PDU is called Protocol Data Unit.It consists of user data and protoc.pdf

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PDU is called Protocol Data Unit.It consists of user data and protocol control information. PDU at layer 4 i.e Networks layer is called Segment. Solution PDU is called Protocol Data Unit.It consists of user data and protocol control information. PDU at layer 4 i.e Networks layer is called Segment..

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You\'re looking for Ka, the dissociation constant, so you measure this by adding the substance to water. 1. Write the balanced equation HBrO + H2O = BrO- + H30+ 2. the pH is 4.95, so you know its acidic and there has to have hydronium as a product. if the pH was >7.0, there would be OH- instead. the pH is a measure of H ions, and to get the concentration, you simply do 10^(-pH). 10^(-4.95) = 1.12E-5 3. The equation for Ka is concentration of products/concentration of reactants. always look at the coefficients infront of the molecules in the reaction. in this case, it is [BrO-)][H3O+]/[HBrO] *liquids and solids are never included in this, only aqueous solutions and gases are* 4. Your concentration of H3O was found to be 1.12E-5, and because the H30 and BrO are in a 1:1 ration in the equation, the concentration for BrO is also 1.12E-5. The concentration for HBrO as a reactant is given, but the final concentration is 0.063-concentration of hydronium...but since the concentration of the hydronium is so small, it doesnt at all affect the final answer. 5. your equation for Ka=((1.12E- 5)(1.12E-5))/(0.063-1.12E-5). The final answer for your Ka is 2.0E-9 Hope this helps, please rate. Solution You\'re looking for Ka, the dissociation constant, so you measure this by adding the substance to water. 1. Write the balanced equation HBrO + H2O = BrO- + H30+ 2. the pH is 4.95, so you know its acidic and there has to have hydronium as a product. if the pH was >7.0, there would be OH- instead. the pH is a measure of H ions, and to get the concentration, you simply do 10^(-pH). 10^(-4.95) = 1.12E-5 3. The equation for Ka is concentration of products/concentration of reactants. always look at the coefficients infront of the molecules in the reaction. in this case, it is [BrO-)][H3O+]/[HBrO] *liquids and solids are never included in this, only aqueous solutions and gases are* 4. Your concentration of H3O was found to be 1.12E-5, and because the H30 and BrO are in a 1:1 ration in the equation, the concentration for BrO is also 1.12E-5. The concentration for HBrO as a reactant is given, but the final concentration is 0.063-concentration of hydronium...but since the concentration of the hydronium is so small, it doesnt at all affect the final answer. 5. your equation for Ka=((1.12E- 5)(1.12E-5))/(0.063-1.12E-5). The final answer for your Ka is 2.0E-9 Hope this helps, please rate..

Youre looking for Ka, the dissociation constant.pdf

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Okay so dipole-dipole interactions would occur between polar molecules such as water. The H has a + charge while O has a - charge. Dipole- induced dipole(or dispersion forces) would be like a molecule that has a polar part to it such as a -COOH group and then a nonpolar part such as a long hydrocarbon chain. The nonpolar part has no charge except the transient charge among the different atoms. I hope that helps Solution Okay so dipole-dipole interactions would occur between polar molecules such as water. The H has a + charge while O has a - charge. Dipole- induced dipole(or dispersion forces) would be like a molecule that has a polar part to it such as a -COOH group and then a nonpolar part such as a long hydrocarbon chain. The nonpolar part has no charge except the transient charge among the different atoms. I hope that helps.

Okay so dipole-dipole interactions would occur be.pdf

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remote operating system installation Remote Installation Services (RIS) ships as a part of the Windows® 2000 Server software system. This document outlines the steps necessary to put in, configure, and use RIS. RIS was designed to cut back the prices incurred by pre-installing or physically visiting every consumer pc to put in the software system (OS). By combining RIS with alternative Windows 2000 IntelliMirror™ management technologies features—User knowledge Management, software package Installation and Maintenance, and User Settings Management—companies get pleasure from higher disaster recovery with easier OS and application management. Prerequisites See Appendix A below to confirm that each your server and consumer hardware meet the remote installation hardware necessities. The drive on the server wherever you decide on to put in RIS should be formatted with the NTFS filing system. RIS needs a big quantity of space and can\'t be put in on constant drive or partition on that Windows 2000 Server is put in. make sure that the chosen drive contains enough free space for a minimum of one full Windows 2000 skilled compact disc—a minimum of roughly 800 megabytes (MB)–1 GB (GB). RIS needs many alternative services that conjointly ship as a part of Windows 2000 Server. These services will be put in on individual servers, or all on one server, counting on your network design: Domain Name Service (DNS) Server. Remote installation depends on DNS for locating the directory service and consumer machine accounts. you\'ll use any Windows 2000 Active DirectoryTM service-compliant DNS server, otherwise you will use the DNS server given Windows 2000 Server. Dynamic Host Configuration Protocol (DHCP) Server. RIS needs a DHCP server to be gift and active on the network. The remote boot-enabled consumer computers receive AN information processing address from the DHCP server before contacting RIS. Active Directory. RIS depends on Windows 2000 Active Directory for locating existing consumer machines further as existing RIS servers. RIS should be put in on a Windows 2000- based server that has access to Active Directory. this will be a site controller or a server that\'s a member of a site with access to the Active Directory. To ensure a in installation, you want to install and tack together the extra services delineate higher than so as for RIS to perform. additionally, certify that you just have each the Windows 2000 Server and skilled CDs on the market. Before starting this gradual guide, you wish to create the common infrastructure, that specifies a specific hardware and software package configuration. The common infrastructure is roofed within the Common Infrastructure gradual guide, \"Part 1: putting in a Windows 2000 Server as a site Controller\" http://www.microsoft.com/windows2000/techinfo/planning/server/serversteps.asp. If you\'re not victimization the common infrastructure, you wish to create the suitable changes to the current instruction set. Alth.

remote operating system installationRemote Installation Services (.pdf

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Title of this process is \"The Project Life Cycle (Phases)\" The project manager and project team have one shared goal: to carry out the work of the project for the purpose of meeting the project’s objectives. Every project has a beginning, a middle period during which activities move the project toward completion, and an ending. A standard project typically has the following four major phases: initiation, planning, implementation, and closure. Taken together, these phases represent the path a project takes from the beginning to its end and are generally referred to as the project “life cycle.” Initiation Phase During the first phases, the initiation phase, the project objective is identified; this can be a business problem or opportunity. An appropriate response to the objective of the project is documented with recommended solution options. A feasibility study is conducted to investigate whether each option addresses the project objective and a final recommended solution is determined. Issues of feasibility (“can we do the project?”) and justification (“should we do the project?”) are addressed. Once the recommended solution is approved, a project is initiated to deliver the approved solution and a project manager is appointed. Approval is then sought by the project manager to move onto the detailed planning phase. Planning Phase The planning phase, is where the project solution is further developed in as much detail as possible and the steps necessary to meet the project’s objective are planned. In this step, the team identifies all of the work to be done. The project’s tasks and resource requirements are identified, along with the strategy for producing them. A project plan is created outlining the activities, tasks, dependencies, and timeframes. The project manager coordinates the preparation of a project budget by providing cost estimates for the labor, equipment, and materials costs. The budget is used to monitor and control cost expenditures during project implementation. Implementation (Execution) Phase During the third phase, the implementation phase, the project plan is put into motion and the work of the project is performed. It is important to maintain control and communicate as needed during implementation. Progress is continuously monitored and appropriate adjustments are made and recorded as variances from the original plan. Throughout this step, project sponsors and other key stakeholders should be kept informed of the project’s status according to the agreed-on frequency and format of communication. The plan should be updated and published on a regular basis. Status reports should always emphasize the anticipated end point in terms of cost, schedule, and quality of deliverables. Each project deliverable produced should be reviewed for quality and measured against the acceptance criteria. Once all of the deliverables have been produced and the customer has accepted the final solution, the project is ready for closure. Closing.

Title of this process is The Project Life Cycle (Phases)The pr.pdf

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The answer is A) Pol II, with twelve subunits on its own, is capable of initiating transcription. because it is composed with 12 subunits and due to incapable of specific promoter recognition it can\'t initiate transcription. Solution The answer is A) Pol II, with twelve subunits on its own, is capable of initiating transcription. because it is composed with 12 subunits and due to incapable of specific promoter recognition it can\'t initiate transcription..

The answer is A) Pol II, with twelve subunits on its own, is capable.pdf

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solution: A= 1 1 0 4 3 1 2 0 5 Finding the determinant of the matrix ? det A = -3 The determinant of A is non-zero, hence the inverse matrix A-1 exists. To find the inverse matrix was calculated the minors and cofactors of the matrix ? M1,1 = 3 1 0 5 = 15 C1,1 = (-1)1+1M1,1 = 15 M1,2 = 4 1 2 5 = 18 C1,2 = (-1)1+2M1,2 = -18 M1,3 = 4 3 2 0 = -6 C1,3 = (-1)1+3M1,3 = -6 M2,1 = 1 0 0 5 = 5 C2,1 = (-1)2+1M2,1 = -5 M2,2 = 1 0 2 5 = 5 C2,2 = (-1)2+2M2,2 = 5 M2,3 = 1 1 2 0 = -2 C2,3 = (-1)2+3M2,3 = 2 M3,1 = 1 0 3 1 = 1 C3,1 = (-1)3+1M3,1 = 1 M3,2 = 1 0 4 1 = 1 C3,2 = (-1)3+2M3,2 = -1 M3,3 = 1 1 4 3 = -1 C3,3 = (-1)3+3M3,3 = -1 Matrix of cofactors: C = 15 -18 -6 -5 5 2 1 -1 -1 Transpose of cofactor matrix: CT = 15 -5 1 -18 5 -1 -6 2 -1 Finding the inverse matrix. A-1 = CT = det A -5 5/3 -1/3 6 -5/3 1/3 2 -2/3 1/3 B) First, let\'s add the Identity Matrix to the right of our matrix 1 1 0 1 0 0 4 3 1 0 1 0 2 0 5 0 0 1 Now, let\'s do Gauss-Jordan Elimination on our new matrix... Add (-4 * row1) to row2 1 1 0 1 0 0 0 -1 1 -4 1 0 2 0 5 0 0 1 Add (-2 * row1) to row3 1 1 0 1 0 0 0 -1 1 -4 1 0 0 -2 5 -2 0 1 Divide row2 by -1 1 1 0 1 0 0 0 1 -1 4 -1 0 0 -2 5 -2 0 1 Add (2 * row2) to row3 1 1 0 1 0 0 0 1 -1 4 -1 0 0 0 3 6 -2 1 Divide row3 by 3 1 1 0 1 0 0 0 1 -1 4 -1 0 0 0 1 2 -2/3 1/3 Add (1 * row3) to row2 1 1 0 1 0 0 0 1 0 6 -5/3 1/3 0 0 1 2 -2/3 1/3 Add (-1 * row2) to row1 1 0 0 -5 5/3 -1/3 0 1 0 6 -5/3 1/3 0 0 1 2 -2/3 1/3 The inverse matrix can now be found in the right 3 columns of our reduced row echelon matrix Here is the inverse matrix: -5 5/3 -1/3 6 -5/3 1/3 2 -2/3 1/3 A= 1 1 0 4 3 1 2 0 5 Solution solution: A= 1 1 0 4 3 1 2 0 5 Finding the determinant of the matrix ? det A = -3 The determinant of A is non-zero, hence the inverse matrix A-1 exists. To find the inverse matrix was calculated the minors and cofactors of the matrix ? M1,1 = 3 1 0 5 = 15 C1,1 = (-1)1+1M1,1 = 15 M1,2 = 4 1 2 5 = 18 C1,2 = (-1)1+2M1,2 = -18 M1,3 = 4 3 2 0 = -6 C1,3 = (-1)1+3M1,3 = -6 M2,1 = 1 0 0 5 = 5 C2,1 = (-1)2+1M2,1 = -5 M2,2 = 1 0 2 5 = 5 C2,2 = (-1)2+2M2,2 = 5 M2,3 = 1 1 2 0 = -2 C2,3 = (-1)2+3M2,3 = 2 M3,1 = 1 0 3 1 = 1 C3,1 = (-1)3+1M3,1 = 1 M3,2 = 1 0 4 1 = 1 C3,2 = (-1)3+2M3,2 = -1 M3,3 = 1 1 4 3 = -1 C3,3 = (-1)3+3M3,3 = -1 Matrix of cofactors: C = 15 -18 -6 -5 5 2 1 -1 -1 Transpose of cofactor matrix: CT = 15 -5 1 -18 5 -1 -6 2 -1 Finding the inverse matrix. A-1 = CT = det A -5 5/3 -1/3 6 -5/3 1/3 2 -2/3 1/3 B) First, let\'s add the Identity Matrix to the right of our matrix 1 1 0 1 0 0 4 3 1 0 1 0 2 0 5 0 0 1 Now, let\'s do Gauss-Jordan Elimination on our new matrix... Add (-4 * row1) to row2 1 1 0 1 0 0 0 -1 1 -4 1 0 2 0 5 0 0 1 Add (-2 * row1) to row3 1 1 0 1 0 0 0 -1 1 -4 1 0 0 -2 5 -2 0 1 Divide row2 by -1 1 1 0 1 0 0 0 1 -1 4 -1 0 0 -2 5 -2 0 1 Add (2 * row2) to row3 1 1 0 1 0 0 0 1 -1 4 -1 0 0 0 3 6 -2 1 Divide row3 by 3 1 1 0 1 0 0 0 1 -1 4 -1 0 0 0 1 2 -2/3 1/3 Add (1 * row3) to row2 1 1 0 1 0 0 0 1 0 6 -5/3 1/3 0 0 1 2 -2/3 1/3 Add (-1 * row2) to row1 1.

solutionA= 1 1 0 4 3 1 2 0 .pdf

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Q1). Gene therapy is an experimental approach to treat the disease by inserting therapeutic genes or changing gene expression levels in the patient’s cells. Genes are nucleic acid polymers; these therapeutic genes used in gene therapy generally encode a functional gene that replaces mutated gene. Potential treatment for single gene genetic disorders such as haemophilia, muscular dystrophy, cystic fibrosis, etc. was demonstrated. Treatment of cancer, AIDS (acquired immune deficiency syndrome), Alzheimer’s disease and rheumatoid arthritis are also under active investigation. All the diseases cannot be cured by gene therapy. Some of hte good targets of the gene therapy include, 1). Diseases with known genetic defects or mutations 2). Diseases with point mutations on one or more genes 3). The possibility of delivering genes to the affected tissues 4). If the substitution of defective gene correct the disease status Solution Q1). Gene therapy is an experimental approach to treat the disease by inserting therapeutic genes or changing gene expression levels in the patient’s cells. Genes are nucleic acid polymers; these therapeutic genes used in gene therapy generally encode a functional gene that replaces mutated gene. Potential treatment for single gene genetic disorders such as haemophilia, muscular dystrophy, cystic fibrosis, etc. was demonstrated. Treatment of cancer, AIDS (acquired immune deficiency syndrome), Alzheimer’s disease and rheumatoid arthritis are also under active investigation. All the diseases cannot be cured by gene therapy. Some of hte good targets of the gene therapy include, 1). Diseases with known genetic defects or mutations 2). Diseases with point mutations on one or more genes 3). The possibility of delivering genes to the affected tissues 4). If the substitution of defective gene correct the disease status.

Q1). Gene therapy is an experimental approach to treat the disease b.pdf

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Solution : Three modes of DNA replication 1) Semi conservative: => Each daughter strand is composed of one old DNA strand and one new DNA strand. 2) Conservative: => The original double helix remains as complete unit. => The new DNA double helix is produced as a single unit. => The old DNA is completly conserved. 3) Dispersive mode: =>Separation of the individual covalent phosphodiester bonds is required for this mode of replication..

Solution Three modes of DNA replication 1) Semi conservative.pdf

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package s3; // Copy paste this Java Template and save it as \"EmergencyRoom.java\" import java.util.*; import java.io.*; // write your matric number here: // write your name here: // write list of collaborators here: // year 2016 hash code: XAbyuzR78fXeaMHBdLan (do NOT delete this line) class patientlist { private String PatientName; private int emergencyLvl; public patientlist(String PatientName, int emergencyLvl) { this.PatientName = PatientName; this.emergencyLvl = emergencyLvl; } public String getPatientName() { return PatientName; } public void setPatientName(String PatientName) { this.PatientName = PatientName; } public int getEmergencyLvl() { return emergencyLvl; } public void setEmergencyLvl(int emergencyLvl) { this.emergencyLvl = emergencyLvl; } } class EmergencyRoom { // if needed, declare a private data structure here that // is accessible to all methods in this class private List patientlist; public EmergencyRoom() { // Write necessary code during construction patientlist = new ArrayList(); } void ArriveAtHospital(String patientName, int emergencyLvl) { // You have to insert the information (patientName, emergencyLvl) // into your chosen data structure if (patientName.length() > 15 || patientName.length() < 1) { System.out.println(\"patient name is either too long or too short.\ Please enter a name between 1 to 15 characters.\"); } else if (emergencyLvl > 100 || emergencyLvl < 30) { System.out.println(\"Emergency level is either too high or too low.\ Please enter a valid level between 30 to 100.\"); } else { int i; for (i = 0; i < patientlist.size(); i++) { if (patientlist.get(i).getPatientName().compareToIgnoreCase(patientName) == 0) { System.out.println(\"Patient already admitted\"); break; } } if (i == patientlist.size()) { patientlist p = new patientlist(patientName.toUpperCase(), emergencyLvl); patientlist.add(p); } } } void UpdateEmergencyLvl(String patientName, int incEmergencyLvl) { // You have to update the emergencyLvl of patientName to // emergencyLvl += incEmergencyLvl // and modify your chosen data structure (if needed) int i; for (i = 0; i < patientlist.size(); i++) { if (patientlist.get(i).getPatientName().compareToIgnoreCase(patientName) == 0) { //System.out.println(\"Patient already admitted\"); //break; if (incEmergencyLvl > 70 || incEmergencyLvl < 0) { System.out.println(\"Emergency level is either too high or too low.\ Please enter a valid increment between 0 to 70.\"); break; } else { patientlist.get(i).setEmergencyLvl(patientlist.get(i).getEmergencyLvl() + incEmergencyLvl); } } } } void Treat(String patientName) { // This patientName is treated by the doctor // remove him/her from your chosen data structure int i; for (i = 0; i < patientlist.size(); i++) { if (patientlist.get(i).getPatientName().compareToIgnoreCase(patientName) == 0) { patientlist.remove(i); } } } String Query() { String ans = \"The emergency room is empty\"; int i; if(patientlist.isEmpty()) return ans; else{ int max=patientlist.get(0).getEme.

package s3; Copy paste this Java Template and save it as Emer.pdf

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Our body is having two line defence system against pathogens.Pathogens are micro-organisms such like virus,bactria,parasite,fungi. First line of defence system:- It is a outside defence system which protect body from the outside ,it includes skin ,tear,mucus,cilia,stomach acid,urine flow. Skin acts as a barrier between pathogens and your body. it stops all microarganisms to enter in to your body unless your skin get brokken.The pathogens can get enter through your nose ,eye,mouth, but through that if pathogen passes .Nose ,eye,mouth contains mucus,tear .they stop pathogen and break down their cell wall because they contain enzyme.those that are not killed immediately are trapped in mucus swallowed. cilia is present in our windpipe.it helps to move mucus and traped pathogens from lungs.Urine flow help to flush out pathogens from the bladder.stomach acid is present in stomach and helps to kill pathogens.this is the fine defence. Second line of defence system :- once pathogen cross your first line of defence second line of defence active for example ,through the cut pathogen enter into body and develope infection then second line defence active .Through a sequence of steps called immune response ,the immune system attacks these pathogens. In the second line of deffence system is a group of cells ,tissues,organs work together to protect the body .This is called immune system. The cells involved in the immune system called white blood cells is helps to destoys the desease causing organisms.there are different types of white blood cells but function is same that protect you from the organisms. They are neutrophiles,T-helper cells,Cytotoxic cells,marcophases,B-cells. These cells work together and kill the micro organism. Tissues and Organs :- This are involved in the immune system are the lymphatic system ,lymph nodes and lymp fluid.This are having specific function. Lymphatic system is a system of thin tube present through out the body .These tubes are called lymph vessels.they contain lymph. Lymph is a fluid that contains white blood cells. lymph helps to protect body form micro organisms. This is a second line of defence system. Solution Our body is having two line defence system against pathogens.Pathogens are micro-organisms such like virus,bactria,parasite,fungi. First line of defence system:- It is a outside defence system which protect body from the outside ,it includes skin ,tear,mucus,cilia,stomach acid,urine flow. Skin acts as a barrier between pathogens and your body. it stops all microarganisms to enter in to your body unless your skin get brokken.The pathogens can get enter through your nose ,eye,mouth, but through that if pathogen passes .Nose ,eye,mouth contains mucus,tear .they stop pathogen and break down their cell wall because they contain enzyme.those that are not killed immediately are trapped in mucus swallowed. cilia is present in our windpipe.it helps to move mucus and traped pathogens from lungs.Urine flow help to flush out.

Our body is having two line defence system against pathogens.Pathoge.pdf

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Once neurons are produced, they migrate and modify to form six layered structure of cortex. The earliest formed neurons form the preplate which is then split into marginal plate and subplate by invading newly formed neurons. The neurons can have two origin: cortical ventricular zone and ganglionic eminences. The corticular neurons exhibit radial migration and form projection neurons while the ganglionic neurons exhibit tangantial migration and form interneurons. Both types arrange in inside out sequence and the early neurons form the deep layers while the late born neuron form the outer layers of cortex. Solution Once neurons are produced, they migrate and modify to form six layered structure of cortex. The earliest formed neurons form the preplate which is then split into marginal plate and subplate by invading newly formed neurons. The neurons can have two origin: cortical ventricular zone and ganglionic eminences. The corticular neurons exhibit radial migration and form projection neurons while the ganglionic neurons exhibit tangantial migration and form interneurons. Both types arrange in inside out sequence and the early neurons form the deep layers while the late born neuron form the outer layers of cortex..

Once neurons are produced, they migrate and modify to form six layer.pdf

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1.A hole is the absence of an electron in a particular place in an atom.Electrons and holes do not move as rapidly through the solid materials as electrons do in a vaccum.The holes move more slowly than the electrons.The holes and electrons move in opposite direction. 2.Here if the electric current moving from left to right ,than the holes will be moving in opposite direction that is from right to left. Regardless of the semiconductor type ,with the potencial applied there will be net effect.Both electrons and holes contribute to total current flow,altough they are moving in opposite directions,they are of opposite charge. Solution 1.A hole is the absence of an electron in a particular place in an atom.Electrons and holes do not move as rapidly through the solid materials as electrons do in a vaccum.The holes move more slowly than the electrons.The holes and electrons move in opposite direction. 2.Here if the electric current moving from left to right ,than the holes will be moving in opposite direction that is from right to left. Regardless of the semiconductor type ,with the potencial applied there will be net effect.Both electrons and holes contribute to total current flow,altough they are moving in opposite directions,they are of opposite charge..

1.A hole is the absence of an electron in a particular place in an a.pdf

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Mitochondrial DNA (mtDNA )is passed only from the females to both their children. So we see that only the members from the 1&2 i.e. the females can contribult their mtDNA for solving the mystery. From 1&2 (1), it is passed to 1-female, 3-female, 5-female, 6-male 1 gets married to 2, and has2 children-Female (1)and Male (3), sine mtDNA is passed only from the females, therefore 3 will no mor pass it on. It will be passed only by 1 to her chldren, and then to their children (they are girls too).For Generation IV, generation I is their great grand parents and 2 is their great-grand mother. For generation V-generation II is their great grand parents and in II-1 is greatgrandmother for 1&2 and ^3&4, 5&6-Generatyion V. 3-Generation II is greatgrand mother for 7, 9,11,12,14. 7-generation II is greatgrand mother for 15,17,19,20,22 (Generation V) 1&2-Generation VI Generation III-1 is greatgrandmother, and for 3 generation VI 4-generation III is greatgrand mother, for 4&5 &6 generation VI -8 is greatgrandmother-noe dead. For 7&8, generation VI-10 Generation 3 is their greatgrandmother.for 9, generation VI 13 is their gteatgrandmother. Solution Mitochondrial DNA (mtDNA )is passed only from the females to both their children. So we see that only the members from the 1&2 i.e. the females can contribult their mtDNA for solving the mystery. From 1&2 (1), it is passed to 1-female, 3-female, 5-female, 6-male 1 gets married to 2, and has2 children-Female (1)and Male (3), sine mtDNA is passed only from the females, therefore 3 will no mor pass it on. It will be passed only by 1 to her chldren, and then to their children (they are girls too).For Generation IV, generation I is their great grand parents and 2 is their great-grand mother. For generation V-generation II is their great grand parents and in II-1 is greatgrandmother for 1&2 and ^3&4, 5&6-Generatyion V. 3-Generation II is greatgrand mother for 7, 9,11,12,14. 7-generation II is greatgrand mother for 15,17,19,20,22 (Generation V) 1&2-Generation VI Generation III-1 is greatgrandmother, and for 3 generation VI 4-generation III is greatgrand mother, for 4&5 &6 generation VI -8 is greatgrandmother-noe dead. For 7&8, generation VI-10 Generation 3 is their greatgrandmother.for 9, generation VI 13 is their gteatgrandmother..

Mitochondrial DNA (mtDNA )is passed only from the females to both th.pdf

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1. What roles does succession and phytoremediation play in ecology, evolution and the ecosystems? Succession plays a major role to change in environmental conditions that causes a pronounced change in an ecosystem. Disturbances often act quickly and with great effect, to alter the physical structure or arrangement of biotic and abiotic elements. Disturbance can also occur over a long period of time and can impact the diversity within an ecosystem. Major ecological disturbances may include fires, flooding, windstorms, insect outbreaks and trampling. Earthquakes, various types of volcanic eruptions, tsunami, firestorms, impact events, climate change, and the devastating effects of human impact on the environment (anthropogenic disturbances) such as clear cutting, forest clearing and the introduction of invasive species can be considered major disturbances. This can be attributed to physical changes in the biotic and abiotic conditions of an ecosystem. Because of this, a disturbance force can change an ecosystem for significantly longer than the period over which the immediate effects persist. With the passage of time following a disturbance, shifts in dominance, shifts in dominance may occur with ephemeral herbaceous life- forms progressively becoming over topped by taller perennials herbs, shrubs and trees. However, in the absence of further disturbance forces, many ecosystems will trend back toward pre- disturbance conditions. Long lived species and those which can regenerate in the presence of their own adults will finally become dominant. Such alteration, accompanied by changes in the abundance of different species over time, is called ecological succession. Succession often leads to conditions that will once again predispose an ecosystem to disturbance. Phytoremediation play a major role to clean up contaminated environments including metals, pesticides, explosives, and oil. Also help prevent wind, rain, and groundwater flow from carrying contaminants away from the site to surrounding areas or deeper underground. Certain plants are able to remove or break down harmful chemicals from the ground when their roots take in water and nutrients from the contaminated soil, sediment, or groundwater. Plants can help clean up contaminants as deep as their roots can reach using natural processes to: • Store the contaminants in the roots, stems, or leaves. • Convert them to less harmful chemicals within the plant or, more commonly, the root zone. • Convert them to vapors, which are released into the air. • Sorb (stick) contaminants onto their roots where very small organisms called “microbes” (such as bacteria) that live in the soil break down the sorbed contaminants to less harmful chemicals. 2. Biogeochemical cycles, succession and phytoremediation. Explain how all three work together for a positive outcome. Biological diversity is dependent on natural disturbance. The success of a wide range of species from all taxonomic groups is closely tied to na.

1. What roles does succession and phytoremediation play in ecology, .pdf

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1. JUnit test for method toString() in SalariedEmployee - import org.junit.Assert; import org.junit.Test; import org.junit.runner.RunWith; import org.mockito.runners.MockitoJUnitRunner; @RunWith(MockitoJUnitRunner.class) public class SalariedEmployeeTest { @Test public void testToString(){ String name=\"name\"; double salary=3000; double bonus=1000; SalariedEmployee employee= new SalariedEmployee(name,salary,bonus); employee.toString(); } } 2. JAva program to replace \"is\" with \"was\" - import java.io.BufferedReader; import java.io.BufferedWriter; import java.io.File; import java.io.FileReader; import java.io.FileWriter; import java.io.IOException; public class StringReplace { public static void main(String[] args) throws IOException { BufferedReader br = null; File file = new File(\"/My Stuffs/Root/Workspace/TestJava/src/com/test/output.txt\"); FileWriter fw = new FileWriter(file.getAbsoluteFile()); BufferedWriter bw = new BufferedWriter(fw); if (!file.exists()) { file.createNewFile(); } try { String sCurrentLine; br = new BufferedReader(new FileReader(\"/My Stuffs/Root/Workspace/TestJava/src/com/test/testing.txt\")); while ((sCurrentLine = br.readLine()) != null) { System.out.println(sCurrentLine); String rep=sCurrentLine.replaceAll(\" is \",\" was \"); System.out.println(rep); bw.write(rep); } } catch (IOException e) { e.printStackTrace(); } finally { try { if (br != null)br.close(); if(bw!=null)bw.close(); } catch (IOException ex) { ex.printStackTrace(); } } } } Solution 1. JUnit test for method toString() in SalariedEmployee - import org.junit.Assert; import org.junit.Test; import org.junit.runner.RunWith; import org.mockito.runners.MockitoJUnitRunner; @RunWith(MockitoJUnitRunner.class) public class SalariedEmployeeTest { @Test public void testToString(){ String name=\"name\"; double salary=3000; double bonus=1000; SalariedEmployee employee= new SalariedEmployee(name,salary,bonus); employee.toString(); } } 2. JAva program to replace \"is\" with \"was\" - import java.io.BufferedReader; import java.io.BufferedWriter; import java.io.File; import java.io.FileReader; import java.io.FileWriter; import java.io.IOException; public class StringReplace { public static void main(String[] args) throws IOException { BufferedReader br = null; File file = new File(\"/My Stuffs/Root/Workspace/TestJava/src/com/test/output.txt\"); FileWriter fw = new FileWriter(file.getAbsoluteFile()); BufferedWriter bw = new BufferedWriter(fw); if (!file.exists()) { file.createNewFile(); } try { String sCurrentLine; br = new BufferedReader(new FileReader(\"/My Stuffs/Root/Workspace/TestJava/src/com/test/testing.txt\")); while ((sCurrentLine = br.readLine()) != null) { System.out.println(sCurrentLine); String rep=sCurrentLine.replaceAll(\" is \",\" was \"); System.out.println(rep); bw.write(rep); } } catch (IOException e) { e.printStackTrace(); } finally { try { if (br != null)br.close(); if(bw!=null)bw.close(); } catch (IOException ex) { ex.printStackTrace(); } } } }.

1. JUnit test for method toString() in SalariedEmployee -import or.pdf

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Lee De Forest, also known as \"Father Of Radio\", was an American inventor with more than 180 patents. During 1890s when Marconi was working on Radio his machine has drawbacks it was slow, insensitive, and not very reliable. De Forest at the same time was determined to build much reliable radio that can work fast as well. He tried getting job under Tesla and Marconi but after being rejected he started on his own, his first job after garduating from Yale was in Western Electric Company\'s telephone lab in Chicago where he started building his first Radio reciever. With Radio research in his mind he went to Lewis institute for a night teaching position to get chance to conduct his experiments at Armour Institute.During 1900 he has extended the range of his Radio transmitter upto 7km by using a Spark Coil Transmitter. De Forest went to New York in 1901 to compete directly with Marconi but actually failed to do so as during the demonstration of his machine, it stopped working and it was a complete failure. Despite of this he didn\'t lose hope and stayed in New York to raise other\'s interest in his idea and in 1902 he did got his promoter. During the time of 1902 and 1903 he developed many types of Radio detectors each was an improvement upon the other. In the coming years he also made a company called DeForest Wireless Telegraph Company but has some conflict with the company\'s management complaining limited support for research and ultimately he resigned from the company. During year 1906 and 1907 he developed new Sparkless Arc Transmitter which in an experimental transsmission used to transmit music.Many of his initial broadcasting experiments were successful but not good enough yet to commercialise. Solution Lee De Forest, also known as \"Father Of Radio\", was an American inventor with more than 180 patents. During 1890s when Marconi was working on Radio his machine has drawbacks it was slow, insensitive, and not very reliable. De Forest at the same time was determined to build much reliable radio that can work fast as well. He tried getting job under Tesla and Marconi but after being rejected he started on his own, his first job after garduating from Yale was in Western Electric Company\'s telephone lab in Chicago where he started building his first Radio reciever. With Radio research in his mind he went to Lewis institute for a night teaching position to get chance to conduct his experiments at Armour Institute.During 1900 he has extended the range of his Radio transmitter upto 7km by using a Spark Coil Transmitter. De Forest went to New York in 1901 to compete directly with Marconi but actually failed to do so as during the demonstration of his machine, it stopped working and it was a complete failure. Despite of this he didn\'t lose hope and stayed in New York to raise other\'s interest in his idea and in 1902 he did got his promoter. During the time of 1902 and 1903 he developed many types of Radio detectors each was an improvement upon.

Lee De Forest, also known as Father Of Radio, was an American in.pdf

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import java.io.; import java.util.Scanner; to run on Athena .pdf

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//g++ -o simpleVector.exe simpleVector.cpp #include #include #include using namespace std; /****************************************************** * Template class named SimpleVector * An array that can hold any specified * data type. Use a dynamic array. ******************************************************/ template class SimpleVector { public: //default constructor SimpleVector(); //constructor with size of array SimpleVector(int size); //copy constructor SimpleVector(const SimpleVector& object); //setter method void SetArray(int size); //getter method void ShowArray(int size); //destructor void FreeMemory(); //accessor method int GetSizeOfArray(); //getter method T GetElementAt(int index); //Overload the [] operator. The argument is a subscript. //This function returns a reference //to the element in the array indexed by the subscript. T operator [] (int subscript); private: //private memeber variable T *array; //to dynamically allocate an array T *copy; }; //These function are going to be used in switch statement //in order to test each case(integer, double, string). void caseOne(int userInput, int size); void caseTwo(int userInput, int size); void caseThree(int userInput, int size); int main() { int userInput; int size; char keepInput; do { cout << \"What type of data do you want?\ \"; cout <<\"1. intger\ \"; cout <<\"2. double\ \"; cout <<\"3. strings\ \"; cin >> userInput; cout << \"Enter the number of data \"; cin >> size; switch(userInput) { case 1: { caseOne(userInput, size); break; } case 2: { caseTwo(userInput, size); break; } case 3: { caseThree(userInput, size); break; } default: cout << \"invalid input\"; } cout << \"\ Do you want to enter the data again?(Y/N)\ \"; cin >> keepInput; } while(toupper(keepInput) == \'Y\'); return 0; } //This funciton is called in the switch satatement //The purpose of this funtion is test SimpleVecor calss //funciton for \'int\' type. void caseOne(int userInput, int size) { SimpleVectorintegerData(size); integerData.SetArray(size); integerData.ShowArray(size); cout << \"\ Enter an index to retrieve the data\ \"; cin >> userInput; cout << \"\ Getting an element of array by \ \"; cout << \"1. GetElementAt function : \"; cout << integerData.GetElementAt(userInput); cout << \"\ 2. overloaded operator [] : \"; cout << integerData[userInput]; cout << \"\ \ Copying the array\"; SimpleVectornewIntegerData(integerData); newIntegerData.ShowArray(size); newIntegerData.FreeMemory(); } //This funciton is called in the switch satatement //The purpose of this funtion is test SimpleVecor calss //funciton for \'double\' type. void caseTwo(int userInput, int size) { SimpleVectordoubleData(size); doubleData.SetArray(size); doubleData.ShowArray(size); cout << \"\ Enter an index to retrieve the data\ \"; cin >> userInput; cout << \"\ Getting an element of array by \ \"; cout << \"1. GetElementAt function : \"; cout << doubleData.GetElementAt(userInput); cout << \"\ 2. overloaded operator [] : \"; cout << doubleData[.

g++ -o simpleVector.exe simpleVector.cpp #include stdio.h #i.pdf

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Differences in the global strategies organizations can use to their global Business. Strategies are. Differences of each strategies Shown in below. Domestic strategy is characterized by heavy centralization of corporate activities in the home country of origin. International sales were sometimes dispersed using agency agency agreements or subsideries but even foreign marketing relies on the domestic homes base for marketing themes. Caterpilar corporation and other heavy capital equipment manufactures fall into this category of firm. Multinational, This strategy concentrates financial management and control out of a central home base while decentralizing production and marketing operations to units in other countries. Many financial service firms along with a host of manufacturers, such as General Motors, Chryler and Intel, fit this pattern. Franchisers - : It was an interesting mix of old and new. One hand, the product was maded, designed, financed and initially produced in the home country. McDonald’s created a new form of fast-food chain in the United States and continues to rely largely on the US for inspiration of new products, strategic management and financing. Transnational -: Stateless, truly globally managed firms that may represent a larger part of international business in the future. The transnational strategy, nearly all the value adding activities are managed from a global perspective without reference to national borders. Business Function Domestic Exporter Multinational Franchiser Transnational Production Centralized Dispersed Coordinated Coordinated Finance/Accounting Centralized Centralized Centralized Coordinated Sales/Marketing Mixed Dispersed Coordinated Coordinated Human Resources Centralized Centralized Coordinated Coordinated Strategic Management Centralized Centralized Centralized Coordinated Information system strategies work best for each business strategy *) The ‘a’ show the dominant patterns and the ‘b’ show the emerging patterns. For instance, domestic exportes rely predominantly on centralized systems and some development of decentralized systems in local marketing regions. Centralized system-: system development and operation occur totally at the domestic home base. Duplicated system - : Development occurs at the home base but operations are handed over to autonomous units in foreign locations. Decentralized system - : Which each foreign unit designs its own unique solutions and systems. Networked System -: System developments and operations occur in an integrated and coordinated fashion across all units. Business Function Domestic Exporter Multinational Franchiser Transnational Production Centralized Dispersed Coordinated Coordinated Finance/Accounting Centralized Centralized Centralized Coordinated Sales/Marketing Mixed Dispersed Coordinated Coordinated Human Resources Centralized Centralized Coordinated Coordinated Strategic Management Centralized Centralized Centralized Coordinated Solution Differences in t.

Differences in the global strategies organizations can use to their .pdf

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Dengue fever, a mosquito borne viral disease is caused by 1 of 4 closely related serotypes of dengue virus. These serotypes are antigenically distinct namely DENV-1 through DEN-4. Infection with any one type confers a very short period of partial heterotypic immunity but lifelong homotypic immunity. During an epidemic several serotypes can be in circulation. In severe infection, lactic acidosis, hypoglycaemia, hypocalcemia and eventual multiple organ failure occurs. Viral damage and subsequent tissue inflammation culminates myocarditis, encephalopathy and liver necrosis. Solution Dengue fever, a mosquito borne viral disease is caused by 1 of 4 closely related serotypes of dengue virus. These serotypes are antigenically distinct namely DENV-1 through DEN-4. Infection with any one type confers a very short period of partial heterotypic immunity but lifelong homotypic immunity. During an epidemic several serotypes can be in circulation. In severe infection, lactic acidosis, hypoglycaemia, hypocalcemia and eventual multiple organ failure occurs. Viral damage and subsequent tissue inflammation culminates myocarditis, encephalopathy and liver necrosis..

Dengue fever, a mosquito borne viral disease is caused by 1 of 4 clo.pdf

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