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to minimise the length, term with \'m\' should be reduced so first to absolute values must be opened with negative sign S2 = -4m+1 - 5m+6 + 10m-3 S2 = m+4 Solution to minimise the length, term with \'m\' should be reduced so first to absolute values must be opened with negative sign S2 = -4m+1 - 5m+6 + 10m-3 S2 = m+4.
H2SO4 may act as 1) an acidexample 2NaOH + H2SO4 - Na2SO4 +.pdf
H2SO4 may act as 1) an acidexample 2NaOH + H2SO4 - Na2SO4 +.pdf
ANJALIENTERPRISES1
no. of sand grains in 1 truck = 10/0.1 x 10 ^-6=10^8 sand grains total time taken= (10 X 6.02x10^23 x 10 )/10^8= 6.02 x 10^17 min= 6.872 x 10^ 13 years Solution no. of sand grains in 1 truck = 10/0.1 x 10 ^-6=10^8 sand grains total time taken= (10 X 6.02x10^23 x 10 )/10^8= 6.02 x 10^17 min= 6.872 x 10^ 13 years.
Computer graphics are pictures and movies produced use computers fre.pdf
Computer graphics are pictures and movies produced use computers fre.pdf
ANJALIENTERPRISES1
NaCl - sodium chloride MgO - magnesium oxide AgI - silver iodiode check its not L CuSO4 - copper sulphate Solution NaCl - sodium chloride MgO - magnesium oxide AgI - silver iodiode check its not L CuSO4 - copper sulphate.
Circle.javaimport java.text.DecimalFormat;public class Circle {.pdf
Circle.javaimport java.text.DecimalFormat;public class Circle {.pdf
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.pdf
C is correct. Only the host that the unicast message is addressed to.pdf
ANJALIENTERPRISES1
empirical formula of C2H4 is CH2 empirical formula mass = 12+2 = 14 Solution empirical formula of C2H4 is CH2 empirical formula mass = 12+2 = 14.
According to the seriousness level, the following organs areThymu.pdf
According to the seriousness level, the following organs areThymu.pdf
ANJALIENTERPRISES1
First substance = Ionic solid Second substance = Covalent solid Solution First substance = Ionic solid Second substance = Covalent solid.
Among 40 subjects randomly choose 20 subjects and assign themSol.pdf
Among 40 subjects randomly choose 20 subjects and assign themSol.pdf
ANJALIENTERPRISES1
ex-1-enol: CH3 - CH2 - CH2 - CH2 - CH2 = CH2-OH hexanal CH3 - CH2 - CH2 - CH2 - CH2 - CH2=O Solution ex-1-enol: CH3 - CH2 - CH2 - CH2 - CH2 = CH2-OH hexanal CH3 - CH2 - CH2 - CH2 - CH2 - CH2=O.
AdvantagesThe main objective of business combination is to elimina.pdf
AdvantagesThe main objective of business combination is to elimina.pdf
ANJALIENTERPRISES1
coordination no is the no of ligands in the coordination sphere and in this case it is 6 Solution coordination no is the no of ligands in the coordination sphere and in this case it is 6.
a. Germ cell speciication in Drosophila is primarily a maternlly con.pdf
a. Germ cell speciication in Drosophila is primarily a maternlly con.pdf
ANJALIENTERPRISES1
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to minimise the length, term with \'m\' should be reduced so first to absolute values must be opened with negative sign S2 = -4m+1 - 5m+6 + 10m-3 S2 = m+4 Solution to minimise the length, term with \'m\' should be reduced so first to absolute values must be opened with negative sign S2 = -4m+1 - 5m+6 + 10m-3 S2 = m+4.
H2SO4 may act as 1) an acidexample 2NaOH + H2SO4 - Na2SO4 +.pdf
H2SO4 may act as 1) an acidexample 2NaOH + H2SO4 - Na2SO4 +.pdf
ANJALIENTERPRISES1
no. of sand grains in 1 truck = 10/0.1 x 10 ^-6=10^8 sand grains total time taken= (10 X 6.02x10^23 x 10 )/10^8= 6.02 x 10^17 min= 6.872 x 10^ 13 years Solution no. of sand grains in 1 truck = 10/0.1 x 10 ^-6=10^8 sand grains total time taken= (10 X 6.02x10^23 x 10 )/10^8= 6.02 x 10^17 min= 6.872 x 10^ 13 years.
Computer graphics are pictures and movies produced use computers fre.pdf
Computer graphics are pictures and movies produced use computers fre.pdf
ANJALIENTERPRISES1
NaCl - sodium chloride MgO - magnesium oxide AgI - silver iodiode check its not L CuSO4 - copper sulphate Solution NaCl - sodium chloride MgO - magnesium oxide AgI - silver iodiode check its not L CuSO4 - copper sulphate.
Circle.javaimport java.text.DecimalFormat;public class Circle {.pdf
Circle.javaimport java.text.DecimalFormat;public class Circle {.pdf
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.pdf
C is correct. Only the host that the unicast message is addressed to.pdf
ANJALIENTERPRISES1
empirical formula of C2H4 is CH2 empirical formula mass = 12+2 = 14 Solution empirical formula of C2H4 is CH2 empirical formula mass = 12+2 = 14.
According to the seriousness level, the following organs areThymu.pdf
According to the seriousness level, the following organs areThymu.pdf
ANJALIENTERPRISES1
First substance = Ionic solid Second substance = Covalent solid Solution First substance = Ionic solid Second substance = Covalent solid.
Among 40 subjects randomly choose 20 subjects and assign themSol.pdf
Among 40 subjects randomly choose 20 subjects and assign themSol.pdf
ANJALIENTERPRISES1
ex-1-enol: CH3 - CH2 - CH2 - CH2 - CH2 = CH2-OH hexanal CH3 - CH2 - CH2 - CH2 - CH2 - CH2=O Solution ex-1-enol: CH3 - CH2 - CH2 - CH2 - CH2 = CH2-OH hexanal CH3 - CH2 - CH2 - CH2 - CH2 - CH2=O.
AdvantagesThe main objective of business combination is to elimina.pdf
AdvantagesThe main objective of business combination is to elimina.pdf
ANJALIENTERPRISES1
coordination no is the no of ligands in the coordination sphere and in this case it is 6 Solution coordination no is the no of ligands in the coordination sphere and in this case it is 6.
a. Germ cell speciication in Drosophila is primarily a maternlly con.pdf
a. Germ cell speciication in Drosophila is primarily a maternlly con.pdf
ANJALIENTERPRISES1
At the beginning of the 11th year, there would be 1,024,000 rabbits. Solution At the beginning of the 11th year, there would be 1,024,000 rabbits..
A is correct. The packets to be filtered would be heading into the r.pdf
A is correct. The packets to be filtered would be heading into the r.pdf
ANJALIENTERPRISES1
1). A) ipsilateral Below the point of spinal cord, the nerve paths belong to the same side (no crossing over occurs). So, the nociceptive information can be found on the same side of the body. Solution 1). A) ipsilateral Below the point of spinal cord, the nerve paths belong to the same side (no crossing over occurs). So, the nociceptive information can be found on the same side of the body..
1). A) ipsilateralBelow the point of spinal cord, the nerve paths .pdf
1). A) ipsilateralBelow the point of spinal cord, the nerve paths .pdf
ANJALIENTERPRISES1
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.
17-The Y-Chromosome DNA testing helps in the examination of the male.pdf
17-The Y-Chromosome DNA testing helps in the examination of the male.pdf
ANJALIENTERPRISES1
#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.
#includestdio.h#includestring.h#includestdlib.h#define M.pdf
#includestdio.h#includestring.h#includestdlib.h#define M.pdf
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.
The pH and pOH of a solution are defined as pH .pdf
The pH and pOH of a solution are defined as pH .pdf
ANJALIENTERPRISES1
goal_state = [1, 8, 7, 2, 0, 6, 3, 4, 5] #goal_state = [1, 0, 7, 2, 8, 6, 3, 4, 5] import sys global count def display_board( state ): #print \"-------------\" print \" %i %i %i \" % (state[0], state[3], state[6]) #print \"-------------\" print \" %i %i %i \" % (state[1], state[4], state[7]) #print \"-------------\" print \" %i %i %i \" % (state[2], state[5], state[8]) #print \"-------------\" def move_up( state ): new_state = state[:] index = new_state.index( 0 ) if index not in [0, 3, 6]: temp = new_state[index - 1] new_state[index - 1] = new_state[index] new_state[index] = temp return new_state else: return None def move_down( state ): new_state = state[:] index = new_state.index( 0 ) if index not in [2, 5, 8]: temp = new_state[index + 1] new_state[index + 1] = new_state[index] new_state[index] = temp return new_state else: return None def move_left( state ): new_state = state[:] index = new_state.index( 0 ) if index not in [0, 1, 2]: temp = new_state[index - 3] new_state[index - 3] = new_state[index] new_state[index] = temp return new_state else: return None def move_right( state ): new_state = state[:] index = new_state.index( 0 ) if index not in [6, 7, 8]: temp = new_state[index + 3] new_state[index + 3] = new_state[index] new_state[index] = temp return new_state else: return None def create_node( state, parent, operator, depth, cost ): return Node( state, parent, operator, depth, cost ) def expand_node( node, nodes ): expanded_nodes = [] expanded_nodes.append( create_node( move_up( node.state ), node, \"u\", node.depth + 1, 0 ) ) expanded_nodes.append( create_node( move_down( node.state ), node, \"d\", node.depth + 1, 0 ) ) expanded_nodes.append( create_node( move_left( node.state ), node, \"l\", node.depth + 1, 0 ) ) expanded_nodes.append( create_node( move_right( node.state), node, \"r\", node.depth + 1, 0 ) ) # Filter the list and remove the nodes that are impossible (move function returned None) expanded_nodes = [node for node in expanded_nodes if node.state != None] #list comprehension! expanded_nodes = [node for node in expanded_nodes if node not in nodes] #list comprehension! #expanded_nodes = [node for node in expanded_nodes if node.state!= ((node.parent).parent).state] #list comprehension! expanded_nodes1 = [] for node in expanded_nodes: temp=node temp=temp.parent temp=temp.parent if temp==None: expanded_nodes1.extend(expanded_nodes) break if node.state !=temp.state: expanded_nodes1.append(node) return expanded_nodes1 def bfs( start, goal ): nodes = [] count=0 # Create the queue with the root node in it. nodes.append( create_node( start, None, None, 0, 0 ) ) while True: # We\'ve run out of states, no solution. if len( nodes ) == 0: return None # take the node from the front of the queue node = nodes.pop(0) count=count+1 #display_board(node.state) # Append the move we made to moves # if this node is the goal, return the moves it took to get here. if node.state == goal: #display_board(node.state) moves = [] temp = node print cou.
goal_state = [1, 8, 7, 2, 0, 6, 3, 4, 5] #goal_state = [1, 0, 7, 2, .pdf
goal_state = [1, 8, 7, 2, 0, 6, 3, 4, 5] #goal_state = [1, 0, 7, 2, .pdf
ANJALIENTERPRISES1
7 steps are required as 9 is the 7th element and the search is linear Solution 7 steps are required as 9 is the 7th element and the search is linear.
7 steps are required as 9 is the 7th element and the search is line.pdf
7 steps are required as 9 is the 7th element and the search is line.pdf
ANJALIENTERPRISES1
(D) the number of moles of hydroxide ion added and the number of moles of monoprotic acid initially present Solution (D) the number of moles of hydroxide ion added and the number of moles of monoprotic acid initially present.
(D) the number of moles of hydroxide ion added and the number of mol.pdf
(D) the number of moles of hydroxide ion added and the number of mol.pdf
ANJALIENTERPRISES1
Look for changes in oxidation numbers. These occur in reactions (c) and (e) only. Thus, reactions (c) and (e) are REDOX reactions. Solution Look for changes in oxidation numbers. These occur in reactions (c) and (e) only. Thus, reactions (c) and (e) are REDOX reactions..
Look for changes in oxidation numbers. These occ.pdf
Look for changes in oxidation numbers. These occ.pdf
ANJALIENTERPRISES1
D) Insulin Solution D) Insulin.
D) Insulin .pdf
D) Insulin .pdf
ANJALIENTERPRISES1
Use Dalton\'s Law of partial pressures. P(Total)=P(H2)+P(H20) p(H2)=P(Total)- P(H20) Look up the partial pressure of water at the given temperature. It comes out to be 0.0351 atm. P(H2)=1-0.0351 = 0.9649 atm Solution Use Dalton\'s Law of partial pressures. P(Total)=P(H2)+P(H20) p(H2)=P(Total)- P(H20) Look up the partial pressure of water at the given temperature. It comes out to be 0.0351 atm. P(H2)=1-0.0351 = 0.9649 atm.
Use Daltons Law of partial pressures. P(Total).pdf
Use Daltons Law of partial pressures. P(Total).pdf
ANJALIENTERPRISES1
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..
The Br was originally neutral, but picks up an ex.pdf
The Br was originally neutral, but picks up an ex.pdf
ANJALIENTERPRISES1
Step1 firstly balance the Phophorous atoms by multiplying Right hand side by 4 Step 2 now balance oxygen atoms by multiplying H2O by 6 Step3 P4O10 +6 H2O-------> 4H3PO4 Solution Step1 firstly balance the Phophorous atoms by multiplying Right hand side by 4 Step 2 now balance oxygen atoms by multiplying H2O by 6 Step3 P4O10 +6 H2O-------> 4H3PO4.
Step1 firstly balance the Phophorous atoms by mul.pdf
Step1 firstly balance the Phophorous atoms by mul.pdf
ANJALIENTERPRISES1
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..
Year 2000 is the period when the productivity growth rate was at min.pdf
Year 2000 is the period when the productivity growth rate was at min.pdf
ANJALIENTERPRISES1
You have tab characters in your text file which is making console to display following both lines aligned at middle. dress across the porch as the There are two ways of doing it. Solution You have tab characters in your text file which is making console to display following both lines aligned at middle. dress across the porch as the There are two ways of doing it..
You have tab characters in your text file which is making console to.pdf
You have tab characters in your text file which is making console to.pdf
ANJALIENTERPRISES1
will update Solution will update.
will updateSolutionwill update.pdf
will updateSolutionwill update.pdf
ANJALIENTERPRISES1
the frequency of the allele I^A in this population-0.22 the frequency of the allele I^B in this population-0.08 Solution the frequency of the allele I^A in this population-0.22 the frequency of the allele I^B in this population-0.08.
the frequency of the allele I^A in this population-0.22the frequen.pdf
the frequency of the allele I^A in this population-0.22the frequen.pdf
ANJALIENTERPRISES1
The reaction is: Sr(s) + 2 H2O(l) => Sr(OH)2(s) + H2(g) Solution The reaction is: Sr(s) + 2 H2O(l) => Sr(OH)2(s) + H2(g).
The reaction isSr(s) + 2 H2O(l) = Sr(OH)2(s) + H2(g)Solution.pdf
The reaction isSr(s) + 2 H2O(l) = Sr(OH)2(s) + H2(g)Solution.pdf
ANJALIENTERPRISES1
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.
Solution!DOCTYPE html html head style type=textcss.pdf
Solution!DOCTYPE html html head style type=textcss.pdf
ANJALIENTERPRISES1
TCP/IP is mostly uaed network protocol in Linux and Mac os. But appletalk also used in both the os. Solution TCP/IP is mostly uaed network protocol in Linux and Mac os. But appletalk also used in both the os..
TCPIP is mostly uaed network protocol in Linux and Mac os.But app.pdf
TCPIP is mostly uaed network protocol in Linux and Mac os.But app.pdf
ANJALIENTERPRISES1
Photos from our Spring Gala 2024. Thank you for making our 2024 Spring Gala unforgettable! Whether you attended or supported us with a donation, we extend our heartfelt thanks for celebrating School-Community Partnerships alongside ExpandED Schools. Special appreciation to our honorees, Rachel Skaistis and Cravath, Swaine, & Moore LLP, for their unwavering support in advancing educational equity. Your contributions have profoundly impacted countless young New Yorkers, and we’re proud to have you as part of our community. Our gratitude also extends to our Board of Directors, sponsors, city partners, community-based organizations, and the inspiring young people who joined us. Your support fuels our mission, and we’re endlessly thankful for your dedication. While we celebrate our gala’s success, our work continues. As you view the evening’s highlights, consider how you can further empower young New Yorkers and educators with your donation, no matter the size. Donate today: https://www.expandedschools.org/support-our-work/donate/
Spring gala 2024 photo slideshow - Celebrating School-Community Partnerships
Spring gala 2024 photo slideshow - Celebrating School-Community Partnerships
expandedwebsite
This presentation covers the essential parameters of Unit 2 Operations Processes of the subject Operations & Supply Chain Management. Topics Covered: Volume Variety and Flow. Types of Processes and Operations Systems - Continuous Flow system and intermittent flow systems.Job Production, Batch Production, Assembly line and Continuous Flow, Process and Product Layout. Design of Service Systems, Service Blueprinting.
OSCM Unit 2_Operations Processes & Systems
OSCM Unit 2_Operations Processes & Systems
Sandeep D Chaudhary
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At the beginning of the 11th year, there would be 1,024,000 rabbits. Solution At the beginning of the 11th year, there would be 1,024,000 rabbits..
A is correct. The packets to be filtered would be heading into the r.pdf
A is correct. The packets to be filtered would be heading into the r.pdf
ANJALIENTERPRISES1
1). A) ipsilateral Below the point of spinal cord, the nerve paths belong to the same side (no crossing over occurs). So, the nociceptive information can be found on the same side of the body. Solution 1). A) ipsilateral Below the point of spinal cord, the nerve paths belong to the same side (no crossing over occurs). So, the nociceptive information can be found on the same side of the body..
1). A) ipsilateralBelow the point of spinal cord, the nerve paths .pdf
1). A) ipsilateralBelow the point of spinal cord, the nerve paths .pdf
ANJALIENTERPRISES1
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.
17-The Y-Chromosome DNA testing helps in the examination of the male.pdf
17-The Y-Chromosome DNA testing helps in the examination of the male.pdf
ANJALIENTERPRISES1
#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.
#includestdio.h#includestring.h#includestdlib.h#define M.pdf
#includestdio.h#includestring.h#includestdlib.h#define M.pdf
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.
The pH and pOH of a solution are defined as pH .pdf
The pH and pOH of a solution are defined as pH .pdf
ANJALIENTERPRISES1
goal_state = [1, 8, 7, 2, 0, 6, 3, 4, 5] #goal_state = [1, 0, 7, 2, 8, 6, 3, 4, 5] import sys global count def display_board( state ): #print \"-------------\" print \" %i %i %i \" % (state[0], state[3], state[6]) #print \"-------------\" print \" %i %i %i \" % (state[1], state[4], state[7]) #print \"-------------\" print \" %i %i %i \" % (state[2], state[5], state[8]) #print \"-------------\" def move_up( state ): new_state = state[:] index = new_state.index( 0 ) if index not in [0, 3, 6]: temp = new_state[index - 1] new_state[index - 1] = new_state[index] new_state[index] = temp return new_state else: return None def move_down( state ): new_state = state[:] index = new_state.index( 0 ) if index not in [2, 5, 8]: temp = new_state[index + 1] new_state[index + 1] = new_state[index] new_state[index] = temp return new_state else: return None def move_left( state ): new_state = state[:] index = new_state.index( 0 ) if index not in [0, 1, 2]: temp = new_state[index - 3] new_state[index - 3] = new_state[index] new_state[index] = temp return new_state else: return None def move_right( state ): new_state = state[:] index = new_state.index( 0 ) if index not in [6, 7, 8]: temp = new_state[index + 3] new_state[index + 3] = new_state[index] new_state[index] = temp return new_state else: return None def create_node( state, parent, operator, depth, cost ): return Node( state, parent, operator, depth, cost ) def expand_node( node, nodes ): expanded_nodes = [] expanded_nodes.append( create_node( move_up( node.state ), node, \"u\", node.depth + 1, 0 ) ) expanded_nodes.append( create_node( move_down( node.state ), node, \"d\", node.depth + 1, 0 ) ) expanded_nodes.append( create_node( move_left( node.state ), node, \"l\", node.depth + 1, 0 ) ) expanded_nodes.append( create_node( move_right( node.state), node, \"r\", node.depth + 1, 0 ) ) # Filter the list and remove the nodes that are impossible (move function returned None) expanded_nodes = [node for node in expanded_nodes if node.state != None] #list comprehension! expanded_nodes = [node for node in expanded_nodes if node not in nodes] #list comprehension! #expanded_nodes = [node for node in expanded_nodes if node.state!= ((node.parent).parent).state] #list comprehension! expanded_nodes1 = [] for node in expanded_nodes: temp=node temp=temp.parent temp=temp.parent if temp==None: expanded_nodes1.extend(expanded_nodes) break if node.state !=temp.state: expanded_nodes1.append(node) return expanded_nodes1 def bfs( start, goal ): nodes = [] count=0 # Create the queue with the root node in it. nodes.append( create_node( start, None, None, 0, 0 ) ) while True: # We\'ve run out of states, no solution. if len( nodes ) == 0: return None # take the node from the front of the queue node = nodes.pop(0) count=count+1 #display_board(node.state) # Append the move we made to moves # if this node is the goal, return the moves it took to get here. if node.state == goal: #display_board(node.state) moves = [] temp = node print cou.
goal_state = [1, 8, 7, 2, 0, 6, 3, 4, 5] #goal_state = [1, 0, 7, 2, .pdf
goal_state = [1, 8, 7, 2, 0, 6, 3, 4, 5] #goal_state = [1, 0, 7, 2, .pdf
ANJALIENTERPRISES1
7 steps are required as 9 is the 7th element and the search is linear Solution 7 steps are required as 9 is the 7th element and the search is linear.
7 steps are required as 9 is the 7th element and the search is line.pdf
7 steps are required as 9 is the 7th element and the search is line.pdf
ANJALIENTERPRISES1
(D) the number of moles of hydroxide ion added and the number of moles of monoprotic acid initially present Solution (D) the number of moles of hydroxide ion added and the number of moles of monoprotic acid initially present.
(D) the number of moles of hydroxide ion added and the number of mol.pdf
(D) the number of moles of hydroxide ion added and the number of mol.pdf
ANJALIENTERPRISES1
Look for changes in oxidation numbers. These occur in reactions (c) and (e) only. Thus, reactions (c) and (e) are REDOX reactions. Solution Look for changes in oxidation numbers. These occur in reactions (c) and (e) only. Thus, reactions (c) and (e) are REDOX reactions..
Look for changes in oxidation numbers. These occ.pdf
Look for changes in oxidation numbers. These occ.pdf
ANJALIENTERPRISES1
D) Insulin Solution D) Insulin.
D) Insulin .pdf
D) Insulin .pdf
ANJALIENTERPRISES1
Use Dalton\'s Law of partial pressures. P(Total)=P(H2)+P(H20) p(H2)=P(Total)- P(H20) Look up the partial pressure of water at the given temperature. It comes out to be 0.0351 atm. P(H2)=1-0.0351 = 0.9649 atm Solution Use Dalton\'s Law of partial pressures. P(Total)=P(H2)+P(H20) p(H2)=P(Total)- P(H20) Look up the partial pressure of water at the given temperature. It comes out to be 0.0351 atm. P(H2)=1-0.0351 = 0.9649 atm.
Use Daltons Law of partial pressures. P(Total).pdf
Use Daltons Law of partial pressures. P(Total).pdf
ANJALIENTERPRISES1
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..
The Br was originally neutral, but picks up an ex.pdf
The Br was originally neutral, but picks up an ex.pdf
ANJALIENTERPRISES1
Step1 firstly balance the Phophorous atoms by multiplying Right hand side by 4 Step 2 now balance oxygen atoms by multiplying H2O by 6 Step3 P4O10 +6 H2O-------> 4H3PO4 Solution Step1 firstly balance the Phophorous atoms by multiplying Right hand side by 4 Step 2 now balance oxygen atoms by multiplying H2O by 6 Step3 P4O10 +6 H2O-------> 4H3PO4.
Step1 firstly balance the Phophorous atoms by mul.pdf
Step1 firstly balance the Phophorous atoms by mul.pdf
ANJALIENTERPRISES1
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..
Year 2000 is the period when the productivity growth rate was at min.pdf
Year 2000 is the period when the productivity growth rate was at min.pdf
ANJALIENTERPRISES1
You have tab characters in your text file which is making console to display following both lines aligned at middle. dress across the porch as the There are two ways of doing it. Solution You have tab characters in your text file which is making console to display following both lines aligned at middle. dress across the porch as the There are two ways of doing it..
You have tab characters in your text file which is making console to.pdf
You have tab characters in your text file which is making console to.pdf
ANJALIENTERPRISES1
will update Solution will update.
will updateSolutionwill update.pdf
will updateSolutionwill update.pdf
ANJALIENTERPRISES1
the frequency of the allele I^A in this population-0.22 the frequency of the allele I^B in this population-0.08 Solution the frequency of the allele I^A in this population-0.22 the frequency of the allele I^B in this population-0.08.
the frequency of the allele I^A in this population-0.22the frequen.pdf
the frequency of the allele I^A in this population-0.22the frequen.pdf
ANJALIENTERPRISES1
The reaction is: Sr(s) + 2 H2O(l) => Sr(OH)2(s) + H2(g) Solution The reaction is: Sr(s) + 2 H2O(l) => Sr(OH)2(s) + H2(g).
The reaction isSr(s) + 2 H2O(l) = Sr(OH)2(s) + H2(g)Solution.pdf
The reaction isSr(s) + 2 H2O(l) = Sr(OH)2(s) + H2(g)Solution.pdf
ANJALIENTERPRISES1
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.
Solution!DOCTYPE html html head style type=textcss.pdf
Solution!DOCTYPE html html head style type=textcss.pdf
ANJALIENTERPRISES1
TCP/IP is mostly uaed network protocol in Linux and Mac os. But appletalk also used in both the os. Solution TCP/IP is mostly uaed network protocol in Linux and Mac os. But appletalk also used in both the os..
TCPIP is mostly uaed network protocol in Linux and Mac os.But app.pdf
TCPIP is mostly uaed network protocol in Linux and Mac os.But app.pdf
ANJALIENTERPRISES1
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