C is correct. Only the host that the unicast message is addressed to.pdf

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A

ANJALIENTERPRISES1

mol L-1 s-1 is the answer Solution mol L-1 s-1 is the answer.

C is correct. Only the host that the unicast message is addressed to will open and process the
message.
A, B, and D are incorrect. All hosts receive the message since they are all connected to a Layer 1
device. Only Host B will open and process the message since the destination address field
matches its MAC address.
Solution
C is correct. Only the host that the unicast message is addressed to will open and process the
message.
A, B, and D are incorrect. All hosts receive the message since they are all connected to a Layer 1
device. Only Host B will open and process the message since the destination address field
matches its MAC address.

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17-The Y-Chromosome DNA testing helps in the examination of the male specific portion of the biological evidence. This is useful where small amount of male DNA is found amidst a large amount of female DNA , as in the sexual assault cases.It can also help in the missing person investigation as it extends the range of potential reference samples. Sine the Y chromosome is passed on from the fathers to their sone (unchanged, except in the case of some mutations), therefore all the males of the same paternal lineage will possess common Y chromosome haplotype. A set of Short tandem repeats (STR) are found on the male specific Y chromosome.The coding genes found on the short arm of the Ychromosome are vital to male sex determination, spermatogenesis andother male related functions. The Y-STRs are polymorphic among the nonrelated males. Sometimes when sepoarating the mal ecomponent and the female component becomes difficult, then the YSTRs help in the determination. Since The Y-STRs are not present in the females, therefore in a case where the male and female are involved, it becomes clear that as the Y-STRs are present the male component is easily detected as only this part will be amplified in the PCR. The YSTRs are also helpful when a number of males are involved. As regular STRs cause masking effect the YSTR will help to identify all the males who have contributed to the evidence. ADVANTAGES- 1. extremely discriminating. 2. Extremely sensitive.e. low amounts of DNA can be used for determination. 3. Not onlu body fluids, but alsp touched materials or handled surfaces, where skinn cells ahev been left can be used for DNA profiling. 4. Multiple testing systems ar available, which increases the likelyhood of successfully generating DNA profiles. 5. Apart from standart=d STR enem mt DNA or YSTRs are used for profiling. 6. With the presence of the current technologies, from amidst mixed DNA samples, the DNA of a particular individual can be profiled. LIMITATIONS 1. Random match probability cannot be excluded. 2. Sometimes even when crime is not committed, we do get ,low levels of DNA on surfaces. 3.The important question while dealing with low DNA quantity is that whether the DNA that was deposited was deposited during the crime or at some other time. 4.Searching evidence is challenging. 5.Sometimes finding the evidence is altogether impossibl. 6. In case of mtDNA /T STRs , as they are passed on from generation to generation either from fathers to sons or mothers to daughters and son, therefore sometimes unrelated persons are also share the same profile. 7. DNA mixtures are very complex. 8.Interpretation is required in many cases. 16- from Cyanobacterial to humans, many terristrial organisms have acquired the circadian rhythms to adapt to sunlight to increase their survival rate.Circadian Rhythm is a biological clock that displays an endogenous, entrainable oscillation of 24 hours.Thesse oscillations are driven by circadian clocks. PULSE CHASE EXPERIME.

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#include #include #include #define MAX_STR_LEN 80 //decalring the structure struct link_node { //declaring the variables char node_str[ MAX_STR_LEN ]; struct link_node *next; }; //method declaration typedef struct link_node link_node; struct link_node *add_node( struct link_node *list,struct link_node *node ); int compare_node( struct link_node *n1, struct link_node *n2 ); void display_list( struct link_node *head ); //implementing the method main int main() { //declaring the main variables link_node * head=NULL,*temp; char str[MAX_STR_LEN]; do { //getting the STRING input printf(\"Enter the string : \"); gets(str); //malloc() is implemented for memory temp=(link_node*)malloc(sizeof(link_node)); temp->next=NULL; strcpy(temp->node_str,str); head=add_node(head,temp); }while(strlen(str)>1); display_list(head); return 0; } //method definition for compare node int compare_node( struct link_node *n1, struct link_node *n2 ) { if(strcmp(n1->node_str,n2->node_str)==0) return 0; if(strcmp(n1->node_str,n2->node_str)<0) return -1; else return 1; } //method definition for adding the node struct link_node *add_node( struct link_node *list,struct link_node *node ) { link_node *temp=list; if(list==NULL) { return node; } if(compare_node(node,list)==-1) { node->next=list; list=node; return list; } else { link_node *prev=list; while(temp!=NULL&&compare_node(node,temp)>=0) { prev=temp; temp=temp->next; } prev->next=node; node->next=temp; return list; } } //method definition for displaying the list void display_list( struct link_node *head ) { link_node *temp; while(head) { printf(\"%s\ \",head->node_str); temp=head; head=head->next; free(temp); } } Sample output: Enter the string: Geneva America Solution #include #include #include #define MAX_STR_LEN 80 //decalring the structure struct link_node { //declaring the variables char node_str[ MAX_STR_LEN ]; struct link_node *next; }; //method declaration typedef struct link_node link_node; struct link_node *add_node( struct link_node *list,struct link_node *node ); int compare_node( struct link_node *n1, struct link_node *n2 ); void display_list( struct link_node *head ); //implementing the method main int main() { //declaring the main variables link_node * head=NULL,*temp; char str[MAX_STR_LEN]; do { //getting the STRING input printf(\"Enter the string : \"); gets(str); //malloc() is implemented for memory temp=(link_node*)malloc(sizeof(link_node)); temp->next=NULL; strcpy(temp->node_str,str); head=add_node(head,temp); }while(strlen(str)>1); display_list(head); return 0; } //method definition for compare node int compare_node( struct link_node *n1, struct link_node *n2 ) { if(strcmp(n1->node_str,n2->node_str)==0) return 0; if(strcmp(n1->node_str,n2->node_str)<0) return -1; else return 1; } //method definition for adding the node struct link_node *add_node( struct link_node *list,struct link_node *node ) { link_node *temp=list; if(list==NULL) { return node; } if(compare_node(node,list)==-1) { node->next=list; list=no.

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ANJALIENTERPRISES1

The pH and pOH of a solution are defined as: pH = -log[H+] pOH = -log[OH-] and pH + pOH = 14.00 [H+] = 0.055 Conversions: [OH-] = 1.00 x 10^-14 / [H+] = 1.00 x 10-14 /0.055= 1.8x10^-13 pOH = -log[OH-]=12.74 Solution The pH and pOH of a solution are defined as: pH = -log[H+] pOH = -log[OH-] and pH + pOH = 14.00 [H+] = 0.055 Conversions: [OH-] = 1.00 x 10^-14 / [H+] = 1.00 x 10-14 /0.055= 1.8x10^-13 pOH = -log[OH-]=12.74.

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The Br was originally neutral, but picks up an extra electron from the N and becomes Br-. The Br- leaves the molecule The O was originally negative, but gives up an electron to form a double bond with C1 and becomes neutral. Leaving O=C=N-C as a neutral molecule. Keeping the labeling of the two carbons it would be O=C1=N-C2. With the double bonds on both sides of C1, the O, C1, N backbone is linear. Solution The Br was originally neutral, but picks up an extra electron from the N and becomes Br-. The Br- leaves the molecule The O was originally negative, but gives up an electron to form a double bond with C1 and becomes neutral. Leaving O=C=N-C as a neutral molecule. Keeping the labeling of the two carbons it would be O=C1=N-C2. With the double bonds on both sides of C1, the O, C1, N backbone is linear..

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Year 2000 is the period when the productivity growth rate was at minimum ie., 0.7. Degrowth refers to those years where productivity growth rate was in negative. Minimum growth rate refers to the minimum positive growth rate achieved. Solution Year 2000 is the period when the productivity growth rate was at minimum ie., 0.7. Degrowth refers to those years where productivity growth rate was in negative. Minimum growth rate refers to the minimum positive growth rate achieved..

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Solution IDENameCostOpen SourceDestinationNETBEANSFreeyescross platform IDEINTELLIJ IDEA$199noit has great moderen UIECLIPSEfreeyesit features huge library of pluginsJDEVELOPERfreeyesit has fully loades tool set for creating javaDR. JAVAFreeyesit supports cross platformBLUEJFreeyesit suitable for small software projectsJCREATORfreeyesit is fast and efficient ideJGRASPfreeyesit is super lightweight ideGREENFOOTfreeyesit supports designing for 2D games CODEENVY$1(month)nocloud based ide.

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UChicago CMSC 23320 - The Best Commit Messages of 2024

UChicago CMSC 23320 - The Best Commit Messages of 2024

UChicago CMSC 23320 - The Best Commit Messages of 2024

TỔNG HỢP HƠN 100 ĐỀ THI THỬ TỐT NGHIỆP THPT TOÁN 2024 - TỪ CÁC TRƯỜNG, TRƯỜNG...

TỔNG HỢP HƠN 100 ĐỀ THI THỬ TỐT NGHIỆP THPT TOÁN 2024 - TỪ CÁC TRƯỜNG, TRƯỜNG...

TỔNG HỢP HƠN 100 ĐỀ THI THỬ TỐT NGHIỆP THPT TOÁN 2024 - TỪ CÁC TRƯỜNG, TRƯỜNG...

OS-operating systems- ch05 (CPU Scheduling) ...

OS-operating systems- ch05 (CPU Scheduling) ...

OS-operating systems- ch05 (CPU Scheduling) ...

DEMONSTRATION LESSON IN ENGLISH 4 MATATAG CURRICULUM

DEMONSTRATION LESSON IN ENGLISH 4 MATATAG CURRICULUM

DEMONSTRATION LESSON IN ENGLISH 4 MATATAG CURRICULUM

Graduate Outcomes Presentation Slides - English (v3).pptx

Graduate Outcomes Presentation Slides - English (v3).pptx

Graduate Outcomes Presentation Slides - English (v3).pptx

Major project report on Tata Motors and its marketing strategies

Major project report on Tata Motors and its marketing strategies

Major project report on Tata Motors and its marketing strategies

diagnosting testing bsc 2nd sem.pptx....

diagnosting testing bsc 2nd sem.pptx....

diagnosting testing bsc 2nd sem.pptx....

Đề tieng anh thpt 2024 danh cho cac ban hoc sinh

Đề tieng anh thpt 2024 danh cho cac ban hoc sinh

Đề tieng anh thpt 2024 danh cho cac ban hoc sinh

SPLICE Working Group:Reusable Code Examples

SPLICE Working Group:Reusable Code Examples

SPLICE Working Group:Reusable Code Examples

SURVEY I created for uni project research

SURVEY I created for uni project research

SURVEY I created for uni project research

Observing-Correct-Grammar-in-Making-Definitions.pptx

Observing-Correct-Grammar-in-Making-Definitions.pptx

Observing-Correct-Grammar-in-Making-Definitions.pptx

會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽

會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽

會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽

An Overview of the Odoo 17 Knowledge App

An Overview of the Odoo 17 Knowledge App

An Overview of the Odoo 17 Knowledge App

Including Mental Health Support in Project Delivery, 14 May.pdf

Including Mental Health Support in Project Delivery, 14 May.pdf

Including Mental Health Support in Project Delivery, 14 May.pdf

male presentation...pdf.................

male presentation...pdf.................

male presentation...pdf.................

Mattingly "AI & Prompt Design: Named Entity Recognition"

Mattingly "AI & Prompt Design: Named Entity Recognition"

Mattingly "AI & Prompt Design: Named Entity Recognition"

e-Sealing at EADTU by Kamakshi Rajagopal

e-Sealing at EADTU by Kamakshi Rajagopal

e-Sealing at EADTU by Kamakshi Rajagopal

When Quality Assurance Meets Innovation in Higher Education - Report launch w...

When Quality Assurance Meets Innovation in Higher Education - Report launch w...

When Quality Assurance Meets Innovation in Higher Education - Report launch w...

C is correct. Only the host that the unicast message is addressed to.pdf

1. C is correct. Only the host that the unicast message is addressed to will open and process the message. A, B, and D are incorrect. All hosts receive the message since they are all connected to a Layer 1 device. Only Host B will open and process the message since the destination address field matches its MAC address. Solution C is correct. Only the host that the unicast message is addressed to will open and process the message. A, B, and D are incorrect. All hosts receive the message since they are all connected to a Layer 1 device. Only Host B will open and process the message since the destination address field matches its MAC address.