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When two data sets are pooled together the pooled mean will always be more accurate, between original mean values and more precise. Ans: (e) Solution When two data sets are pooled together the pooled mean will always be more accurate, between original mean values and more precise. Ans: (e).
When two data sets are pooled together the pooled mean will alway.pdf
When two data sets are pooled together the pooled mean will alway.pdf
aparnatiwari291
DELTOID : flexes and medially rotates arm; prime mover of arm abduction; extends and laterally rotates arm. LATISSIMUS DORI : prime mover of arm extension; adducts and medially rotates arm (big muscle). PECTORALIS MAJOR : prime mover of arm flexion, adducts and medially rotates arm ( big muscle). TERES MAJOR : Extends, adducts and medially rotates arm. Solution DELTOID : flexes and medially rotates arm; prime mover of arm abduction; extends and laterally rotates arm. LATISSIMUS DORI : prime mover of arm extension; adducts and medially rotates arm (big muscle). PECTORALIS MAJOR : prime mover of arm flexion, adducts and medially rotates arm ( big muscle). TERES MAJOR : Extends, adducts and medially rotates arm..
DELTOID flexes and medially rotates arm; prime mover of arm abduct.pdf
DELTOID flexes and medially rotates arm; prime mover of arm abduct.pdf
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Event Cash Flow Assets Claims Income Statement OA/IA/FA Cash + Supplies stock + Fixed asset = Note payable + Bank loan short term + Accounts payable + Common Stock + Retained Earnings Revenue - Expense = Net Income a FA 75,000 + + = + + + 75,000 + - = b OA (5,000) + 5,000 + = + + + + - = c IA (10,000) + + 80,000 = 70,000 + + + + - = d OA 27,000 + + = + 27,000 + + + - = e FA (3,500) + + = + + + + (3,500) - = f OA 17,000 + + = + + + + 17,000 - = 17,000 g OA (4,000) + + = + + + + - 4,000 = (4,000) h OA + 1,000 + = + + 1,000 + + - = + + = + + + + - = + + = + + + + - = + + = + + + + - = + + = + + + + - = + + = + + + + - = 96,500 + 6,000 + 80,000 = 70,000 + 27,000 + 1,000 + 75,000 + (3,500) 17,000 - 4,000 = 13,000 Solution Event Cash Flow Assets Claims Income Statement OA/IA/FA Cash + Supplies stock + Fixed asset = Note payable + Bank loan short term + Accounts payable + Common Stock + Retained Earnings Revenue - Expense = Net Income a FA 75,000 + + = + + + 75,000 + - = b OA (5,000) + 5,000 + = + + + + - = c IA (10,000) + + 80,000 = 70,000 + + + + - = d OA 27,000 + + = + 27,000 + + + - = e FA (3,500) + + = + + + + (3,500) - = f OA 17,000 + + = + + + + 17,000 - = 17,000 g OA (4,000) + + = + + + + - 4,000 = (4,000) h OA + 1,000 + = + + 1,000 + + - = + + = + + + + - = + + = + + + + - = + + = + + + + - = + + = + + + + - = + + = + + + + - = 96,500 + 6,000 + 80,000 = 70,000 + 27,000 + 1,000 + 75,000 + (3,500) 17,000 - 4,000 = 13,000.
Event Cash Flow Assets Claims Income Statement OAIAFA Cash.pdf
Event Cash Flow Assets Claims Income Statement OAIAFA Cash.pdf
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We know that Molarity = No . of moles / Volume of solution in L 0.5 M = n / 1.0 L ---> n = 0.5 mol Li2SO4 ---> 2Li+ + SO4 2- So 1 mole of Li2SO4 produces 1 mole of SO4 2- & 2 moles of Li+ So No . of moles of SO4 2- ions present in 0.5 mol of Li2SO4 is 1 * 0.5 = 0.5 mol Solution We know that Molarity = No . of moles / Volume of solution in L 0.5 M = n / 1.0 L ---> n = 0.5 mol Li2SO4 ---> 2Li+ + SO4 2- So 1 mole of Li2SO4 produces 1 mole of SO4 2- & 2 moles of Li+ So No . of moles of SO4 2- ions present in 0.5 mol of Li2SO4 is 1 * 0.5 = 0.5 mol.
We know that Molarity = No . of moles Volume of.pdf
We know that Molarity = No . of moles Volume of.pdf
aparnatiwari291
there are no molecules given..pls repost and post on my profile Solution there are no molecules given..pls repost and post on my profile.
there are no molecules given..pls repost and post.pdf
there are no molecules given..pls repost and post.pdf
aparnatiwari291
The false statement is Oxygen forms binary compounds with nonmetals called acid anhydrides. Solution The false statement is Oxygen forms binary compounds with nonmetals called acid anhydrides..
The false statement is Oxygen forms binary compou.pdf
The false statement is Oxygen forms binary compou.pdf
aparnatiwari291
the concentration is 0.1550 Solution the concentration is 0.1550.
the concentration is 0.1550 .pdf
the concentration is 0.1550 .pdf
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The area is how many Hydrogens are on the graph. You would need a correlation table to determine what groups are on the figure. First, calculate the degrees of saturation. (18- 10)/2 = 4 degrees this could be a ring wth 3 pi bonds maybe benxene Solution The area is how many Hydrogens are on the graph. You would need a correlation table to determine what groups are on the figure. First, calculate the degrees of saturation. (18- 10)/2 = 4 degrees this could be a ring wth 3 pi bonds maybe benxene.
The area is how many Hydrogens are on the graph. .pdf
The area is how many Hydrogens are on the graph. .pdf
aparnatiwari291
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When two data sets are pooled together the pooled mean will always be more accurate, between original mean values and more precise. Ans: (e) Solution When two data sets are pooled together the pooled mean will always be more accurate, between original mean values and more precise. Ans: (e).
When two data sets are pooled together the pooled mean will alway.pdf
When two data sets are pooled together the pooled mean will alway.pdf
aparnatiwari291
DELTOID : flexes and medially rotates arm; prime mover of arm abduction; extends and laterally rotates arm. LATISSIMUS DORI : prime mover of arm extension; adducts and medially rotates arm (big muscle). PECTORALIS MAJOR : prime mover of arm flexion, adducts and medially rotates arm ( big muscle). TERES MAJOR : Extends, adducts and medially rotates arm. Solution DELTOID : flexes and medially rotates arm; prime mover of arm abduction; extends and laterally rotates arm. LATISSIMUS DORI : prime mover of arm extension; adducts and medially rotates arm (big muscle). PECTORALIS MAJOR : prime mover of arm flexion, adducts and medially rotates arm ( big muscle). TERES MAJOR : Extends, adducts and medially rotates arm..
DELTOID flexes and medially rotates arm; prime mover of arm abduct.pdf
DELTOID flexes and medially rotates arm; prime mover of arm abduct.pdf
aparnatiwari291
Event Cash Flow Assets Claims Income Statement OA/IA/FA Cash + Supplies stock + Fixed asset = Note payable + Bank loan short term + Accounts payable + Common Stock + Retained Earnings Revenue - Expense = Net Income a FA 75,000 + + = + + + 75,000 + - = b OA (5,000) + 5,000 + = + + + + - = c IA (10,000) + + 80,000 = 70,000 + + + + - = d OA 27,000 + + = + 27,000 + + + - = e FA (3,500) + + = + + + + (3,500) - = f OA 17,000 + + = + + + + 17,000 - = 17,000 g OA (4,000) + + = + + + + - 4,000 = (4,000) h OA + 1,000 + = + + 1,000 + + - = + + = + + + + - = + + = + + + + - = + + = + + + + - = + + = + + + + - = + + = + + + + - = 96,500 + 6,000 + 80,000 = 70,000 + 27,000 + 1,000 + 75,000 + (3,500) 17,000 - 4,000 = 13,000 Solution Event Cash Flow Assets Claims Income Statement OA/IA/FA Cash + Supplies stock + Fixed asset = Note payable + Bank loan short term + Accounts payable + Common Stock + Retained Earnings Revenue - Expense = Net Income a FA 75,000 + + = + + + 75,000 + - = b OA (5,000) + 5,000 + = + + + + - = c IA (10,000) + + 80,000 = 70,000 + + + + - = d OA 27,000 + + = + 27,000 + + + - = e FA (3,500) + + = + + + + (3,500) - = f OA 17,000 + + = + + + + 17,000 - = 17,000 g OA (4,000) + + = + + + + - 4,000 = (4,000) h OA + 1,000 + = + + 1,000 + + - = + + = + + + + - = + + = + + + + - = + + = + + + + - = + + = + + + + - = + + = + + + + - = 96,500 + 6,000 + 80,000 = 70,000 + 27,000 + 1,000 + 75,000 + (3,500) 17,000 - 4,000 = 13,000.
Event Cash Flow Assets Claims Income Statement OAIAFA Cash.pdf
Event Cash Flow Assets Claims Income Statement OAIAFA Cash.pdf
aparnatiwari291
We know that Molarity = No . of moles / Volume of solution in L 0.5 M = n / 1.0 L ---> n = 0.5 mol Li2SO4 ---> 2Li+ + SO4 2- So 1 mole of Li2SO4 produces 1 mole of SO4 2- & 2 moles of Li+ So No . of moles of SO4 2- ions present in 0.5 mol of Li2SO4 is 1 * 0.5 = 0.5 mol Solution We know that Molarity = No . of moles / Volume of solution in L 0.5 M = n / 1.0 L ---> n = 0.5 mol Li2SO4 ---> 2Li+ + SO4 2- So 1 mole of Li2SO4 produces 1 mole of SO4 2- & 2 moles of Li+ So No . of moles of SO4 2- ions present in 0.5 mol of Li2SO4 is 1 * 0.5 = 0.5 mol.
We know that Molarity = No . of moles Volume of.pdf
We know that Molarity = No . of moles Volume of.pdf
aparnatiwari291
there are no molecules given..pls repost and post on my profile Solution there are no molecules given..pls repost and post on my profile.
there are no molecules given..pls repost and post.pdf
there are no molecules given..pls repost and post.pdf
aparnatiwari291
The false statement is Oxygen forms binary compounds with nonmetals called acid anhydrides. Solution The false statement is Oxygen forms binary compounds with nonmetals called acid anhydrides..
The false statement is Oxygen forms binary compou.pdf
The false statement is Oxygen forms binary compou.pdf
aparnatiwari291
the concentration is 0.1550 Solution the concentration is 0.1550.
the concentration is 0.1550 .pdf
the concentration is 0.1550 .pdf
aparnatiwari291
The area is how many Hydrogens are on the graph. You would need a correlation table to determine what groups are on the figure. First, calculate the degrees of saturation. (18- 10)/2 = 4 degrees this could be a ring wth 3 pi bonds maybe benxene Solution The area is how many Hydrogens are on the graph. You would need a correlation table to determine what groups are on the figure. First, calculate the degrees of saturation. (18- 10)/2 = 4 degrees this could be a ring wth 3 pi bonds maybe benxene.
The area is how many Hydrogens are on the graph. .pdf
The area is how many Hydrogens are on the graph. .pdf
aparnatiwari291
SO4{2-} (aq) + Sn{2+} (aq) + 4 H{+} (aq)============> H2SO3(aq) + Sn{4+} (aq)+ H2O(l) Solution SO4{2-} (aq) + Sn{2+} (aq) + 4 H{+} (aq)============> H2SO3(aq) + Sn{4+} (aq)+ H2O(l).
SO4{2-} (aq) + Sn{2+} (aq) + 4 H{+} (aq)=========.pdf
SO4{2-} (aq) + Sn{2+} (aq) + 4 H{+} (aq)=========.pdf
aparnatiwari291
oxygen double bond oxygen Solution oxygen double bond oxygen.
oxygen double bond oxygen .pdf
oxygen double bond oxygen .pdf
aparnatiwari291
Once HCl is added in solution, the ions would dissociate into their component parts (H+ and Cl-). As per the second equation, adding more H+ would initially increase the quantity of products, thusly shifting the reaction towards the reactants by LeChatlier\'s Principle. If NaOH was added, the dissociation would create OH- anions that would soak up H+ cations. By removing products, the equlibrium is shifted towards the products by LeChatlier\'s Principle. Best of wishes! (: Solution Once HCl is added in solution, the ions would dissociate into their component parts (H+ and Cl-). As per the second equation, adding more H+ would initially increase the quantity of products, thusly shifting the reaction towards the reactants by LeChatlier\'s Principle. If NaOH was added, the dissociation would create OH- anions that would soak up H+ cations. By removing products, the equlibrium is shifted towards the products by LeChatlier\'s Principle. Best of wishes! (:.
Once HCl is added in solution, the ions would dis.pdf
Once HCl is added in solution, the ions would dis.pdf
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its e1 OMe attacks at 3-position by transfer of anion to stable form Solution its e1 OMe attacks at 3-position by transfer of anion to stable form.
its e1 OMe attacks at 3-position by transfer of a.pdf
its e1 OMe attacks at 3-position by transfer of a.pdf
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H and F are identical, you just need to rotate H until it matches F, I is also identical to G. H and F are enantiomers of I and G, this also means that they are Chiral (non super imposable on their mirror image) to each other. Solution H and F are identical, you just need to rotate H until it matches F, I is also identical to G. H and F are enantiomers of I and G, this also means that they are Chiral (non super imposable on their mirror image) to each other..
H and F are identical, you just need to rotate H .pdf
H and F are identical, you just need to rotate H .pdf
aparnatiwari291
From weakest to strongest: A, B, C Butanal is an aldehyde which means it has the C=O so it cannot hydrogen bond, only dipole-dipole interactions. However, butanol can hydrogen-bond because of the OH group so it would have the strongest intermolecular force. Pentane would have weakest because it has no electronegative atoms. Solution From weakest to strongest: A, B, C Butanal is an aldehyde which means it has the C=O so it cannot hydrogen bond, only dipole-dipole interactions. However, butanol can hydrogen-bond because of the OH group so it would have the strongest intermolecular force. Pentane would have weakest because it has no electronegative atoms..
From weakest to strongest A, B, C Butanal is an .pdf
From weakest to strongest A, B, C Butanal is an .pdf
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Effusion is inversely proportional to the mass. Hence, Methane > HF > Propane > Hydrogen sulfide Solution Effusion is inversely proportional to the mass. Hence, Methane > HF > Propane > Hydrogen sulfide.
Effusion is inversely proportional to the mass. .pdf
Effusion is inversely proportional to the mass. .pdf
aparnatiwari291
Benzoin and methanol both have an alcohol group, but benzoin has a ketone group and phenyl groups within it also. The interesting part about benzoin is that it is a conjugated pi system, meaning that electrons from the phenyl groups can shift around the double bonds throughout the whole molecule without effecting anything; this also impacts the ketone group. This is also known as resonance, you can see an example here http://alevelchem.com/drop/peptide_angles.gif Because of this effect, the whole rest of the benzoin moiety (besides the alcohol group) can be considered one giant add-on (Let\'s call it R) If methanol looks like this... H3C-OH Then our benzoin because of its delocalized resonant electrons will, in comparison, just look like... R-OH When placed side by side this is how both molecules will act with each other. Therefore the polarities of both molecules are similar because electronically both molecules function the same way (Methanol with a CH3, and Benzoin with a conjugated R group). Solution Benzoin and methanol both have an alcohol group, but benzoin has a ketone group and phenyl groups within it also. The interesting part about benzoin is that it is a conjugated pi system, meaning that electrons from the phenyl groups can shift around the double bonds throughout the whole molecule without effecting anything; this also impacts the ketone group. This is also known as resonance, you can see an example here http://alevelchem.com/drop/peptide_angles.gif Because of this effect, the whole rest of the benzoin moiety (besides the alcohol group) can be considered one giant add-on (Let\'s call it R) If methanol looks like this... H3C-OH Then our benzoin because of its delocalized resonant electrons will, in comparison, just look like... R-OH When placed side by side this is how both molecules will act with each other. Therefore the polarities of both molecules are similar because electronically both molecules function the same way (Methanol with a CH3, and Benzoin with a conjugated R group)..
Benzoin and methanol both have an alcohol group, .pdf
Benzoin and methanol both have an alcohol group, .pdf
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When acid and base react together they form salt and water.This reaction is known as Neutralization Reaction. ForExample HCl+NaOH ---------> NaCl + H2O HCl is acid while NaOH is a base, when they react with eachother they form NaCl, that is a salt (table salt) These reactions are also reversible reactions. Solution When acid and base react together they form salt and water.This reaction is known as Neutralization Reaction. ForExample HCl+NaOH ---------> NaCl + H2O HCl is acid while NaOH is a base, when they react with eachother they form NaCl, that is a salt (table salt) These reactions are also reversible reactions..
When acid and base react together they form salt and water.This reac.pdf
When acid and base react together they form salt and water.This reac.pdf
aparnatiwari291
There are 37 classes in java.lang package as per Java SE7 The java.lang package contains classes that are fundamental to the design of the Java language. 1) Object, Class, ClassLoader, System and Compiler classes 2) Wrapper classes:- Boolean, Character, Byte, Short, Integer, Long, Float, and Double 3) Math, StrictMath classes 4) String, StringBuffer and StringBuilder classes 5) Process, ProcessBuilder, Runtime, RuntimePermission classes 6) Enum>, Throwable, Void classes ---------------------------------------------- 1) Object, Class, ClassLoader, System and Compiler classes ---------------------------------------------- Object: The Object class is the root of the class hierarchy. Every class has Object as a superclass. All objects, including arrays, implement the methods of this class. Class: The Class class instance represent classes and interfaces in a running Java application.It has no public constructor. ClassLoader: The ClassLoader class is an object that is responsible for loading classes. This class is an abstract class. It may be used by security managers to indicate security domains. System: The System class contains several useful class fields and methods. It cannot be instantiated. Some of the facilities provided by System class are standard input, output, error output streams and access to externally defined properties and environment variables. This class has a utility method for quickly copying a portion of an array. Also it has methods for loading files and libraries Compiler: The Compiler class is provided to support Java-to-native-code compilers and related services. By design, it serves as a placeholder for a JIT compiler implementation. SecurityManager: The SecurityManager class allows applications to implement a security policy. It allows an application to determine, before performing a possibly unsafe or sensitive operation, what the operation is and whether it is being attempted in a security context that allows the operation to be performed. The application can allow or disallow the operation. ---------------------------------------------- 2) Wrapper classes:- Boolean, Character, Byte, Short, Integer, Long, Float, and Double ---------------------------------------------- Frequently it is necessary to represent a value of primitive type as if it were an object. The wrapper classes serve this purpose. An object of type Double, for example, contains a field whose type is double, representing that value in such a way that a reference to it can be stored in a variable of reference type. These classes also provide a number of methods for converting among primitive values, as well as supporting such standard methods as equals and hashCode. Other classes in support to wrapper classes: ---------------------------------------------- Number: The Number class is the superclass of classes BigDecimal, BigInteger, Byte, Double, Float, Integer, Long, and Short.The Subclasses of Number must provide methods to convert the represented num.
There are 37 classes in java.lang package as per Java SE7 The java.pdf
There are 37 classes in java.lang package as per Java SE7 The java.pdf
aparnatiwari291
The solution is as below: EmployeeDemo.java import java.util.Scanner; public class EmployeeDemo { static Scanner scan = new Scanner(System.in); public static void main(String[] args) { Company clerk = new Company( ); Employee employee = new Employee();// one student int count,dependent; String fname,lname; float hourly_rate,no_of_hours,l_tax,f_tax,s_tax; System.out.println(\"Enter number of employees:\"); int numberOfEmployees = scan.nextInt(); for (count = 0; count < numberOfEmployees; count++) { System.out.println(\"Enter data for employee number \" + count); System.out.println(\"First Name \"); fname = scan.next(); System.out.println(\"Last Name \"); lname = scan.next(); System.out.println(\"Number of dependents \"); dependent = scan.nextInt(); System.out.println(\"Hourly rate \"); hourly_rate = scan.nextFloat(); System.out.println(\"Number of hours worked \"); no_of_hours = scan.nextFloat(); System.out.println(\"Local tax withheld to date \"); l_tax = scan.nextFloat(); System.out.println(\"Federal tax withheld to date \"); f_tax = scan.nextFloat(); System.out.println(\"State tax withheld to date \"); s_tax = scan.nextFloat(); employee.readInput(fname,lname,dependent,hourly_rate,no_of_hours,l_tax,f_tax,s_tax); employee.calculateData(count); employee.writeOutput(count); clerk.colectDataForCompanyReport(employee,count); } clerk.printDataForCompanyReport(); } } Employee.java import java.util.ArrayList; public class Employee { private ArrayList fname,lname; private ArrayList dependent; private ArrayList hourly_rate,no_of_hours,l_tax,f_tax,s_tax,g_wages,cf_tax,cl_tax,cs_tax,t_tax,ct_tax,net_pay; public Employee(){ fname = new ArrayList(); lname = new ArrayList(); dependent = new ArrayList(); hourly_rate = new ArrayList(); no_of_hours = new ArrayList(); l_tax = new ArrayList(); f_tax = new ArrayList(); s_tax = new ArrayList(); g_wages = new ArrayList(); cl_tax = new ArrayList(); cf_tax = new ArrayList(); cs_tax = new ArrayList(); t_tax = new ArrayList(); ct_tax = new ArrayList(); net_pay = new ArrayList(); } public ArrayList getFname() { return fname; } public void setFname(ArrayList fname) { this.fname = fname; } public ArrayList getLname() { return lname; } public void setLname(ArrayList lname) { this.lname = lname; } public ArrayList getDependent() { return dependent; } public void setDependent(ArrayList dependent) { this.dependent = dependent; } public ArrayList getHourly_rate() { return hourly_rate; } public void setHourly_rate(ArrayList hourly_rate) { this.hourly_rate = hourly_rate; } public ArrayList getNo_of_hours() { return no_of_hours; } public void setNo_of_hours(ArrayList no_of_hours) { this.no_of_hours = no_of_hours; } public ArrayList getL_tax() { return l_tax; } public void setL_tax(ArrayList l_tax) { this.l_tax = l_tax; } public ArrayList getF_tax() { return f_tax; } public void setF_tax(ArrayList f_tax) { this.f_tax = f_tax; } public ArrayList getS_tax() { return s_tax; } public void setS_tax(ArrayList s_tax) { this.s_tax = s_ta.
The solution is as belowEmployeeDemo.javaimport java.util.Scann.pdf
The solution is as belowEmployeeDemo.javaimport java.util.Scann.pdf
aparnatiwari291
The precations to be given to the workers are: Solution The precations to be given to the workers are:.
The precations to be given to the workers areSolutionThe prec.pdf
The precations to be given to the workers areSolutionThe prec.pdf
aparnatiwari291
At 1700 cm-1 a strong signal in an IR spectum is the result of a ketone (C=O). (It can shift if the compound is conjugated). Solution At 1700 cm-1 a strong signal in an IR spectum is the result of a ketone (C=O). (It can shift if the compound is conjugated)..
At 1700 cm-1 a strong signal in an IR spectum is .pdf
At 1700 cm-1 a strong signal in an IR spectum is .pdf
aparnatiwari291
Result: best_fit_for_X1 = DistName: \'tlocationscale\' NLogL: 2.1206e+03 BIC: 4.2620e+03 AIC: 4.2472e+03 AICc: 4.2473e+03 ParamNames: {\'mu\' \'sigma\' \'nu\'} ParamDescription: {\'location\' \'scale\' \'degrees of freedom\'} Params: [1.0071 1.9637 37.4643] Paramci: [2x3 double] ParamCov: [3x3 double] Support: [1x1 struct] best_fit_for_X2 = DistName: \'generalized extreme value\' NLogL: 2.4968e+03 BIC: 5.0142e+03 AIC: 4.9995e+03 AICc: 4.9995e+03 ParamNames: {\'k\' \'sigma\' \'mu\'} ParamDescription: {\'shape\' \'scale\' \'location\'} Params: [-0.2802 2.9571 1.0858] Paramci: [2x3 double] ParamCov: [3x3 double] Support: [1x1 struct] correlation_between_X1_X2 = 0.0084 (hence X1 and X2 are independent) meanY = 3.1394 varY = 12.8269 meanZ = 4.1492 varZ = 25.1576 correlation_between_Y_Z = 0.9438 (Hence Y and Z are dependent) Code: function main load X.mat [D PD]=allfitdist(X1,\'NLogL\'); [D2 PD2]=allfitdist(X2,\'NLogL\'); %best fit using maximum likelihood best_fit_for_X1=D(1) best_fit_for_X2=D2(1) correlation_between_X1_X2=correlation(X1,X2) Y=X1+X2; Z=2*X1+X2; meanY=mean(Y) varY=var(Y) meanZ=mean(Z) varZ=var(Z) correlation_between_Y_Z=correlation(Y,Z) end function[r]=correlation(x,y) avx = mean(x); ex2 = sum((x - avx).^2); %Wasteful multiple passes are a pain, avy = mean(y); ey2 = sum((y - avy).^2); %But explicit loops exy = sum((x - avx).*(y - avy)); %Are interpreted slowly. r = exy/sqrt(ex2*ey2); %Division by N cancels; ignoring the N - 1 ritual. end function [D PD] = allfitdist(data,sortby,varargin) %ALLFITDIST Fit all valid parametric probability distributions to data. % [D PD] = ALLFITDIST(DATA) fits all valid parametric probability % distributions to the data in vector DATA, and returns a struct D of % fitted distributions and parameters and a struct of objects PD % representing the fitted distributions. PD is an object in a class % derived from the ProbDist class. % % [...] = ALLFITDIST(DATA,SORTBY) returns the struct of valid distributions % sorted by the parameter SORTBY % NLogL - Negative of the log likelihood % BIC - Bayesian information criterion (default) % AIC - Akaike information criterion % AICc - AIC with a correction for finite sample sizes % % [...] = ALLFITDIST(...,\'DISCRETE\') specifies it is a discrete % distribution and does not attempt to fit a continuous distribution % to the data % % [...] = ALLFITDIST(...,\'PDF\') or (...,\'CDF\') plots either the PDF or CDF % of a subset of the fitted distribution. The distributions are plotted in % order of fit, according to SORTBY. % % List of distributions it will try to fit % Continuous (default) % Beta % Birnbaum-Saunders % Exponential % Extreme value % Gamma % Generalized extreme value % Generalized Pareto % Inverse Gaussian % Logistic % Log-logistic % Lognormal % Nakagami % Normal % Rayleigh % Rician % t location-scale % Weibull % % Discrete (\'DISCRETE\') % Binomial % Negative binomial % Poisson % % Optional inputs: % [...] = ALLFITDIST(...,\'n\',N,...) % For the \'binomial\' .
Resultbest_fit_for_X1 =DistName tlocationscaleNLogL 2.1.pdf
Resultbest_fit_for_X1 =DistName tlocationscaleNLogL 2.1.pdf
aparnatiwari291
Please rate the solution give thumbs if this is helpful. Impact of data quality on business performance: Functions of the database technology: Differences between centralized and distributed database architecture: The role of a master reference file: Solution Please rate the solution give thumbs if this is helpful. Impact of data quality on business performance: Functions of the database technology: Differences between centralized and distributed database architecture: The role of a master reference file:.
Please rate the solution give thumbs if this is helpful.Impact of .pdf
Please rate the solution give thumbs if this is helpful.Impact of .pdf
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Please find the answers and explanations below: Part 1 (True/False) 1. Replication and transcription proceed in 5\' to 3\' direction: TRUE (Both DNA replication and transcription of mRNA proceed in 5\' to 3\' direcition. This is because the firstly, the DNA polymerase retains only 5\' to 3\' directionality and thus, replication begins only in this direction. .Secondarily, there must be sequence similarity between the DNA and the RNA transcribed from it thus the orientation of transcription is necessarily 5\' to 3\' in nature) 2. All cells have telomerase activity : FALSE (Telomerase or terminal transferase is an enzyme which is required for maintenance of terminal positions of chromosomes by self-replication and prevents shortening of chromosomes. It also prevents sticking of chromosomes together and thus clumping of chromosomes. Telomerase acitivty is absent in majority of somatic cells in the body.) 3. Only mRNA is transcribed: TRUE (Only the mRNA which carries the necessary information for translation is transcribed from the DNA template. The tRNA is utilized to transfer the transcribed information and rRNA is utilized to generate the polypeptide on the ribosomal surface during translation) 4. Promoter regions are upstream from the gene: TRUE (A gene promoter is the region of DNA located upstream the gene to be transcribed and carries the consensus sequence which could be recognized by RNA polymerase for its binding and carrying on transcription) 5. A single tRNA can carry several different amino acids: FALSE (The tRNA is specific for only one kind of amino acid which it could carry. It is a highly stringent and specific nature of this biomolecule which determines the specificness for the amino acid which will be synthesized from the codon) Part 2 Please find the definitions below: 1. Semi-conservative replication: The mode of replication of double stranded DNA in which a daughter DNA molecules carries a strand from parent DNA and the complementary strand is synthesized de novo is called semi-conservative replication. 2. Okazaki fragments: Okazaki fragments are short stretches of DNA synthesized discountinuously on the lagging strand on the open frame of the parent DNA. These stretches of DNA are finally conjoined together to form a continous strand of newly synthesized DNA. 3. Consensus sequence: Consensus sequences are those genetically conserved sequences of DNA which remain similar (upto 99%) within organisms of same or different species and are actively engaged in crucial functions such as promoter region binding, transcription sites for transcription factors, enzymatic genes etc. Any change or deterioration in these consesus sequence might be highly deletrious for the cells. 4. Codon: A codon can be defined as a triplet set of nucleotides which come in a specific order and encode for a specific type of amino acid upon translation. For example, the codon AAA encodes for amino acid lysine. Exceptionally, codons UAA, UAG and UGA do not .
Please find the answers and explanations belowPart 1 (TrueFalse).pdf
Please find the answers and explanations belowPart 1 (TrueFalse).pdf
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Ionic product of water Kw = [H+][OH-] = 10-14 [OH-] = 10-14/2.7 x 10-6 = 3.7 x 10-9 M Solution Ionic product of water Kw = [H+][OH-] = 10-14 [OH-] = 10-14/2.7 x 10-6 = 3.7 x 10-9 M.
Ionic product of water Kw = [H+][OH-] = 10-14[OH-] = 10-142.7 x 1.pdf
Ionic product of water Kw = [H+][OH-] = 10-14[OH-] = 10-142.7 x 1.pdf
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Instruction set architecture of a machine fill the semantic gap between the user And the machines. And serve as the early point for the design of a new machine or Modification of existing ones. ISA design must describe how memory addresses are interpret and specified in the Instructions. Most machines are byte addressed. That is a 8 bits is the smallest unique address. There are two different conventions for ordering bytes within a word one is Little Endian and the other Big Endian Example: Big Endian Little Endian Address value Address value 1000 90 1000 CD 1001 AB 1001 12 1002 12 1002 AB 1003 CD 1003 90 Physical register file is an array of computer registers in a CPU more difficult CPUs use register so that the mapping of which physical entry supplies a particular architectural register change dynamically throughout execution Example: Physical register Id LastUseItag --- --- (p-1) 145 Solution Instruction set architecture of a machine fill the semantic gap between the user And the machines. And serve as the early point for the design of a new machine or Modification of existing ones. ISA design must describe how memory addresses are interpret and specified in the Instructions. Most machines are byte addressed. That is a 8 bits is the smallest unique address. There are two different conventions for ordering bytes within a word one is Little Endian and the other Big Endian Example: Big Endian Little Endian Address value Address value 1000 90 1000 CD 1001 AB 1001 12 1002 12 1002 AB 1003 CD 1003 90 Physical register file is an array of computer registers in a CPU more difficult CPUs use register so that the mapping of which physical entry supplies a particular architectural register change dynamically throughout execution Example: Physical register Id LastUseItag --- --- (p-1) 145.
Instruction set architecture of a machine fill the semantic gap betw.pdf
Instruction set architecture of a machine fill the semantic gap betw.pdf
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all real numbers Solution all real numbers.
all real numbers .pdf
all real numbers .pdf
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In the given text the following phrases reflect scientific approach of studying Homo floresiensis. Comparing height with the modern humans: would stand waist high. Studying the features of skull: shape and thickness of the bones were human like. Size of the brain: her brain was the size of a chimpanzee’s. All these reflect a scientific approach where the species is compared with modern humans and conclusions drawn from the observations. Sometimes science cannot answer all questions like the existence of supernatural powers and beings. Moreover evolution denies the role of God in the universe. One theory beyond science is that life on earth is created by gods. Science does not explain the existence of gods or if super natural entities intervene in human affairs. This explanation is beyond the nature and so beyond the scope of what can be studied by science. This theory of origin of human beings is a belief. Science is an activity that seeks to explore the natural world using well-established, clearly- delineated methods. Scientific method is based on the fact of asking a question, formulating a hypothesis, performing experiments, collecting data and drawing conclusions. Studying human evolution in a scientific way enables in better understanding of our origin and evolution. Understanding the process by which Homo sapiens came into existence is crucial for understanding the kind of creature we are and this is possible by studying scientifically. Solution In the given text the following phrases reflect scientific approach of studying Homo floresiensis. Comparing height with the modern humans: would stand waist high. Studying the features of skull: shape and thickness of the bones were human like. Size of the brain: her brain was the size of a chimpanzee’s. All these reflect a scientific approach where the species is compared with modern humans and conclusions drawn from the observations. Sometimes science cannot answer all questions like the existence of supernatural powers and beings. Moreover evolution denies the role of God in the universe. One theory beyond science is that life on earth is created by gods. Science does not explain the existence of gods or if super natural entities intervene in human affairs. This explanation is beyond the nature and so beyond the scope of what can be studied by science. This theory of origin of human beings is a belief. Science is an activity that seeks to explore the natural world using well-established, clearly- delineated methods. Scientific method is based on the fact of asking a question, formulating a hypothesis, performing experiments, collecting data and drawing conclusions. Studying human evolution in a scientific way enables in better understanding of our origin and evolution. Understanding the process by which Homo sapiens came into existence is crucial for understanding the kind of creature we are and this is possible by studying scientifically..
In the given text the following phrases reflect scientific approach .pdf
In the given text the following phrases reflect scientific approach .pdf
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SO4{2-} (aq) + Sn{2+} (aq) + 4 H{+} (aq)============> H2SO3(aq) + Sn{4+} (aq)+ H2O(l) Solution SO4{2-} (aq) + Sn{2+} (aq) + 4 H{+} (aq)============> H2SO3(aq) + Sn{4+} (aq)+ H2O(l).
SO4{2-} (aq) + Sn{2+} (aq) + 4 H{+} (aq)=========.pdf
SO4{2-} (aq) + Sn{2+} (aq) + 4 H{+} (aq)=========.pdf
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oxygen double bond oxygen Solution oxygen double bond oxygen.
oxygen double bond oxygen .pdf
oxygen double bond oxygen .pdf
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Once HCl is added in solution, the ions would dissociate into their component parts (H+ and Cl-). As per the second equation, adding more H+ would initially increase the quantity of products, thusly shifting the reaction towards the reactants by LeChatlier\'s Principle. If NaOH was added, the dissociation would create OH- anions that would soak up H+ cations. By removing products, the equlibrium is shifted towards the products by LeChatlier\'s Principle. Best of wishes! (: Solution Once HCl is added in solution, the ions would dissociate into their component parts (H+ and Cl-). As per the second equation, adding more H+ would initially increase the quantity of products, thusly shifting the reaction towards the reactants by LeChatlier\'s Principle. If NaOH was added, the dissociation would create OH- anions that would soak up H+ cations. By removing products, the equlibrium is shifted towards the products by LeChatlier\'s Principle. Best of wishes! (:.
Once HCl is added in solution, the ions would dis.pdf
Once HCl is added in solution, the ions would dis.pdf
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its e1 OMe attacks at 3-position by transfer of anion to stable form Solution its e1 OMe attacks at 3-position by transfer of anion to stable form.
its e1 OMe attacks at 3-position by transfer of a.pdf
its e1 OMe attacks at 3-position by transfer of a.pdf
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H and F are identical, you just need to rotate H until it matches F, I is also identical to G. H and F are enantiomers of I and G, this also means that they are Chiral (non super imposable on their mirror image) to each other. Solution H and F are identical, you just need to rotate H until it matches F, I is also identical to G. H and F are enantiomers of I and G, this also means that they are Chiral (non super imposable on their mirror image) to each other..
H and F are identical, you just need to rotate H .pdf
H and F are identical, you just need to rotate H .pdf
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From weakest to strongest: A, B, C Butanal is an aldehyde which means it has the C=O so it cannot hydrogen bond, only dipole-dipole interactions. However, butanol can hydrogen-bond because of the OH group so it would have the strongest intermolecular force. Pentane would have weakest because it has no electronegative atoms. Solution From weakest to strongest: A, B, C Butanal is an aldehyde which means it has the C=O so it cannot hydrogen bond, only dipole-dipole interactions. However, butanol can hydrogen-bond because of the OH group so it would have the strongest intermolecular force. Pentane would have weakest because it has no electronegative atoms..
From weakest to strongest A, B, C Butanal is an .pdf
From weakest to strongest A, B, C Butanal is an .pdf
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Effusion is inversely proportional to the mass. Hence, Methane > HF > Propane > Hydrogen sulfide Solution Effusion is inversely proportional to the mass. Hence, Methane > HF > Propane > Hydrogen sulfide.
Effusion is inversely proportional to the mass. .pdf
Effusion is inversely proportional to the mass. .pdf
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Benzoin and methanol both have an alcohol group, but benzoin has a ketone group and phenyl groups within it also. The interesting part about benzoin is that it is a conjugated pi system, meaning that electrons from the phenyl groups can shift around the double bonds throughout the whole molecule without effecting anything; this also impacts the ketone group. This is also known as resonance, you can see an example here http://alevelchem.com/drop/peptide_angles.gif Because of this effect, the whole rest of the benzoin moiety (besides the alcohol group) can be considered one giant add-on (Let\'s call it R) If methanol looks like this... H3C-OH Then our benzoin because of its delocalized resonant electrons will, in comparison, just look like... R-OH When placed side by side this is how both molecules will act with each other. Therefore the polarities of both molecules are similar because electronically both molecules function the same way (Methanol with a CH3, and Benzoin with a conjugated R group). Solution Benzoin and methanol both have an alcohol group, but benzoin has a ketone group and phenyl groups within it also. The interesting part about benzoin is that it is a conjugated pi system, meaning that electrons from the phenyl groups can shift around the double bonds throughout the whole molecule without effecting anything; this also impacts the ketone group. This is also known as resonance, you can see an example here http://alevelchem.com/drop/peptide_angles.gif Because of this effect, the whole rest of the benzoin moiety (besides the alcohol group) can be considered one giant add-on (Let\'s call it R) If methanol looks like this... H3C-OH Then our benzoin because of its delocalized resonant electrons will, in comparison, just look like... R-OH When placed side by side this is how both molecules will act with each other. Therefore the polarities of both molecules are similar because electronically both molecules function the same way (Methanol with a CH3, and Benzoin with a conjugated R group)..
Benzoin and methanol both have an alcohol group, .pdf
Benzoin and methanol both have an alcohol group, .pdf
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When acid and base react together they form salt and water.This reaction is known as Neutralization Reaction. ForExample HCl+NaOH ---------> NaCl + H2O HCl is acid while NaOH is a base, when they react with eachother they form NaCl, that is a salt (table salt) These reactions are also reversible reactions. Solution When acid and base react together they form salt and water.This reaction is known as Neutralization Reaction. ForExample HCl+NaOH ---------> NaCl + H2O HCl is acid while NaOH is a base, when they react with eachother they form NaCl, that is a salt (table salt) These reactions are also reversible reactions..
When acid and base react together they form salt and water.This reac.pdf
When acid and base react together they form salt and water.This reac.pdf
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There are 37 classes in java.lang package as per Java SE7 The java.lang package contains classes that are fundamental to the design of the Java language. 1) Object, Class, ClassLoader, System and Compiler classes 2) Wrapper classes:- Boolean, Character, Byte, Short, Integer, Long, Float, and Double 3) Math, StrictMath classes 4) String, StringBuffer and StringBuilder classes 5) Process, ProcessBuilder, Runtime, RuntimePermission classes 6) Enum>, Throwable, Void classes ---------------------------------------------- 1) Object, Class, ClassLoader, System and Compiler classes ---------------------------------------------- Object: The Object class is the root of the class hierarchy. Every class has Object as a superclass. All objects, including arrays, implement the methods of this class. Class: The Class class instance represent classes and interfaces in a running Java application.It has no public constructor. ClassLoader: The ClassLoader class is an object that is responsible for loading classes. This class is an abstract class. It may be used by security managers to indicate security domains. System: The System class contains several useful class fields and methods. It cannot be instantiated. Some of the facilities provided by System class are standard input, output, error output streams and access to externally defined properties and environment variables. This class has a utility method for quickly copying a portion of an array. Also it has methods for loading files and libraries Compiler: The Compiler class is provided to support Java-to-native-code compilers and related services. By design, it serves as a placeholder for a JIT compiler implementation. SecurityManager: The SecurityManager class allows applications to implement a security policy. It allows an application to determine, before performing a possibly unsafe or sensitive operation, what the operation is and whether it is being attempted in a security context that allows the operation to be performed. The application can allow or disallow the operation. ---------------------------------------------- 2) Wrapper classes:- Boolean, Character, Byte, Short, Integer, Long, Float, and Double ---------------------------------------------- Frequently it is necessary to represent a value of primitive type as if it were an object. The wrapper classes serve this purpose. An object of type Double, for example, contains a field whose type is double, representing that value in such a way that a reference to it can be stored in a variable of reference type. These classes also provide a number of methods for converting among primitive values, as well as supporting such standard methods as equals and hashCode. Other classes in support to wrapper classes: ---------------------------------------------- Number: The Number class is the superclass of classes BigDecimal, BigInteger, Byte, Double, Float, Integer, Long, and Short.The Subclasses of Number must provide methods to convert the represented num.
There are 37 classes in java.lang package as per Java SE7 The java.pdf
There are 37 classes in java.lang package as per Java SE7 The java.pdf
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The solution is as below: EmployeeDemo.java import java.util.Scanner; public class EmployeeDemo { static Scanner scan = new Scanner(System.in); public static void main(String[] args) { Company clerk = new Company( ); Employee employee = new Employee();// one student int count,dependent; String fname,lname; float hourly_rate,no_of_hours,l_tax,f_tax,s_tax; System.out.println(\"Enter number of employees:\"); int numberOfEmployees = scan.nextInt(); for (count = 0; count < numberOfEmployees; count++) { System.out.println(\"Enter data for employee number \" + count); System.out.println(\"First Name \"); fname = scan.next(); System.out.println(\"Last Name \"); lname = scan.next(); System.out.println(\"Number of dependents \"); dependent = scan.nextInt(); System.out.println(\"Hourly rate \"); hourly_rate = scan.nextFloat(); System.out.println(\"Number of hours worked \"); no_of_hours = scan.nextFloat(); System.out.println(\"Local tax withheld to date \"); l_tax = scan.nextFloat(); System.out.println(\"Federal tax withheld to date \"); f_tax = scan.nextFloat(); System.out.println(\"State tax withheld to date \"); s_tax = scan.nextFloat(); employee.readInput(fname,lname,dependent,hourly_rate,no_of_hours,l_tax,f_tax,s_tax); employee.calculateData(count); employee.writeOutput(count); clerk.colectDataForCompanyReport(employee,count); } clerk.printDataForCompanyReport(); } } Employee.java import java.util.ArrayList; public class Employee { private ArrayList fname,lname; private ArrayList dependent; private ArrayList hourly_rate,no_of_hours,l_tax,f_tax,s_tax,g_wages,cf_tax,cl_tax,cs_tax,t_tax,ct_tax,net_pay; public Employee(){ fname = new ArrayList(); lname = new ArrayList(); dependent = new ArrayList(); hourly_rate = new ArrayList(); no_of_hours = new ArrayList(); l_tax = new ArrayList(); f_tax = new ArrayList(); s_tax = new ArrayList(); g_wages = new ArrayList(); cl_tax = new ArrayList(); cf_tax = new ArrayList(); cs_tax = new ArrayList(); t_tax = new ArrayList(); ct_tax = new ArrayList(); net_pay = new ArrayList(); } public ArrayList getFname() { return fname; } public void setFname(ArrayList fname) { this.fname = fname; } public ArrayList getLname() { return lname; } public void setLname(ArrayList lname) { this.lname = lname; } public ArrayList getDependent() { return dependent; } public void setDependent(ArrayList dependent) { this.dependent = dependent; } public ArrayList getHourly_rate() { return hourly_rate; } public void setHourly_rate(ArrayList hourly_rate) { this.hourly_rate = hourly_rate; } public ArrayList getNo_of_hours() { return no_of_hours; } public void setNo_of_hours(ArrayList no_of_hours) { this.no_of_hours = no_of_hours; } public ArrayList getL_tax() { return l_tax; } public void setL_tax(ArrayList l_tax) { this.l_tax = l_tax; } public ArrayList getF_tax() { return f_tax; } public void setF_tax(ArrayList f_tax) { this.f_tax = f_tax; } public ArrayList getS_tax() { return s_tax; } public void setS_tax(ArrayList s_tax) { this.s_tax = s_ta.
The solution is as belowEmployeeDemo.javaimport java.util.Scann.pdf
The solution is as belowEmployeeDemo.javaimport java.util.Scann.pdf
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The precations to be given to the workers are: Solution The precations to be given to the workers are:.
The precations to be given to the workers areSolutionThe prec.pdf
The precations to be given to the workers areSolutionThe prec.pdf
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At 1700 cm-1 a strong signal in an IR spectum is the result of a ketone (C=O). (It can shift if the compound is conjugated). Solution At 1700 cm-1 a strong signal in an IR spectum is the result of a ketone (C=O). (It can shift if the compound is conjugated)..
At 1700 cm-1 a strong signal in an IR spectum is .pdf
At 1700 cm-1 a strong signal in an IR spectum is .pdf
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Result: best_fit_for_X1 = DistName: \'tlocationscale\' NLogL: 2.1206e+03 BIC: 4.2620e+03 AIC: 4.2472e+03 AICc: 4.2473e+03 ParamNames: {\'mu\' \'sigma\' \'nu\'} ParamDescription: {\'location\' \'scale\' \'degrees of freedom\'} Params: [1.0071 1.9637 37.4643] Paramci: [2x3 double] ParamCov: [3x3 double] Support: [1x1 struct] best_fit_for_X2 = DistName: \'generalized extreme value\' NLogL: 2.4968e+03 BIC: 5.0142e+03 AIC: 4.9995e+03 AICc: 4.9995e+03 ParamNames: {\'k\' \'sigma\' \'mu\'} ParamDescription: {\'shape\' \'scale\' \'location\'} Params: [-0.2802 2.9571 1.0858] Paramci: [2x3 double] ParamCov: [3x3 double] Support: [1x1 struct] correlation_between_X1_X2 = 0.0084 (hence X1 and X2 are independent) meanY = 3.1394 varY = 12.8269 meanZ = 4.1492 varZ = 25.1576 correlation_between_Y_Z = 0.9438 (Hence Y and Z are dependent) Code: function main load X.mat [D PD]=allfitdist(X1,\'NLogL\'); [D2 PD2]=allfitdist(X2,\'NLogL\'); %best fit using maximum likelihood best_fit_for_X1=D(1) best_fit_for_X2=D2(1) correlation_between_X1_X2=correlation(X1,X2) Y=X1+X2; Z=2*X1+X2; meanY=mean(Y) varY=var(Y) meanZ=mean(Z) varZ=var(Z) correlation_between_Y_Z=correlation(Y,Z) end function[r]=correlation(x,y) avx = mean(x); ex2 = sum((x - avx).^2); %Wasteful multiple passes are a pain, avy = mean(y); ey2 = sum((y - avy).^2); %But explicit loops exy = sum((x - avx).*(y - avy)); %Are interpreted slowly. r = exy/sqrt(ex2*ey2); %Division by N cancels; ignoring the N - 1 ritual. end function [D PD] = allfitdist(data,sortby,varargin) %ALLFITDIST Fit all valid parametric probability distributions to data. % [D PD] = ALLFITDIST(DATA) fits all valid parametric probability % distributions to the data in vector DATA, and returns a struct D of % fitted distributions and parameters and a struct of objects PD % representing the fitted distributions. PD is an object in a class % derived from the ProbDist class. % % [...] = ALLFITDIST(DATA,SORTBY) returns the struct of valid distributions % sorted by the parameter SORTBY % NLogL - Negative of the log likelihood % BIC - Bayesian information criterion (default) % AIC - Akaike information criterion % AICc - AIC with a correction for finite sample sizes % % [...] = ALLFITDIST(...,\'DISCRETE\') specifies it is a discrete % distribution and does not attempt to fit a continuous distribution % to the data % % [...] = ALLFITDIST(...,\'PDF\') or (...,\'CDF\') plots either the PDF or CDF % of a subset of the fitted distribution. The distributions are plotted in % order of fit, according to SORTBY. % % List of distributions it will try to fit % Continuous (default) % Beta % Birnbaum-Saunders % Exponential % Extreme value % Gamma % Generalized extreme value % Generalized Pareto % Inverse Gaussian % Logistic % Log-logistic % Lognormal % Nakagami % Normal % Rayleigh % Rician % t location-scale % Weibull % % Discrete (\'DISCRETE\') % Binomial % Negative binomial % Poisson % % Optional inputs: % [...] = ALLFITDIST(...,\'n\',N,...) % For the \'binomial\' .
Resultbest_fit_for_X1 =DistName tlocationscaleNLogL 2.1.pdf
Resultbest_fit_for_X1 =DistName tlocationscaleNLogL 2.1.pdf
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Please rate the solution give thumbs if this is helpful. Impact of data quality on business performance: Functions of the database technology: Differences between centralized and distributed database architecture: The role of a master reference file: Solution Please rate the solution give thumbs if this is helpful. Impact of data quality on business performance: Functions of the database technology: Differences between centralized and distributed database architecture: The role of a master reference file:.
Please rate the solution give thumbs if this is helpful.Impact of .pdf
Please rate the solution give thumbs if this is helpful.Impact of .pdf
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Please find the answers and explanations below: Part 1 (True/False) 1. Replication and transcription proceed in 5\' to 3\' direction: TRUE (Both DNA replication and transcription of mRNA proceed in 5\' to 3\' direcition. This is because the firstly, the DNA polymerase retains only 5\' to 3\' directionality and thus, replication begins only in this direction. .Secondarily, there must be sequence similarity between the DNA and the RNA transcribed from it thus the orientation of transcription is necessarily 5\' to 3\' in nature) 2. All cells have telomerase activity : FALSE (Telomerase or terminal transferase is an enzyme which is required for maintenance of terminal positions of chromosomes by self-replication and prevents shortening of chromosomes. It also prevents sticking of chromosomes together and thus clumping of chromosomes. Telomerase acitivty is absent in majority of somatic cells in the body.) 3. Only mRNA is transcribed: TRUE (Only the mRNA which carries the necessary information for translation is transcribed from the DNA template. The tRNA is utilized to transfer the transcribed information and rRNA is utilized to generate the polypeptide on the ribosomal surface during translation) 4. Promoter regions are upstream from the gene: TRUE (A gene promoter is the region of DNA located upstream the gene to be transcribed and carries the consensus sequence which could be recognized by RNA polymerase for its binding and carrying on transcription) 5. A single tRNA can carry several different amino acids: FALSE (The tRNA is specific for only one kind of amino acid which it could carry. It is a highly stringent and specific nature of this biomolecule which determines the specificness for the amino acid which will be synthesized from the codon) Part 2 Please find the definitions below: 1. Semi-conservative replication: The mode of replication of double stranded DNA in which a daughter DNA molecules carries a strand from parent DNA and the complementary strand is synthesized de novo is called semi-conservative replication. 2. Okazaki fragments: Okazaki fragments are short stretches of DNA synthesized discountinuously on the lagging strand on the open frame of the parent DNA. These stretches of DNA are finally conjoined together to form a continous strand of newly synthesized DNA. 3. Consensus sequence: Consensus sequences are those genetically conserved sequences of DNA which remain similar (upto 99%) within organisms of same or different species and are actively engaged in crucial functions such as promoter region binding, transcription sites for transcription factors, enzymatic genes etc. Any change or deterioration in these consesus sequence might be highly deletrious for the cells. 4. Codon: A codon can be defined as a triplet set of nucleotides which come in a specific order and encode for a specific type of amino acid upon translation. For example, the codon AAA encodes for amino acid lysine. Exceptionally, codons UAA, UAG and UGA do not .
Please find the answers and explanations belowPart 1 (TrueFalse).pdf
Please find the answers and explanations belowPart 1 (TrueFalse).pdf
aparnatiwari291
Ionic product of water Kw = [H+][OH-] = 10-14 [OH-] = 10-14/2.7 x 10-6 = 3.7 x 10-9 M Solution Ionic product of water Kw = [H+][OH-] = 10-14 [OH-] = 10-14/2.7 x 10-6 = 3.7 x 10-9 M.
Ionic product of water Kw = [H+][OH-] = 10-14[OH-] = 10-142.7 x 1.pdf
Ionic product of water Kw = [H+][OH-] = 10-14[OH-] = 10-142.7 x 1.pdf
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Instruction set architecture of a machine fill the semantic gap between the user And the machines. And serve as the early point for the design of a new machine or Modification of existing ones. ISA design must describe how memory addresses are interpret and specified in the Instructions. Most machines are byte addressed. That is a 8 bits is the smallest unique address. There are two different conventions for ordering bytes within a word one is Little Endian and the other Big Endian Example: Big Endian Little Endian Address value Address value 1000 90 1000 CD 1001 AB 1001 12 1002 12 1002 AB 1003 CD 1003 90 Physical register file is an array of computer registers in a CPU more difficult CPUs use register so that the mapping of which physical entry supplies a particular architectural register change dynamically throughout execution Example: Physical register Id LastUseItag --- --- (p-1) 145 Solution Instruction set architecture of a machine fill the semantic gap between the user And the machines. And serve as the early point for the design of a new machine or Modification of existing ones. ISA design must describe how memory addresses are interpret and specified in the Instructions. Most machines are byte addressed. That is a 8 bits is the smallest unique address. There are two different conventions for ordering bytes within a word one is Little Endian and the other Big Endian Example: Big Endian Little Endian Address value Address value 1000 90 1000 CD 1001 AB 1001 12 1002 12 1002 AB 1003 CD 1003 90 Physical register file is an array of computer registers in a CPU more difficult CPUs use register so that the mapping of which physical entry supplies a particular architectural register change dynamically throughout execution Example: Physical register Id LastUseItag --- --- (p-1) 145.
Instruction set architecture of a machine fill the semantic gap betw.pdf
Instruction set architecture of a machine fill the semantic gap betw.pdf
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all real numbers Solution all real numbers.
all real numbers .pdf
all real numbers .pdf
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In the given text the following phrases reflect scientific approach of studying Homo floresiensis. Comparing height with the modern humans: would stand waist high. Studying the features of skull: shape and thickness of the bones were human like. Size of the brain: her brain was the size of a chimpanzee’s. All these reflect a scientific approach where the species is compared with modern humans and conclusions drawn from the observations. Sometimes science cannot answer all questions like the existence of supernatural powers and beings. Moreover evolution denies the role of God in the universe. One theory beyond science is that life on earth is created by gods. Science does not explain the existence of gods or if super natural entities intervene in human affairs. This explanation is beyond the nature and so beyond the scope of what can be studied by science. This theory of origin of human beings is a belief. Science is an activity that seeks to explore the natural world using well-established, clearly- delineated methods. Scientific method is based on the fact of asking a question, formulating a hypothesis, performing experiments, collecting data and drawing conclusions. Studying human evolution in a scientific way enables in better understanding of our origin and evolution. Understanding the process by which Homo sapiens came into existence is crucial for understanding the kind of creature we are and this is possible by studying scientifically. Solution In the given text the following phrases reflect scientific approach of studying Homo floresiensis. Comparing height with the modern humans: would stand waist high. Studying the features of skull: shape and thickness of the bones were human like. Size of the brain: her brain was the size of a chimpanzee’s. All these reflect a scientific approach where the species is compared with modern humans and conclusions drawn from the observations. Sometimes science cannot answer all questions like the existence of supernatural powers and beings. Moreover evolution denies the role of God in the universe. One theory beyond science is that life on earth is created by gods. Science does not explain the existence of gods or if super natural entities intervene in human affairs. This explanation is beyond the nature and so beyond the scope of what can be studied by science. This theory of origin of human beings is a belief. Science is an activity that seeks to explore the natural world using well-established, clearly- delineated methods. Scientific method is based on the fact of asking a question, formulating a hypothesis, performing experiments, collecting data and drawing conclusions. Studying human evolution in a scientific way enables in better understanding of our origin and evolution. Understanding the process by which Homo sapiens came into existence is crucial for understanding the kind of creature we are and this is possible by studying scientifically..
In the given text the following phrases reflect scientific approach .pdf
In the given text the following phrases reflect scientific approach .pdf
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Li+ Solu.pdf
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