1 drop per 10 lbs17 drops per 170 lbsequally distributed into fo.pdf

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1 drop per 10 lbs 17 drops per 170 lbs equally distributed into four doses, 17/4=4.25 Answer: 4.25 drops per dose Solution 1 drop per 10 lbs 17 drops per 170 lbs equally distributed into four doses, 17/4=4.25 Answer: 4.25 drops per dose.

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HClO4 in aqueous medium ionizes as Hydrogen(+1)ca.pdf

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Which of the following would be a description of a system unit? A case and internal components used to process data into information A computer’s main memory component The module that interprets and carries out computer instructions A data and information storage device Ans: A case and internal components used to process data into information Question 2 Select one answer. True or false? A tower or minitower PC is a type of all-in-one desktop unit. True False Ans: True Question 3 Fill in the blanks by selecting one option from each menu. A/an --expansion card internal drive bay processor chip is inserted on the motherboard and provides more memory or controls a peripheral device such as a graphics adapter that supports video games. Ans: Expansion card Question 4 Select one answer What is a peripheral? A high-speed storage location that temporarily holds data and instructions for future use A component of the system board that controls the timing of computer activities A slot on the motherboard that holds memory modules and helps speed computer processes A device that connects to the system unit and is controlled by the computer’s processor Ans: A device that connects to the system unit and is controlled by the computer’s processor Question 5 Fill in the blanks by selecting one option from each menu. If you often play the latest computer games on your PC, which of the following would you consider critical components of the PC? Select Yes or No for each option. The weight and size of your PC --YesNo Ans: No Sophisticated graphics and sound adapter cards --YesNo Ans: Yes Processor speed and amount of RAM --YesNo Ans: Yes Question 1 Select one answer. If you use your desktop computer only for doing schoolwork, which input and output device combination would be best from the options below (considering the best combination of cost, screen resolution, and keyboard design)? A gaming keyboard and a 27\" LCD monitor A foldable keyboard and a 15\" LCD monitor A wireless keyboard and a 24\" LED monitor A split-style keyboard and a 30\" LED monitor Ans: A wireless keyboard and a 24\" LED monitor Question 2 Select one answer. You are considering using cloud storage as your primary system for saving your data. What advantages does this storage media have over using your system’s hard drive or some other external media (external hard drive, flash drive, R/W disks) for storing your data? You can always get to your data from anywhere. The data is more secure in cloud storage than on your own storage devices. The cloud storage provider ensures that you have a backup of your data. Cloud storage is definitely less expensive than the cost of a hard drive or external devices. Ans: You can always get to your data from anywhere. Question 3 Select one answer. Which of the following can be considered an output device? Microphone Stylus Wireless mouse Speaker Ans: Speaker Question 4 Select one answer. In this diagram of a simple home network, what appears to be missing? .

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Water is a polar inorganic solvent. Benzene is a nonpolar organic solvent. The solvents will generally dissolve solutes with similar polarity (with certain exceptions). CaCO3 - insoluble in water, insoluble in benzene CaCO3 is a polar ionic compound that is insoluble in water. Ionic compounds are insoluble in nonpolar organic solvents like benzene. NaNO3 - soluble in water, insoluble in benzene NaNO3 is a polar ionic compound that is soluble in water. Ionic compounds are insoluble in nonpolar organic solvents like benzene. K2SO4 - soluble in water, insoluble in benzene K2SO4 is a polar ionic compound that is soluble in water. Ionic compounds are insoluble in nonpolar organic solvents like benzene. Petroleum jelly - insoluble in water, soluble in benzene Petroleum jelly is a paraffin and a nonpolar organic compound. Thus it is insoluble in water but soluble in benzene. Butane - insoluble in water, soluble in benzene Butane is a nonpolar organic compound. Thus it is insoluble in water but soluble in benzene. Acetone - soluble in water, soluble in benzene Acetone is a polar organic compound. Thus it is soluble in both water and benzene. Solution Water is a polar inorganic solvent. Benzene is a nonpolar organic solvent. The solvents will generally dissolve solutes with similar polarity (with certain exceptions). CaCO3 - insoluble in water, insoluble in benzene CaCO3 is a polar ionic compound that is insoluble in water. Ionic compounds are insoluble in nonpolar organic solvents like benzene. NaNO3 - soluble in water, insoluble in benzene NaNO3 is a polar ionic compound that is soluble in water. Ionic compounds are insoluble in nonpolar organic solvents like benzene. K2SO4 - soluble in water, insoluble in benzene K2SO4 is a polar ionic compound that is soluble in water. Ionic compounds are insoluble in nonpolar organic solvents like benzene. Petroleum jelly - insoluble in water, soluble in benzene Petroleum jelly is a paraffin and a nonpolar organic compound. Thus it is insoluble in water but soluble in benzene. Butane - insoluble in water, soluble in benzene Butane is a nonpolar organic compound. Thus it is insoluble in water but soluble in benzene. Acetone - soluble in water, soluble in benzene Acetone is a polar organic compound. Thus it is soluble in both water and benzene..

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viruses which have single strande DNA in their genome come under category of paroviridae which generally includes animal pathogens. These viruses when infect the host, the single stranded DNA is first converted into double stranded DNA, and then using the host machinary for transcription, ds DNA transcribes into viral mRNA. For transcription, RNA polymerase, transcription factors , enzymes are acquired from the host and divided into initiation, elongation and termaination. For gene expression in viruses transcription is generally not coupled with translation in ssDNA virus. The virus instructs transcription machinary to make viral components in the host cell. Solution viruses which have single strande DNA in their genome come under category of paroviridae which generally includes animal pathogens. These viruses when infect the host, the single stranded DNA is first converted into double stranded DNA, and then using the host machinary for transcription, ds DNA transcribes into viral mRNA. For transcription, RNA polymerase, transcription factors , enzymes are acquired from the host and divided into initiation, elongation and termaination. For gene expression in viruses transcription is generally not coupled with translation in ssDNA virus. The virus instructs transcription machinary to make viral components in the host cell..

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This word problem is about a triangle whose perimeter is 47 miles. Since it\'s asking to find the length of each side, then there are 3 unknowns. So here\'s what we do: Let the variables S1 be the length of side 1 in miles, S2 be the length of side 2 in miles, and S3 be the length of side 3 in miles. Now, the next thing to do with these variables is to set up equations that follow the word problem. Here they are as follows: S1+S2+S3=47 the perimeter of the triangle by adding the lengths of all three sides S1=(2*S2)-5 1st side (represented by S1) is (represented by =) 5 less (represented by -5) than twice the 2nd side (represented by 2*S2). S3=S1+2 3rd side (represented by S3) is 2 more (represented by +2) than the 1st side. Next, since we have more than one unknown, we should express one equation in terms of one variable. Since only one variable has a coefficient which is S2 in the term, 2*S2, I suggest that we express the other variable equations of S1 and S3 in terms of S2. Here\'s how it goes: S1=(2*S2)-5 as S1 is already expressed in terms of S2 S3=S1+2=[(2*S2)-5]+2 I plugged in the S1 equation to express S3 in terms of S2. =(2*S2)-5+2 I combined like terms. =(2*S2)-3. Now that I have expressed both S1 and S3 in terms of S2, I can plug those equations into that for the triangle perimeter: [(2*S2)-5]+S2+[(2*S2)-3]=47 (2*S2)-5+S2+(2*S2)-3=47 Again, I combined like terms. (5*S2)-8=47 +8=+8 I added 8 on both sides of the equation to eliminate -8 on the left side. 5*S2=55 5*S2/5=55/5 S2=11 S1=(2*11)-5=22-5=17 Now that I have found the value of S2, I can plug it into the S1 equation. S3=17+2=19 Now that I have found the value of S1, I can plug it into the S3 equation. So now here are the lengths of each side of the triangle. Side 1 is 17 miles, side 2 is 11 miles, and side 3 is 19 miles. Solution This word problem is about a triangle whose perimeter is 47 miles. Since it\'s asking to find the length of each side, then there are 3 unknowns. So here\'s what we do: Let the variables S1 be the length of side 1 in miles, S2 be the length of side 2 in miles, and S3 be the length of side 3 in miles. Now, the next thing to do with these variables is to set up equations that follow the word problem. Here they are as follows: S1+S2+S3=47 the perimeter of the triangle by adding the lengths of all three sides S1=(2*S2)-5 1st side (represented by S1) is (represented by =) 5 less (represented by -5) than twice the 2nd side (represented by 2*S2). S3=S1+2 3rd side (represented by S3) is 2 more (represented by +2) than the 1st side. Next, since we have more than one unknown, we should express one equation in terms of one variable. Since only one variable has a coefficient which is S2 in the term, 2*S2, I suggest that we express the other variable equations of S1 and S3 in terms of S2. Here\'s how it goes: S1=(2*S2)-5 as S1 is already expressed in terms of S2 S3=S1+2=[(2*S2)-5]+2 I plugged in the S1 equation to express S3 in terms of S2. =(2*S2)-5+2 I combined l.

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Program to print the Diamond Shape :- #include int main() { int i, j, n; printf(\"Enter a positive int number greater than one to print the Diamond Shape : \"); scanf(\"%d\", &n); //To Print the upper pyramid for(i=1; i<=n; i++) { for(j=i; j=1; i--) { for(j=i; j<=n; j++) { printf(\" \"); } for(j=2; j<(2*i-1); j++) { printf(\"*\"); } printf(\"\ \"); } return 0; } Solution Program to print the Diamond Shape :- #include int main() { int i, j, n; printf(\"Enter a positive int number greater than one to print the Diamond Shape : \"); scanf(\"%d\", &n); //To Print the upper pyramid for(i=1; i<=n; i++) { for(j=i; j=1; i--) { for(j=i; j<=n; j++) { printf(\" \"); } for(j=2; j<(2*i-1); j++) { printf(\"*\"); } printf(\"\ \"); } return 0; }.

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Isomers which have their atoms connected in the same sequence but differ in the way the atoms are oriented in space - i.e. the difference between two stereoisomers lies only in the three dimensional arrangement of atoms. since Cis-3-hexene to trans-3-hexene differs in three dimensional arrangement of atoms and arrangement is same in 2D space , they are stereo isomers Solution Isomers which have their atoms connected in the same sequence but differ in the way the atoms are oriented in space - i.e. the difference between two stereoisomers lies only in the three dimensional arrangement of atoms. since Cis-3-hexene to trans-3-hexene differs in three dimensional arrangement of atoms and arrangement is same in 2D space , they are stereo isomers.

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import java.util.Scanner; import java.text.DecimalFormat; import java.io.*; public class SavingAccount //MUST match the file name! { public static void main(String[] args)throws IOException { //Create a decimal format for displaying dollars DecimalFormat dollar = new DecimalFormat(\"#,###.##\"); //Constants double depositSum = 0.0; double withdrawalSum = 0.0; double earnedInt = 0.0; double startBalance = 500; //Starting balance // Executables System.out.println(\"This program solves Programming Challenge 6.11\"); System.out.println(); //Create Scanner object for keyboard input. Scanner keyboard = new Scanner(System.in); //Get interest rate System.out.print(\"Enter the annual interest: \"); double testInterest = keyboard.nextDouble(); //Create an object that accept the starting balance and annual interest SavingAccount1 account = new SavingAccount1(startBalance, testInterest); //Open Deposit file. File file = new File (\"BankDeposits.txt\"); Scanner inputFile = new Scanner(file); //Read line in file while (inputFile.hasNext()); { //Read numbers double num = inputFile.nextDouble(); //Add the numbers depositSum += num; } //Deposit the file input. account.deposit(depositSum); //Close the file inputFile.close(); //Open Withdrawal file File file2 = new File(\"BankWithdrawal.txt\"); Scanner inputFile2 = new Scanner(file2); //Read lines in file while (inputFile2.hasNext()); { //Read numbers double num2 = inputFile2.nextDouble(); //Add the numbers withdrawalSum += num2; } //Withdrawal the file input from account. account.withdraw(withdrawalSum); //Close the file inputFile2.close(); //Add the monthly interest account.addInt(); //Get amount of interest earned. earnedInt += account.getInterest(); //Display the data System.out.println(\"Account balance $\" + dollar.format(account.getBalance())); System.out.println(\"Total interest earned $\" + dollar.format(account.getInterest())); } }//end class public class SavingAccount1 //MUST match the file name! { public static void main(String[] args) { System.out.println(\"This program solves Programming Challenge 6.10\"); System.out.println(); } //Fields private double balance; //Account balance private double annualInterest; //annual interest private double monthInt; //monthly interest private double earnedInt; //earned interest double totalWithdraw; double totalDeposit; /** 018 This constructor sets the starting balance 019 and the annual interest at 0.0. 020 */ public void Ward_Tassinda_SavingAccount1() { balance = 0.0; annualInterest = 0.0; } /** 029 This constructor set the starting balance and the annual interest rate 030 to the value passed as an argument. 031 @param startBalance The starting balance. 032 */ public void Ward_Tassinda_SavingAccount1(double startBalance, double interestRate) { balance = startBalance; annualInterest = interestRate; } /** 041 This constructor sets the starting balance to 042 the value in the String argument. 043 @param str The starting balance, as a String. 044 */ public void .

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// Points.java import java.util.ArrayList; import java.util.Scanner; public class Points{ double x; double y; Points(double xco, double yco){ x = xco; y = yco; } } // PolygonArea.java import java.util.ArrayList; import java.util.Scanner; public class PolygonArea { // funtion to determine area of polygon public static double getArea(ArrayList coordinates) { double sumxy = 0; double sumyx = 0; for (int i = 0; i < coordinates.size(); i++) { int index = (i + 1) % coordinates.size(); Points point1 = coordinates.get(i); Points point2 = coordinates.get(index); sumxy += (point1.x * point2.y); sumyx += (point1.y * point2.x); } double areaPolygon = 0.5 * (sumxy - sumyx); if (areaPolygon > 0) return areaPolygon; else return -areaPolygon; } public static void main(String[] args) { Scanner scan = new Scanner(System.in); System.out.print(\"Enter the number of the points: \"); int totalPoints = scan.nextInt(); System.out.print(\"Enter the coordinates of the points: \"); ArrayList coordinates = new ArrayList<>(); for (int i = 0; i < totalPoints; i++) { coordinates.add(new Points(scan.nextDouble(), scan.nextDouble())); } System.out.println(\"The total area is \" + getArea(coordinates)); } } /* output: Enter the number of the points: 3 Enter the coordinates of the points: 0 0 0 3 4 0 The total area is 6.0 */ Solution // Points.java import java.util.ArrayList; import java.util.Scanner; public class Points{ double x; double y; Points(double xco, double yco){ x = xco; y = yco; } } // PolygonArea.java import java.util.ArrayList; import java.util.Scanner; public class PolygonArea { // funtion to determine area of polygon public static double getArea(ArrayList coordinates) { double sumxy = 0; double sumyx = 0; for (int i = 0; i < coordinates.size(); i++) { int index = (i + 1) % coordinates.size(); Points point1 = coordinates.get(i); Points point2 = coordinates.get(index); sumxy += (point1.x * point2.y); sumyx += (point1.y * point2.x); } double areaPolygon = 0.5 * (sumxy - sumyx); if (areaPolygon > 0) return areaPolygon; else return -areaPolygon; } public static void main(String[] args) { Scanner scan = new Scanner(System.in); System.out.print(\"Enter the number of the points: \"); int totalPoints = scan.nextInt(); System.out.print(\"Enter the coordinates of the points: \"); ArrayList coordinates = new ArrayList<>(); for (int i = 0; i < totalPoints; i++) { coordinates.add(new Points(scan.nextDouble(), scan.nextDouble())); } System.out.println(\"The total area is \" + getArea(coordinates)); } } /* output: Enter the number of the points: 3 Enter the coordinates of the points: 0 0 0 3 4 0 The total area is 6.0 */.

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General Instructions: 1. The objective of this check list is to guide the Internal Auditors in the conduct of Internal Audit in a systematic manner so that all the significant points in finance & accounts, procurement, contract & tenders, payment of bills of contractors, planning process, flow of funds and staffing etc. are checked before finalization of Internal Audit Report 2. The Internal Auditors should familiarize themselves with the Manual on Financial Management & Procurement thoroughly before conducting Internal Audit. Para 100 to 105 of the Manual on Financial Management & Procurement provides broad guidelines for Internal Audit of SSA and Paras 51 to 99.7 of the Manual contains guidelines for maintenance of accounts . 3. Scope and coverage of Internal Audit and report – Internal Audit of District Project Offices and sub -district units selected on a percentage basis should be conducted so as to cover all districts and sub- districts units at least once in 3 years . The Internal Audit report should indicate name of unit covered, date of Audit, review of planning process to ascertain whether the activities under SSA have been planned as per prescribed norms, review of financial systems to check the efficacy of internal checks and controls, review of establishment matters and staffing, review of Procurement and status of Statutory Audit, C& AG Audit and Internal Audit. 4 Finance & Accounts – 4.1 To check whether 22 Registers( Annexure-I is attached for ready reference) indicated in para 52 of FM&P Manual are maintained. 4.2 To verify about the number of savings Bank Accounts maintained for SSA at SPO /DPO /Block & VEC Levels. The matter should be reported if more then one savings bank accounts are maintained at any levels. The reasons for maintenance of multiple Bank Accounts should also be mentioned in the Internal Audit Report 4.3 Cash book - To check cash book for ascertaining whether cash book is written daily as envisaged in para 79 of FM&P Manual monthly on regular basis. The Internal Auditor must also conduct physical verification of cash and report heavy cash balance if any. 4.4 To check whether Bank Reconciliation is done monthly on regular basis or not 4.5 Payment of bills of contractors Annexure-P To verify whether 4.5.1 Original bill duly signed by contractors is submitted. 4.5.2 Contractor has put his initials in all cuttings and corrections in the bill. 4.5.3 All Supporting documents are attached with the bills. 4.5.4 The rates, security deposit, and deductions are as per terms and conditions specified in the agreement 4.5.5 The variations in quantities and completion period etc. have been authorized by the competent authority 4.5.6 Job completion certificate has been processed by the dealing assistant of SSA. 4.5.7 Bills passed for payment are as per rules of the society. 4.5.8 All terms and conditions of the contract are fulfilled before passing the bills. 4.5.9 Every final bill is checked in detail with measurement boo.

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falseSolutionfalse.pdf

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Computer security, additionally called cybersecurity or IT security, is that the protection of knowledge systems from thieving or harm to the hardware, the package, and to the knowledge on them, moreover as from disruption or misdirection of the services they supply.[1] It includes dominant physical access to the hardware, moreover as protective against damage that will come back via network access, information and code injection,[2] and thanks to malpractice by operators, whether or not intentional, accidental, or thanks to them being tricked into deviating from secure procedures.[3] The field is of growing importance thanks to the increasing reliance on laptop systems and therefore the web in most societies,[4] wireless networks like Bluetooth and Wi-Fi - and therefore the growth of \"smart\" devices, as well as smartphones, televisions and small devices as a part of the net of Things. A vulnerability may be a system susceptibleness or flaw. several vulnerabilities ar documented within the Common Vulnerabilities and Exposures (CVE) information. AN exploitable vulnerability is one that a minimum of one operating attack or \"exploit\" exists.[5] To secure a automatic data processing system, it\'s vital to know the attacks that may be created against it, and these threats will generally be classified into one among the classes below: Backdoors[edit] A backdoor in an exceedingly automatic data processing system, a cryptosystem or AN algorithmic rule, is any secret methodology of bypassing traditional authentication or security controls. they will exist for variety of reasons, as well as by original style or from poor configuration. they will are accessorial by a licensed party to permit some legitimate access, or by AN assailant for malicious reasons; however notwithstanding the motives for his or her existence, they produce a vulnerability. Denial-of-service attack[edit] Denial of service attacks ar designed to create a machine or network resource unavailable to its supposed users.[6] Attackers will deny service to individual victims, like by deliberately coming into a wrong countersign enough consecutive times to cause the victim account to be fast, or they will overload the capabilities of a machine or network and block all users directly. whereas a network attack from one IP address may be blocked by adding a brand new firewall rule, several sorts of Distributed denial of service (DDoS) attacks ar doable, wherever the attack comes from an outsized variety of points – and defensive is far harder. Such attacks will originate from the zombie computers of a botnet, however a spread of alternative techniques ar doable as well as reflection and amplification attacks, wherever innocent systems ar fooled into causing traffic to the victim. Solution Computer security, additionally called cybersecurity or IT security, is that the protection of knowledge systems from thieving or harm to the hardware, the package, and to the knowledge on them, moreover.

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C. Nitrosomonas, Nitrosolobus, and Nitrosovibrio Nitrifiers are chemolithotropic organisms which obtains energy by oxidating inorganic nitrogen compounds. These three organisms convert Ammonia to Nitrites which inturn converted into nitrate by other organisms such as Nitrobacter or Nitrospina. False: Neisseria gonorrhoeae - responsible for causing sexually transmitted disease Gonorrhea Neisseria meningitidis- responsible for causing meningitis and sepsis Thiobacillus spp- It belongs to Hydrogenophilaceae family and mostly responsible for oxidizing hydrogen. Solution C. Nitrosomonas, Nitrosolobus, and Nitrosovibrio Nitrifiers are chemolithotropic organisms which obtains energy by oxidating inorganic nitrogen compounds. These three organisms convert Ammonia to Nitrites which inturn converted into nitrate by other organisms such as Nitrobacter or Nitrospina. False: Neisseria gonorrhoeae - responsible for causing sexually transmitted disease Gonorrhea Neisseria meningitidis- responsible for causing meningitis and sepsis Thiobacillus spp- It belongs to Hydrogenophilaceae family and mostly responsible for oxidizing hydrogen..

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C is correct. Routing and logical addressing occur at the Internet layer of the TCP/IP stack. A, B, and D are incorrect. A is incorrect because the Network layer is an OSI model layer. B is incorrect because Data Link is an OSI model layer. D is incorrect because framing, error checking, and media access occur at the Network Access layer of the TCP/IP stack. Solution C is correct. Routing and logical addressing occur at the Internet layer of the TCP/IP stack. A, B, and D are incorrect. A is incorrect because the Network layer is an OSI model layer. B is incorrect because Data Link is an OSI model layer. D is incorrect because framing, error checking, and media access occur at the Network Access layer of the TCP/IP stack..

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