to the environment and are highly homozygous with favorable alleles at high frequency.
But .
2. Write the equilibrium constant expression forSolution2. Wr.pdf
2. Write the equilibrium constant expression for:
Solution
2. Write the equilibrium constant expression for:.
More from deepakangel (20)
Halfway titration Half of the base is neutralise.pdf
Halfway titration Half of the base is neutralise.pdf
The HTML canvas element is used to draw graphics on a web page.T.pdf
The HTML canvas element is used to draw graphics on a web page.T.pdf
Fluorescein has a OH group and a COOH group. If a.pdf
Fluorescein has a OH group and a COOH group. If a.pdf
Solution 3Capital = Total Assets – Current Liabilities .pdf
Solution 3Capital = Total Assets – Current Liabilities .pdf
Q1ROMThe boot loader basically put Hex code into already there .pdf
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