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L/2^1/2 Solution L/2^1/2.
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#!/bin/bash # Decode a UDP packet raw binary dataDone let offs=40 declare -i ws declare -i end declare -i rmndr declare -i byteval declare -i bitws NetworkOrder16Bit() { ws=`hexdump -s $offs -n 1 -e \'1/1 \"%u\"\' test2.pkts` ((offs+=1)) ((ws*=256)) byteval=`hexdump -s $offs -n 1 -e \'1/1 \"%u\"\' test2.pkts` ((offs+=1)) ((ws+=byteval)) } NetworkOrder32Bit() { ws=`hexdump -s $offs -n 1 -e \'1/1 \"%u\"\' test2.pkts` ((offs+=1)) ((ws*=256)) byteval=`hexdump -s $offs -n 1 -e \'1/1 \"%u\"\' test2.pkts` ((offs+=1)) ((ws+=byteval)) ((ws*=256)) byteval=`hexdump -s $offs -n 1 -e \'1/1 \"%u\"\' test2.pkts` ((offs+=1)) ((ws+=byteval)) ((ws*=256)) byteval=`hexdump -s $offs -n 1 -e \'1/1 \"%u\"\' test2.pkts` ((offs+=1)) ((ws+=byteval)) } echo echo end=$offs+74 #Start with the destination MAC address hexdump -s $offs -n 6 -e \'\"Ether Dest: \" 5/1 \"%02x:\" 1/1 \"%02x\ \"\' test2.pkts ((offs+=6)) # Then the source hexdump -s $offs -n 6 -e \'\"Ether Src: \" 5/1 \"%02x:\" 1/1 \"%02x\ \"\' test2.pkts ((offs+=6)) # Next is the packet type, but its in network byte order so we can\'t # use \'1/2 \"%04x\"\' - we may be run on a system with another byte order and you\'ll # get different output hexdump -s $offs -n 2 -e \'\"Ether Type: \" 2/1 \"%02x\" \"\ \"\' test2.pkts ((offs+=2)) # IP header # Get the version/len value into a variable we can do arithmetic with byteval=`hexdump -s $offs -n 1 -e \'1/1 \"%u\"\' test2.pkts` ((offs+=1)) # The most significant 4 bits are the IP version ((bitws=byteval&0xF0)) ((bitws/=16)) echo \"IP Version: $bitws\" # Least significant 4 bits are the header length ((bitws=byteval&0x0F)) ((bitws*=4)) echo \"Header Length: $bitws\" # Type of service byteval=`hexdump -s $offs -n 1 -e \'1/1 \"%u\"\' test2.pkts` ((offs+=1)) # Precedence ((bitws=byteval&0xE0)) ((bitws/=32)) echo \"Precedence: $bitws\" # Delay ((bitws=byteval&0x10)) if (( bitws )) then echo \"Minimize delay\" fi # Throughput ((bitws=byteval&0x08)) if (( bitws )) then echo \"High throughput\" fi # Reliability ((bitws=byteval&0x08)) if (( bitws )) then echo \"High reliability\" fi # Cost ((bitws=byteval&0x08)) if (( bitws )) then echo \"Minimize cost\" fi # Message length - in network byte order NetworkOrder16Bit echo \"Message Length: $ws\" # ID hexdump -s $offs -n 2 -e \'\"ID: \" 2/1 \"%02x\" \"\ \"\' test2.pkts ((offs+=2)) # Flags byteval=`hexdump -s $offs -n 1 -e \'1/1 \"%u\"\' test2.pkts` ((offs+=1)) # DF ((bitws=byteval&0x40)) if (( bitws != 0)) then echo \"Don\'t fragment\" else # More fragments ((bitws=byteval&0x20)) if (( bitws )) then echo \"More fragments\" fi # Fragment number ((ws=bitws&0x1F)) ((ws*=256)) byteval=`hexdump -s $offs -n 1 -e \'1/1 \"%u\"\' test2.pkts` ((ws+=byteval)) echo \"Fragment number: $ws\" fi ((offs+=1)) hexdump -s $offs -n 1 -e \'\"TTL: \" 1/1 \"%u\ \"\' test2.pkts ((offs+=1)) hexdump -s $offs -n 1 -e \'\"Protocol: \" 1/1 \"%u\ \"\' test2.pkts ((offs+=1)) hexdump -s $offs -n 2 -e \'\"Header checksum: \" 2/1 \"%02x\" \"\ \"\' test2.pkts ((offs.
#!binbash# Decode a UDP packet raw binary dataDonelet offs=40.pdf
#!binbash# Decode a UDP packet raw binary dataDonelet offs=40.pdf
aesalem06
You\'d only need SOCl2 So d then na Solution You\'d only need SOCl2 So d then na.
Youd only need SOCl2 So d then na .pdf
Youd only need SOCl2 So d then na .pdf
aesalem06
you can see the difference b/w IE5 and IE6 is very high.. it means the element is going to inert gas configuration after removing five e-s hence the element is nitrogen N and its configuration is 1S2 2S2 2P3 Solution you can see the difference b/w IE5 and IE6 is very high.. it means the element is going to inert gas configuration after removing five e-s hence the element is nitrogen N and its configuration is 1S2 2S2 2P3.
you can see the difference bw IE5 and IE6 is ver.pdf
you can see the difference bw IE5 and IE6 is ver.pdf
aesalem06
they have the same vapour pressure Solution they have the same vapour pressure.
they have the same vapour pressure .pdf
they have the same vapour pressure .pdf
aesalem06
The elements As, Sb and Bi show no tendency to form anion by gaining electrons . This is probably due to the fact that increasing size of the atom makes it difficult for the nucleus to hold extra electrons. Solution The elements As, Sb and Bi show no tendency to form anion by gaining electrons . This is probably due to the fact that increasing size of the atom makes it difficult for the nucleus to hold extra electrons..
The elements As, Sb and Bi show no tendency to fo.pdf
The elements As, Sb and Bi show no tendency to fo.pdf
aesalem06
taking the distance from point P in water for length = 14 m time taken is = 14/5 = 2.8 hr and taking diagonal path in grassy plain = sqrt(15^2+11^2) time taken = 6.2003 hr therefore total time taken is = 9.000358413 hr Solution taking the distance from point P in water for length = 14 m time taken is = 14/5 = 2.8 hr and taking diagonal path in grassy plain = sqrt(15^2+11^2) time taken = 6.2003 hr therefore total time taken is = 9.000358413 hr.
taking the distance from point P in water for len.pdf
taking the distance from point P in water for len.pdf
aesalem06
PH3: Phosphine H3N: ammonia H3PO2: Hypophosphorus Acid CuSO4: Copper Sulphate P2O2: Phosphorus Dioxide H2SO5:Peroxymonosulfuric acid Titanium(IV) Oxide: TiO2 Solution PH3: Phosphine H3N: ammonia H3PO2: Hypophosphorus Acid CuSO4: Copper Sulphate P2O2: Phosphorus Dioxide H2SO5:Peroxymonosulfuric acid Titanium(IV) Oxide: TiO2.
PH3 Phosphine H3N ammonia H3PO2 Hypophosphorus.pdf
PH3 Phosphine H3N ammonia H3PO2 Hypophosphorus.pdf
aesalem06
order is 3 and it is a linear Solution order is 3 and it is a linear.
order is 3 and it is a linear .pdf
order is 3 and it is a linear .pdf
aesalem06
Recommended
#!/bin/bash # Decode a UDP packet raw binary dataDone let offs=40 declare -i ws declare -i end declare -i rmndr declare -i byteval declare -i bitws NetworkOrder16Bit() { ws=`hexdump -s $offs -n 1 -e \'1/1 \"%u\"\' test2.pkts` ((offs+=1)) ((ws*=256)) byteval=`hexdump -s $offs -n 1 -e \'1/1 \"%u\"\' test2.pkts` ((offs+=1)) ((ws+=byteval)) } NetworkOrder32Bit() { ws=`hexdump -s $offs -n 1 -e \'1/1 \"%u\"\' test2.pkts` ((offs+=1)) ((ws*=256)) byteval=`hexdump -s $offs -n 1 -e \'1/1 \"%u\"\' test2.pkts` ((offs+=1)) ((ws+=byteval)) ((ws*=256)) byteval=`hexdump -s $offs -n 1 -e \'1/1 \"%u\"\' test2.pkts` ((offs+=1)) ((ws+=byteval)) ((ws*=256)) byteval=`hexdump -s $offs -n 1 -e \'1/1 \"%u\"\' test2.pkts` ((offs+=1)) ((ws+=byteval)) } echo echo end=$offs+74 #Start with the destination MAC address hexdump -s $offs -n 6 -e \'\"Ether Dest: \" 5/1 \"%02x:\" 1/1 \"%02x\ \"\' test2.pkts ((offs+=6)) # Then the source hexdump -s $offs -n 6 -e \'\"Ether Src: \" 5/1 \"%02x:\" 1/1 \"%02x\ \"\' test2.pkts ((offs+=6)) # Next is the packet type, but its in network byte order so we can\'t # use \'1/2 \"%04x\"\' - we may be run on a system with another byte order and you\'ll # get different output hexdump -s $offs -n 2 -e \'\"Ether Type: \" 2/1 \"%02x\" \"\ \"\' test2.pkts ((offs+=2)) # IP header # Get the version/len value into a variable we can do arithmetic with byteval=`hexdump -s $offs -n 1 -e \'1/1 \"%u\"\' test2.pkts` ((offs+=1)) # The most significant 4 bits are the IP version ((bitws=byteval&0xF0)) ((bitws/=16)) echo \"IP Version: $bitws\" # Least significant 4 bits are the header length ((bitws=byteval&0x0F)) ((bitws*=4)) echo \"Header Length: $bitws\" # Type of service byteval=`hexdump -s $offs -n 1 -e \'1/1 \"%u\"\' test2.pkts` ((offs+=1)) # Precedence ((bitws=byteval&0xE0)) ((bitws/=32)) echo \"Precedence: $bitws\" # Delay ((bitws=byteval&0x10)) if (( bitws )) then echo \"Minimize delay\" fi # Throughput ((bitws=byteval&0x08)) if (( bitws )) then echo \"High throughput\" fi # Reliability ((bitws=byteval&0x08)) if (( bitws )) then echo \"High reliability\" fi # Cost ((bitws=byteval&0x08)) if (( bitws )) then echo \"Minimize cost\" fi # Message length - in network byte order NetworkOrder16Bit echo \"Message Length: $ws\" # ID hexdump -s $offs -n 2 -e \'\"ID: \" 2/1 \"%02x\" \"\ \"\' test2.pkts ((offs+=2)) # Flags byteval=`hexdump -s $offs -n 1 -e \'1/1 \"%u\"\' test2.pkts` ((offs+=1)) # DF ((bitws=byteval&0x40)) if (( bitws != 0)) then echo \"Don\'t fragment\" else # More fragments ((bitws=byteval&0x20)) if (( bitws )) then echo \"More fragments\" fi # Fragment number ((ws=bitws&0x1F)) ((ws*=256)) byteval=`hexdump -s $offs -n 1 -e \'1/1 \"%u\"\' test2.pkts` ((ws+=byteval)) echo \"Fragment number: $ws\" fi ((offs+=1)) hexdump -s $offs -n 1 -e \'\"TTL: \" 1/1 \"%u\ \"\' test2.pkts ((offs+=1)) hexdump -s $offs -n 1 -e \'\"Protocol: \" 1/1 \"%u\ \"\' test2.pkts ((offs+=1)) hexdump -s $offs -n 2 -e \'\"Header checksum: \" 2/1 \"%02x\" \"\ \"\' test2.pkts ((offs.
#!binbash# Decode a UDP packet raw binary dataDonelet offs=40.pdf
#!binbash# Decode a UDP packet raw binary dataDonelet offs=40.pdf
aesalem06
You\'d only need SOCl2 So d then na Solution You\'d only need SOCl2 So d then na.
Youd only need SOCl2 So d then na .pdf
Youd only need SOCl2 So d then na .pdf
aesalem06
you can see the difference b/w IE5 and IE6 is very high.. it means the element is going to inert gas configuration after removing five e-s hence the element is nitrogen N and its configuration is 1S2 2S2 2P3 Solution you can see the difference b/w IE5 and IE6 is very high.. it means the element is going to inert gas configuration after removing five e-s hence the element is nitrogen N and its configuration is 1S2 2S2 2P3.
you can see the difference bw IE5 and IE6 is ver.pdf
you can see the difference bw IE5 and IE6 is ver.pdf
aesalem06
they have the same vapour pressure Solution they have the same vapour pressure.
they have the same vapour pressure .pdf
they have the same vapour pressure .pdf
aesalem06
The elements As, Sb and Bi show no tendency to form anion by gaining electrons . This is probably due to the fact that increasing size of the atom makes it difficult for the nucleus to hold extra electrons. Solution The elements As, Sb and Bi show no tendency to form anion by gaining electrons . This is probably due to the fact that increasing size of the atom makes it difficult for the nucleus to hold extra electrons..
The elements As, Sb and Bi show no tendency to fo.pdf
The elements As, Sb and Bi show no tendency to fo.pdf
aesalem06
taking the distance from point P in water for length = 14 m time taken is = 14/5 = 2.8 hr and taking diagonal path in grassy plain = sqrt(15^2+11^2) time taken = 6.2003 hr therefore total time taken is = 9.000358413 hr Solution taking the distance from point P in water for length = 14 m time taken is = 14/5 = 2.8 hr and taking diagonal path in grassy plain = sqrt(15^2+11^2) time taken = 6.2003 hr therefore total time taken is = 9.000358413 hr.
taking the distance from point P in water for len.pdf
taking the distance from point P in water for len.pdf
aesalem06
PH3: Phosphine H3N: ammonia H3PO2: Hypophosphorus Acid CuSO4: Copper Sulphate P2O2: Phosphorus Dioxide H2SO5:Peroxymonosulfuric acid Titanium(IV) Oxide: TiO2 Solution PH3: Phosphine H3N: ammonia H3PO2: Hypophosphorus Acid CuSO4: Copper Sulphate P2O2: Phosphorus Dioxide H2SO5:Peroxymonosulfuric acid Titanium(IV) Oxide: TiO2.
PH3 Phosphine H3N ammonia H3PO2 Hypophosphorus.pdf
PH3 Phosphine H3N ammonia H3PO2 Hypophosphorus.pdf
aesalem06
order is 3 and it is a linear Solution order is 3 and it is a linear.
order is 3 and it is a linear .pdf
order is 3 and it is a linear .pdf
aesalem06
Nope. When an atom is ionized it only affects the electrons in the electron cloud, which is outside the nucleus. Let me know if this doesn\'t clear things up. Solution Nope. When an atom is ionized it only affects the electrons in the electron cloud, which is outside the nucleus. Let me know if this doesn\'t clear things up..
Nope. When an atom is ionized it only affects the.pdf
Nope. When an atom is ionized it only affects the.pdf
aesalem06
HOCl (aq) + HCI(aq) -> H2O(l) + Cl2 (g) 6Cl2(g) + 12NaOH(aq) -> 10NaCl (aq) + 2NaCIO3(aq) + 6H2O(l) MgCO3(aq) +2HCI (aq) -> MgCl2(aq) + H2O(l) + CO2 (g) Solution HOCl (aq) + HCI(aq) -> H2O(l) + Cl2 (g) 6Cl2(g) + 12NaOH(aq) -> 10NaCl (aq) + 2NaCIO3(aq) + 6H2O(l) MgCO3(aq) +2HCI (aq) -> MgCl2(aq) + H2O(l) + CO2 (g).
HOCl (aq) + HCI(aq) - H2O(l) + Cl2 (g) 6Cl2(g) +.pdf
HOCl (aq) + HCI(aq) - H2O(l) + Cl2 (g) 6Cl2(g) +.pdf
aesalem06
In Atmospehere : Pressure = 105/760 = 0.138 Atmosphere In Pascals: Pressure = 0.138 x 1.013 x 10^5 = 13995.4Pa Solution In Atmospehere : Pressure = 105/760 = 0.138 Atmosphere In Pascals: Pressure = 0.138 x 1.013 x 10^5 = 13995.4Pa.
In Atmospehere Pressure = 105760 = 0.138 Atmo.pdf
In Atmospehere Pressure = 105760 = 0.138 Atmo.pdf
aesalem06
H2SO4 is a strong electrolyte. It dissociates almost completely in water while rest four compounds don\'t dissolve in water completely. Solution H2SO4 is a strong electrolyte. It dissociates almost completely in water while rest four compounds don\'t dissolve in water completely..
H2SO4 is a strong electrolyte. It dissociates alm.pdf
H2SO4 is a strong electrolyte. It dissociates alm.pdf
aesalem06
DE is positive in exothermic reactions Solution DE is positive in exothermic reactions.
DE is positive in exothermic reactions .pdf
DE is positive in exothermic reactions .pdf
aesalem06
The yield to maturity (YTM) is the measure of bonds rate of return that considers both the interest income and any capital gain or loss. it should be noted that yield to maturity is not same as current yield as current yield is the annual interest dividend by the bonds current value and does not account for capital gain or loss. YTM is important because it compare a bonds expected return with those of other securities available in the market. YTM considers the three sources of potential return from a bond (coupon payments, capital gains, and reinvestment returns which helps investors in Understanding how yields vary with market prices (that as bond prices fall, yields rise; and as bond prices rise, yields fall).it also helps investors to anticipate the effects of market changes on their portfolios. The yield to maturity is expressed as an annual percentage rate The following formula can be used to find out its approximate value: APPROX YTM=C+F-P/n/F+P/2 Where C=coupon value F=face value P=market price And n=years to maturity Solution The yield to maturity (YTM) is the measure of bonds rate of return that considers both the interest income and any capital gain or loss. it should be noted that yield to maturity is not same as current yield as current yield is the annual interest dividend by the bonds current value and does not account for capital gain or loss. YTM is important because it compare a bonds expected return with those of other securities available in the market. YTM considers the three sources of potential return from a bond (coupon payments, capital gains, and reinvestment returns which helps investors in Understanding how yields vary with market prices (that as bond prices fall, yields rise; and as bond prices rise, yields fall).it also helps investors to anticipate the effects of market changes on their portfolios. The yield to maturity is expressed as an annual percentage rate The following formula can be used to find out its approximate value: APPROX YTM=C+F-P/n/F+P/2 Where C=coupon value F=face value P=market price And n=years to maturity.
The yield to maturity (YTM) is the measure of bonds rate of return t.pdf
The yield to maturity (YTM) is the measure of bonds rate of return t.pdf
aesalem06
TDS (Total Dissolved Solids ) is a measure of the combined content of all inorganic and organic substances contained in a liquid, Gravimetric methods are the most accurate and involve evaporating the liquid solvent and measuring the mass of residues left. Hence TDS = 37 - 15 = 22g TSS (Total Suspended Solid) are the suspended or colloidal particles TSS = 25gm Solution TDS (Total Dissolved Solids ) is a measure of the combined content of all inorganic and organic substances contained in a liquid, Gravimetric methods are the most accurate and involve evaporating the liquid solvent and measuring the mass of residues left. Hence TDS = 37 - 15 = 22g TSS (Total Suspended Solid) are the suspended or colloidal particles TSS = 25gm.
TDS (Total Dissolved Solids ) is a measure of the combined content o.pdf
TDS (Total Dissolved Solids ) is a measure of the combined content o.pdf
aesalem06
magnesium oxide + water => magnesium hydroxide MgO (s) + H2O (l) => Mg(OH)2 (s) Solution magnesium oxide + water => magnesium hydroxide MgO (s) + H2O (l) => Mg(OH)2 (s).
magnesium oxide + water = magnesium hydroxideMgO (s) + H2O (l) =.pdf
magnesium oxide + water = magnesium hydroxideMgO (s) + H2O (l) =.pdf
aesalem06
Meiosis, inheritance and variation: The uniqueness and similarities within the closely related family members occurs because of inherit traits in the form of genes are passing from one generation to another generation. This gene passing from our ancestral parents to our parents is called as heredity. Replication is a process of duplication of DNA. That occurs during the S phase of Eukaryotic cell cycle. In identical small organisms these genes pass from one generation to next by duplication of their genetic information and then splitting into two identical organisms. In complex organisms specialized sex cells called gametes carry half of the genetic information. These gametes from both male and female organisms combine each other and produce new organisms. The gametes produced process is termed as meiosis. Meiosis: During this process 1 diploid cell which contains 46 chromosomes undergoes cell division into 2 cycles. In which only once DNA replication is takes place. it results 4 haploid daughter cells. The zygote is formed from the combination of both an egg and a sperm cells. Zygotes have same number of chromosomes it is due to the halving nature of the number of chromosomes in sexual reproduction. This critical character stabilize chromosome numbers from one generation to next generation. Phases of meiosis: Interphase: In this the DNA is prepared for replication. After completion of replication process the structure of each chromosomes compressed into 2 identical chromatids. Prophase I : In this chromosomes structure appeared as in X - Shaped. The homologous chromosomes are pair up each other and crossing over takes placed. In this the DNA was exchanged. the nuclear membrane was completely dissolved at end of this stage. The Microtubules can attach to the chromosomes. Metaphase I: The synopsized chromosomes are aliened at equator of the cell. The chromosomes are aliened randomly in which different combinations each time meiosis occurs. Anaphase I: During this phase the molecules of chromosomes separates and migrates towards to 2 poles of cell. Telophase I or Cytokinesis: In this the cell divides into two daughter cells. Prophase II : Same as Prophase I. But the DNA does not replicate again. Metaphase II: The chromosomes aliened at equator, again the alignment is random. Anaphase II: The chromosomes separates and migrates towards to 2 poles of cell. Telophase II: The nucleoplasm is reformed and the cytoplasm is divides into 2 haploid daughter cells and finally 4 new haploid daughter cells is formed. Genetic variation is increased by meiosis due to the independent assortment nature in meiosis. The reshuffling process of genes in meiosis is increasing the genetic variation in a population. Solution Meiosis, inheritance and variation: The uniqueness and similarities within the closely related family members occurs because of inherit traits in the form of genes are passing from one generation to another generation. This gene passing from our an.
Meiosis, inheritance and variationThe uniqueness and similarities.pdf
Meiosis, inheritance and variationThe uniqueness and similarities.pdf
aesalem06
Inter-Process-Communication (or IPC for short) are mechanisms provided by the kernel to allow processes to communicate with each other. The types of inter process communication on Linux OS are: The following IPC mechanisms are supported by Windows: 1. Clipboard - The clipboard acts as a central depository for data sharing among applications. When a user performs a cut or copy operation in an application, the application puts the selected data on the clipboard in one or more standard or application-defined formats. Any other application can then retrieve the data from the clipboard, choosing from the available formats that it understands. 2. File Mapping - File mapping enables a process to treat the contents of a file as if they were a block of memory in the process\'s address space. The process can use simple pointer operations to examine and modify the contents of the file. When two or more processes access the same file mapping, each process receives a pointer to memory in its own address space that it can use to read or modify the contents of the file. 3. Mailslot - Mailslots provide one-way communication. Any process that creates a mailslot is a mailslot server. Other processes, called mailslot clients, send messages to the mailslot server by writing a message to its mailslot. 4. RPC - RPC enables applications to call functions remotely. Therefore, RPC makes IPC as easy as calling a function. RPC operates between processes on a single computer or on different computers on a network. 5. Windows Socket - Windows Sockets is a protocol-independent interface capable of supporting current and emerging networking capabilities. The following IPC mechanisms are supported by Mac OS: 1. Mach Ports : Mach 3.0 is capable of running as a stand-alone kernel, with other traditional OS- services like IO, file systems and networking stack running as user mode.It is much faster to make a direct call between linked components than it is to send messages or do RPC between separate tasks. 2. Apple Events : Universally supported by GUI applications on Mac OS for remote control.Operations like opening or telling an application to open a file or quit etc can be done using these. 3. Pasteboard - Copy Paste , Drag and drop done between applications is performed using this technique. 4. Distributed Objects - Remote messaging feature of Cocoa to call an object in different Cocoa applicaton. Windows server uses the best technique to manage IPC because a) It provides an efficient way for two or more processes on the same computer to share data. b) It is capable of supporting current and emerging networking capabilities, such as quality of service monitoring, robust asynchronous communication, I/O completion ports for superior performance, and protocol-specific network features. => Multiprocessing : refers to the use of two or morecentral processing units (CPU) within a single computer system.All the operating systems provide support for multiprocessing. Windows manages.
Inter-Process-Communication (or IPC for short) are mechanisms provid.pdf
Inter-Process-Communication (or IPC for short) are mechanisms provid.pdf
aesalem06
B)120.0° note: So3 is sp2-hybridized and adopts a trigonal planar geometry. Solution B)120.0° note: So3 is sp2-hybridized and adopts a trigonal planar geometry..
B)120.0° note So3 is sp2-hybridized and adopts .pdf
B)120.0° note So3 is sp2-hybridized and adopts .pdf
aesalem06
imple cellular phones that do very little over voice and text line of work generally include one process unit that handles everything - computer programme (keyboard, display), RF process, battery management, etc. These ar SoC - Systems on Chip and ar nearer to microcontrollers than microprocessors as a result of they decide to do everything - from process, to device interfacing, to memory, to program and knowledge storage, etc - on one chip. Smartphones have multiple microprocessors and microcontrollers in them. the most processor could be a chip with a bus to speak with memory on separate chips (though usually contained within the same IC package), and busses to speak with the remainder of the devices. they typically contain a number of the controllers, like the show controller, so that they have a number of the options of microcontrollers, however they\'re still additional chip than microcontroller. The cellular chipset typically includes a microprocessor/microcontroller that basically blurs the lines. the newest generation of smartphones usually tend additional toward microprocessors for the RF chipset, thus on provide the manufacturer flexibility by mistreatment software package style instead of hardware style for a few options, however they are doing have additional items of the hardware interface (rf, etc) on the chip itself. The rest of the phone has many microcontrollers. These management the touchscreen, audio, sensors, cameras, etc. So it depends on the kind of phone you are thinking of. an easy phone uses a microcontroller. a posh phone uses each. Solution imple cellular phones that do very little over voice and text line of work generally include one process unit that handles everything - computer programme (keyboard, display), RF process, battery management, etc. These ar SoC - Systems on Chip and ar nearer to microcontrollers than microprocessors as a result of they decide to do everything - from process, to device interfacing, to memory, to program and knowledge storage, etc - on one chip. Smartphones have multiple microprocessors and microcontrollers in them. the most processor could be a chip with a bus to speak with memory on separate chips (though usually contained within the same IC package), and busses to speak with the remainder of the devices. they typically contain a number of the controllers, like the show controller, so that they have a number of the options of microcontrollers, however they\'re still additional chip than microcontroller. The cellular chipset typically includes a microprocessor/microcontroller that basically blurs the lines. the newest generation of smartphones usually tend additional toward microprocessors for the RF chipset, thus on provide the manufacturer flexibility by mistreatment software package style instead of hardware style for a few options, however they are doing have additional items of the hardware interface (rf, etc) on the chip itself. The rest of the phone has many microcontrollers..
imple cellular phones that do very little over voice and text line o.pdf
imple cellular phones that do very little over voice and text line o.pdf
aesalem06
Given Data: Consider MTU in Bit IP header:16-bit Header:320 bit Appended header:420 bit Target maximum packet size of:1500 bit 420 header bits for the transport layer, 320 header bits for the IP layer. so total header=740 Bit to delivered at destination=1500-740=760 Solution Given Data: Consider MTU in Bit IP header:16-bit Header:320 bit Appended header:420 bit Target maximum packet size of:1500 bit 420 header bits for the transport layer, 320 header bits for the IP layer. so total header=740 Bit to delivered at destination=1500-740=760.
Given DataConsider MTU in Bit IP header16-bit Header320 bit.pdf
Given DataConsider MTU in Bit IP header16-bit Header320 bit.pdf
aesalem06
Features that PL/SQL shares with other standard programming languages: Solution Features that PL/SQL shares with other standard programming languages:.
Features that PLSQL shares with other standard programming language.pdf
Features that PLSQL shares with other standard programming language.pdf
aesalem06
Draw two electron arrow towards Zn. somewhere in the middle between zinc and oxygen ...make sure it connects the zinc in the templet. Solution Draw two electron arrow towards Zn. somewhere in the middle between zinc and oxygen ...make sure it connects the zinc in the templet..
Draw two electron arrow towards Zn. somewhere in the middle between .pdf
Draw two electron arrow towards Zn. somewhere in the middle between .pdf
aesalem06
Confidence Interval CI = x Solution Confidence Interval CI = x.
Confidence IntervalCI = xSolutionConfidence IntervalCI = x.pdf
Confidence IntervalCI = xSolutionConfidence IntervalCI = x.pdf
aesalem06
A)2moles of calcium reacts with 1mole of O2 to form 2moles of calcium oxide. no.of terms of particular element on right side should be equal to left side here it is equal hence the equation is balanced B)1mole of methane reacts with 2moles of O2 to form 2moles of carbondioxide and 2moles of water Solution A)2moles of calcium reacts with 1mole of O2 to form 2moles of calcium oxide. no.of terms of particular element on right side should be equal to left side here it is equal hence the equation is balanced B)1mole of methane reacts with 2moles of O2 to form 2moles of carbondioxide and 2moles of water.
A)2moles of calcium reacts with 1mole of O2 to fo.pdf
A)2moles of calcium reacts with 1mole of O2 to fo.pdf
aesalem06
Answer: Two Yellow colored and two ebony colored fruit flies are provided as mentioned in the question. It is not known to us, of them, which one is the dominant trait for the body color character. Case 1: Cross #1 Yellow colored fruit flies are crossed with ebony colored fruit flies and as a result all progeny obtained in the F1 generation are of a single color- Ebony. This observation leads to the impression that Ebony coloration is the dominant trait whereas Yellow is the recessive trait. Both the parents are homozygous for the trait and are pure breeding varieties. If that would not have been the case, in the F1 generation itself two variants- both Ebony and Yellow colored progeny would have appeared. According to Mendelian laws, Let, \'Y\' be the factor responsible for Ebony coloration and \'y\' be the factor responsible for Yellow coloration. Thus, in the parental generation, Genotype of the parent having Ebony coloration (homozygous pure breeding variety) - YY Genotype of the parent having Yellow coloration (homozygous pure breeding variety) - yy Hence, genotype of the progeny in F1 generation - Yy Case 2 : Cross #2 In the present case it was observed that 113 Yellow fruit flies were obtained and 120 Ebony fruit flies were obtained in the F1 generation as a result of a cross between an ebony & another yellow fruit fly. So, in the F1 generation, both the variants were obtained almost approximately in 1:1 phenotypic ratio. Since both the variants were obtained, it may be concluded that parent bearing ebony coloration was heterozygous whereas the yellow colored fly was a pure breeding variety. Thus, Genotype of the parent having Ebony coloration - Yy (heterozygous) Genotype of the parent having Yellow coloration -yy (homozygous) Ebony colored parent gives rise to two types of gametes- one bearing \'Y\' factor and the other bearing \'y\' factor. Yellow colored parent gives rise to only gametes bearing \'y\' factor. Gametes Y y y Yy (heterozygous ebony) yy (homozygous yellow) y Yy yy Progeny obtained in the F1 generation have genotype Yy ( heterozygous ebony) and yy(homozygous yellow). Case 3: Cross #3 Two ebony offspring are crossed having genotype \'Yy\'. Genotype of the progeny in F2 generation can be obtained by using checkerboard- Gametes Y y Y YY (homozygous Ebony) Yy (heterozygous ebony) y Yy (heterozygous ebony) yy (homozygous yellow) Thus, phenotypic rato between Ebony & Yellow colored fruit fly -3:1 (Ebony: Yellow) Genotypic ratio in F2 generation - YY: Yy : yy - 1: 2: 1 Case 4 :Cross #4 Mating is performed between Ebony colored fruit fly #3 having genotype \'Yy\' and Ebony colored fruit fly #2 having genotype \'YY\'. #3 ebony colored fly gives rise to to different types of gametes- one bearing \'Y\' allele and the other bearing \'y\' allele for body color character. The genotype and the phenotype of the progeny in this case can be obtained using checkerboard. Gametes Y y Y YY ( homozygous ebony) Yy (heterozygous ebony) Y YY Yy Th.
AnswerTwo Yellow colored and two ebony colored fruit flies are pr.pdf
AnswerTwo Yellow colored and two ebony colored fruit flies are pr.pdf
aesalem06
answer B) Cl - F F creates strong dipole moment for a, trigonal planar, dipole moment is distributed equally. C, linear, also distributed equally D) linear E) tetrahedral, distributed equally Solution answer B) Cl - F F creates strong dipole moment for a, trigonal planar, dipole moment is distributed equally. C, linear, also distributed equally D) linear E) tetrahedral, distributed equally.
answer B) Cl - FF creates strong dipole momentfor a, trigonal pl.pdf
answer B) Cl - FF creates strong dipole momentfor a, trigonal pl.pdf
aesalem06
ANSWER: The events occuring in Complex I in order are 3, 2, 1, 4 NADH is oxidized to NAD+ by reducing FMN to FMNH2 while oxidising Fe+2 to Fe+3 for the electron transport. Finally the CoQ is reduced to CoQH2 after which the electron enters in Complex II Solution ANSWER: The events occuring in Complex I in order are 3, 2, 1, 4 NADH is oxidized to NAD+ by reducing FMN to FMNH2 while oxidising Fe+2 to Fe+3 for the electron transport. Finally the CoQ is reduced to CoQH2 after which the electron enters in Complex II.
ANSWER The events occuring in Complex I in order are 3, 2, 1, 4NA.pdf
ANSWER The events occuring in Complex I in order are 3, 2, 1, 4NA.pdf
aesalem06
For more information about my speaking and training work, visit: https://www.pookyknightsmith.com/speaking/
Trauma-Informed Leadership - Five Practical Principles
Trauma-Informed Leadership - Five Practical Principles
Pooky Knightsmith
LESSON PLAN IN ENGLISH 4 MATATAG CURRICULUM
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Nope. When an atom is ionized it only affects the electrons in the electron cloud, which is outside the nucleus. Let me know if this doesn\'t clear things up. Solution Nope. When an atom is ionized it only affects the electrons in the electron cloud, which is outside the nucleus. Let me know if this doesn\'t clear things up..
Nope. When an atom is ionized it only affects the.pdf
Nope. When an atom is ionized it only affects the.pdf
aesalem06
HOCl (aq) + HCI(aq) -> H2O(l) + Cl2 (g) 6Cl2(g) + 12NaOH(aq) -> 10NaCl (aq) + 2NaCIO3(aq) + 6H2O(l) MgCO3(aq) +2HCI (aq) -> MgCl2(aq) + H2O(l) + CO2 (g) Solution HOCl (aq) + HCI(aq) -> H2O(l) + Cl2 (g) 6Cl2(g) + 12NaOH(aq) -> 10NaCl (aq) + 2NaCIO3(aq) + 6H2O(l) MgCO3(aq) +2HCI (aq) -> MgCl2(aq) + H2O(l) + CO2 (g).
HOCl (aq) + HCI(aq) - H2O(l) + Cl2 (g) 6Cl2(g) +.pdf
HOCl (aq) + HCI(aq) - H2O(l) + Cl2 (g) 6Cl2(g) +.pdf
aesalem06
In Atmospehere : Pressure = 105/760 = 0.138 Atmosphere In Pascals: Pressure = 0.138 x 1.013 x 10^5 = 13995.4Pa Solution In Atmospehere : Pressure = 105/760 = 0.138 Atmosphere In Pascals: Pressure = 0.138 x 1.013 x 10^5 = 13995.4Pa.
In Atmospehere Pressure = 105760 = 0.138 Atmo.pdf
In Atmospehere Pressure = 105760 = 0.138 Atmo.pdf
aesalem06
H2SO4 is a strong electrolyte. It dissociates almost completely in water while rest four compounds don\'t dissolve in water completely. Solution H2SO4 is a strong electrolyte. It dissociates almost completely in water while rest four compounds don\'t dissolve in water completely..
H2SO4 is a strong electrolyte. It dissociates alm.pdf
H2SO4 is a strong electrolyte. It dissociates alm.pdf
aesalem06
DE is positive in exothermic reactions Solution DE is positive in exothermic reactions.
DE is positive in exothermic reactions .pdf
DE is positive in exothermic reactions .pdf
aesalem06
The yield to maturity (YTM) is the measure of bonds rate of return that considers both the interest income and any capital gain or loss. it should be noted that yield to maturity is not same as current yield as current yield is the annual interest dividend by the bonds current value and does not account for capital gain or loss. YTM is important because it compare a bonds expected return with those of other securities available in the market. YTM considers the three sources of potential return from a bond (coupon payments, capital gains, and reinvestment returns which helps investors in Understanding how yields vary with market prices (that as bond prices fall, yields rise; and as bond prices rise, yields fall).it also helps investors to anticipate the effects of market changes on their portfolios. The yield to maturity is expressed as an annual percentage rate The following formula can be used to find out its approximate value: APPROX YTM=C+F-P/n/F+P/2 Where C=coupon value F=face value P=market price And n=years to maturity Solution The yield to maturity (YTM) is the measure of bonds rate of return that considers both the interest income and any capital gain or loss. it should be noted that yield to maturity is not same as current yield as current yield is the annual interest dividend by the bonds current value and does not account for capital gain or loss. YTM is important because it compare a bonds expected return with those of other securities available in the market. YTM considers the three sources of potential return from a bond (coupon payments, capital gains, and reinvestment returns which helps investors in Understanding how yields vary with market prices (that as bond prices fall, yields rise; and as bond prices rise, yields fall).it also helps investors to anticipate the effects of market changes on their portfolios. The yield to maturity is expressed as an annual percentage rate The following formula can be used to find out its approximate value: APPROX YTM=C+F-P/n/F+P/2 Where C=coupon value F=face value P=market price And n=years to maturity.
The yield to maturity (YTM) is the measure of bonds rate of return t.pdf
The yield to maturity (YTM) is the measure of bonds rate of return t.pdf
aesalem06
TDS (Total Dissolved Solids ) is a measure of the combined content of all inorganic and organic substances contained in a liquid, Gravimetric methods are the most accurate and involve evaporating the liquid solvent and measuring the mass of residues left. Hence TDS = 37 - 15 = 22g TSS (Total Suspended Solid) are the suspended or colloidal particles TSS = 25gm Solution TDS (Total Dissolved Solids ) is a measure of the combined content of all inorganic and organic substances contained in a liquid, Gravimetric methods are the most accurate and involve evaporating the liquid solvent and measuring the mass of residues left. Hence TDS = 37 - 15 = 22g TSS (Total Suspended Solid) are the suspended or colloidal particles TSS = 25gm.
TDS (Total Dissolved Solids ) is a measure of the combined content o.pdf
TDS (Total Dissolved Solids ) is a measure of the combined content o.pdf
aesalem06
magnesium oxide + water => magnesium hydroxide MgO (s) + H2O (l) => Mg(OH)2 (s) Solution magnesium oxide + water => magnesium hydroxide MgO (s) + H2O (l) => Mg(OH)2 (s).
magnesium oxide + water = magnesium hydroxideMgO (s) + H2O (l) =.pdf
magnesium oxide + water = magnesium hydroxideMgO (s) + H2O (l) =.pdf
aesalem06
Meiosis, inheritance and variation: The uniqueness and similarities within the closely related family members occurs because of inherit traits in the form of genes are passing from one generation to another generation. This gene passing from our ancestral parents to our parents is called as heredity. Replication is a process of duplication of DNA. That occurs during the S phase of Eukaryotic cell cycle. In identical small organisms these genes pass from one generation to next by duplication of their genetic information and then splitting into two identical organisms. In complex organisms specialized sex cells called gametes carry half of the genetic information. These gametes from both male and female organisms combine each other and produce new organisms. The gametes produced process is termed as meiosis. Meiosis: During this process 1 diploid cell which contains 46 chromosomes undergoes cell division into 2 cycles. In which only once DNA replication is takes place. it results 4 haploid daughter cells. The zygote is formed from the combination of both an egg and a sperm cells. Zygotes have same number of chromosomes it is due to the halving nature of the number of chromosomes in sexual reproduction. This critical character stabilize chromosome numbers from one generation to next generation. Phases of meiosis: Interphase: In this the DNA is prepared for replication. After completion of replication process the structure of each chromosomes compressed into 2 identical chromatids. Prophase I : In this chromosomes structure appeared as in X - Shaped. The homologous chromosomes are pair up each other and crossing over takes placed. In this the DNA was exchanged. the nuclear membrane was completely dissolved at end of this stage. The Microtubules can attach to the chromosomes. Metaphase I: The synopsized chromosomes are aliened at equator of the cell. The chromosomes are aliened randomly in which different combinations each time meiosis occurs. Anaphase I: During this phase the molecules of chromosomes separates and migrates towards to 2 poles of cell. Telophase I or Cytokinesis: In this the cell divides into two daughter cells. Prophase II : Same as Prophase I. But the DNA does not replicate again. Metaphase II: The chromosomes aliened at equator, again the alignment is random. Anaphase II: The chromosomes separates and migrates towards to 2 poles of cell. Telophase II: The nucleoplasm is reformed and the cytoplasm is divides into 2 haploid daughter cells and finally 4 new haploid daughter cells is formed. Genetic variation is increased by meiosis due to the independent assortment nature in meiosis. The reshuffling process of genes in meiosis is increasing the genetic variation in a population. Solution Meiosis, inheritance and variation: The uniqueness and similarities within the closely related family members occurs because of inherit traits in the form of genes are passing from one generation to another generation. This gene passing from our an.
Meiosis, inheritance and variationThe uniqueness and similarities.pdf
Meiosis, inheritance and variationThe uniqueness and similarities.pdf
aesalem06
Inter-Process-Communication (or IPC for short) are mechanisms provided by the kernel to allow processes to communicate with each other. The types of inter process communication on Linux OS are: The following IPC mechanisms are supported by Windows: 1. Clipboard - The clipboard acts as a central depository for data sharing among applications. When a user performs a cut or copy operation in an application, the application puts the selected data on the clipboard in one or more standard or application-defined formats. Any other application can then retrieve the data from the clipboard, choosing from the available formats that it understands. 2. File Mapping - File mapping enables a process to treat the contents of a file as if they were a block of memory in the process\'s address space. The process can use simple pointer operations to examine and modify the contents of the file. When two or more processes access the same file mapping, each process receives a pointer to memory in its own address space that it can use to read or modify the contents of the file. 3. Mailslot - Mailslots provide one-way communication. Any process that creates a mailslot is a mailslot server. Other processes, called mailslot clients, send messages to the mailslot server by writing a message to its mailslot. 4. RPC - RPC enables applications to call functions remotely. Therefore, RPC makes IPC as easy as calling a function. RPC operates between processes on a single computer or on different computers on a network. 5. Windows Socket - Windows Sockets is a protocol-independent interface capable of supporting current and emerging networking capabilities. The following IPC mechanisms are supported by Mac OS: 1. Mach Ports : Mach 3.0 is capable of running as a stand-alone kernel, with other traditional OS- services like IO, file systems and networking stack running as user mode.It is much faster to make a direct call between linked components than it is to send messages or do RPC between separate tasks. 2. Apple Events : Universally supported by GUI applications on Mac OS for remote control.Operations like opening or telling an application to open a file or quit etc can be done using these. 3. Pasteboard - Copy Paste , Drag and drop done between applications is performed using this technique. 4. Distributed Objects - Remote messaging feature of Cocoa to call an object in different Cocoa applicaton. Windows server uses the best technique to manage IPC because a) It provides an efficient way for two or more processes on the same computer to share data. b) It is capable of supporting current and emerging networking capabilities, such as quality of service monitoring, robust asynchronous communication, I/O completion ports for superior performance, and protocol-specific network features. => Multiprocessing : refers to the use of two or morecentral processing units (CPU) within a single computer system.All the operating systems provide support for multiprocessing. Windows manages.
Inter-Process-Communication (or IPC for short) are mechanisms provid.pdf
Inter-Process-Communication (or IPC for short) are mechanisms provid.pdf
aesalem06
B)120.0° note: So3 is sp2-hybridized and adopts a trigonal planar geometry. Solution B)120.0° note: So3 is sp2-hybridized and adopts a trigonal planar geometry..
B)120.0° note So3 is sp2-hybridized and adopts .pdf
B)120.0° note So3 is sp2-hybridized and adopts .pdf
aesalem06
imple cellular phones that do very little over voice and text line of work generally include one process unit that handles everything - computer programme (keyboard, display), RF process, battery management, etc. These ar SoC - Systems on Chip and ar nearer to microcontrollers than microprocessors as a result of they decide to do everything - from process, to device interfacing, to memory, to program and knowledge storage, etc - on one chip. Smartphones have multiple microprocessors and microcontrollers in them. the most processor could be a chip with a bus to speak with memory on separate chips (though usually contained within the same IC package), and busses to speak with the remainder of the devices. they typically contain a number of the controllers, like the show controller, so that they have a number of the options of microcontrollers, however they\'re still additional chip than microcontroller. The cellular chipset typically includes a microprocessor/microcontroller that basically blurs the lines. the newest generation of smartphones usually tend additional toward microprocessors for the RF chipset, thus on provide the manufacturer flexibility by mistreatment software package style instead of hardware style for a few options, however they are doing have additional items of the hardware interface (rf, etc) on the chip itself. The rest of the phone has many microcontrollers. These management the touchscreen, audio, sensors, cameras, etc. So it depends on the kind of phone you are thinking of. an easy phone uses a microcontroller. a posh phone uses each. Solution imple cellular phones that do very little over voice and text line of work generally include one process unit that handles everything - computer programme (keyboard, display), RF process, battery management, etc. These ar SoC - Systems on Chip and ar nearer to microcontrollers than microprocessors as a result of they decide to do everything - from process, to device interfacing, to memory, to program and knowledge storage, etc - on one chip. Smartphones have multiple microprocessors and microcontrollers in them. the most processor could be a chip with a bus to speak with memory on separate chips (though usually contained within the same IC package), and busses to speak with the remainder of the devices. they typically contain a number of the controllers, like the show controller, so that they have a number of the options of microcontrollers, however they\'re still additional chip than microcontroller. The cellular chipset typically includes a microprocessor/microcontroller that basically blurs the lines. the newest generation of smartphones usually tend additional toward microprocessors for the RF chipset, thus on provide the manufacturer flexibility by mistreatment software package style instead of hardware style for a few options, however they are doing have additional items of the hardware interface (rf, etc) on the chip itself. The rest of the phone has many microcontrollers..
imple cellular phones that do very little over voice and text line o.pdf
imple cellular phones that do very little over voice and text line o.pdf
aesalem06
Given Data: Consider MTU in Bit IP header:16-bit Header:320 bit Appended header:420 bit Target maximum packet size of:1500 bit 420 header bits for the transport layer, 320 header bits for the IP layer. so total header=740 Bit to delivered at destination=1500-740=760 Solution Given Data: Consider MTU in Bit IP header:16-bit Header:320 bit Appended header:420 bit Target maximum packet size of:1500 bit 420 header bits for the transport layer, 320 header bits for the IP layer. so total header=740 Bit to delivered at destination=1500-740=760.
Given DataConsider MTU in Bit IP header16-bit Header320 bit.pdf
Given DataConsider MTU in Bit IP header16-bit Header320 bit.pdf
aesalem06
Features that PL/SQL shares with other standard programming languages: Solution Features that PL/SQL shares with other standard programming languages:.
Features that PLSQL shares with other standard programming language.pdf
Features that PLSQL shares with other standard programming language.pdf
aesalem06
Draw two electron arrow towards Zn. somewhere in the middle between zinc and oxygen ...make sure it connects the zinc in the templet. Solution Draw two electron arrow towards Zn. somewhere in the middle between zinc and oxygen ...make sure it connects the zinc in the templet..
Draw two electron arrow towards Zn. somewhere in the middle between .pdf
Draw two electron arrow towards Zn. somewhere in the middle between .pdf
aesalem06
Confidence Interval CI = x Solution Confidence Interval CI = x.
Confidence IntervalCI = xSolutionConfidence IntervalCI = x.pdf
Confidence IntervalCI = xSolutionConfidence IntervalCI = x.pdf
aesalem06
A)2moles of calcium reacts with 1mole of O2 to form 2moles of calcium oxide. no.of terms of particular element on right side should be equal to left side here it is equal hence the equation is balanced B)1mole of methane reacts with 2moles of O2 to form 2moles of carbondioxide and 2moles of water Solution A)2moles of calcium reacts with 1mole of O2 to form 2moles of calcium oxide. no.of terms of particular element on right side should be equal to left side here it is equal hence the equation is balanced B)1mole of methane reacts with 2moles of O2 to form 2moles of carbondioxide and 2moles of water.
A)2moles of calcium reacts with 1mole of O2 to fo.pdf
A)2moles of calcium reacts with 1mole of O2 to fo.pdf
aesalem06
Answer: Two Yellow colored and two ebony colored fruit flies are provided as mentioned in the question. It is not known to us, of them, which one is the dominant trait for the body color character. Case 1: Cross #1 Yellow colored fruit flies are crossed with ebony colored fruit flies and as a result all progeny obtained in the F1 generation are of a single color- Ebony. This observation leads to the impression that Ebony coloration is the dominant trait whereas Yellow is the recessive trait. Both the parents are homozygous for the trait and are pure breeding varieties. If that would not have been the case, in the F1 generation itself two variants- both Ebony and Yellow colored progeny would have appeared. According to Mendelian laws, Let, \'Y\' be the factor responsible for Ebony coloration and \'y\' be the factor responsible for Yellow coloration. Thus, in the parental generation, Genotype of the parent having Ebony coloration (homozygous pure breeding variety) - YY Genotype of the parent having Yellow coloration (homozygous pure breeding variety) - yy Hence, genotype of the progeny in F1 generation - Yy Case 2 : Cross #2 In the present case it was observed that 113 Yellow fruit flies were obtained and 120 Ebony fruit flies were obtained in the F1 generation as a result of a cross between an ebony & another yellow fruit fly. So, in the F1 generation, both the variants were obtained almost approximately in 1:1 phenotypic ratio. Since both the variants were obtained, it may be concluded that parent bearing ebony coloration was heterozygous whereas the yellow colored fly was a pure breeding variety. Thus, Genotype of the parent having Ebony coloration - Yy (heterozygous) Genotype of the parent having Yellow coloration -yy (homozygous) Ebony colored parent gives rise to two types of gametes- one bearing \'Y\' factor and the other bearing \'y\' factor. Yellow colored parent gives rise to only gametes bearing \'y\' factor. Gametes Y y y Yy (heterozygous ebony) yy (homozygous yellow) y Yy yy Progeny obtained in the F1 generation have genotype Yy ( heterozygous ebony) and yy(homozygous yellow). Case 3: Cross #3 Two ebony offspring are crossed having genotype \'Yy\'. Genotype of the progeny in F2 generation can be obtained by using checkerboard- Gametes Y y Y YY (homozygous Ebony) Yy (heterozygous ebony) y Yy (heterozygous ebony) yy (homozygous yellow) Thus, phenotypic rato between Ebony & Yellow colored fruit fly -3:1 (Ebony: Yellow) Genotypic ratio in F2 generation - YY: Yy : yy - 1: 2: 1 Case 4 :Cross #4 Mating is performed between Ebony colored fruit fly #3 having genotype \'Yy\' and Ebony colored fruit fly #2 having genotype \'YY\'. #3 ebony colored fly gives rise to to different types of gametes- one bearing \'Y\' allele and the other bearing \'y\' allele for body color character. The genotype and the phenotype of the progeny in this case can be obtained using checkerboard. Gametes Y y Y YY ( homozygous ebony) Yy (heterozygous ebony) Y YY Yy Th.
AnswerTwo Yellow colored and two ebony colored fruit flies are pr.pdf
AnswerTwo Yellow colored and two ebony colored fruit flies are pr.pdf
aesalem06
answer B) Cl - F F creates strong dipole moment for a, trigonal planar, dipole moment is distributed equally. C, linear, also distributed equally D) linear E) tetrahedral, distributed equally Solution answer B) Cl - F F creates strong dipole moment for a, trigonal planar, dipole moment is distributed equally. C, linear, also distributed equally D) linear E) tetrahedral, distributed equally.
answer B) Cl - FF creates strong dipole momentfor a, trigonal pl.pdf
answer B) Cl - FF creates strong dipole momentfor a, trigonal pl.pdf
aesalem06
ANSWER: The events occuring in Complex I in order are 3, 2, 1, 4 NADH is oxidized to NAD+ by reducing FMN to FMNH2 while oxidising Fe+2 to Fe+3 for the electron transport. Finally the CoQ is reduced to CoQH2 after which the electron enters in Complex II Solution ANSWER: The events occuring in Complex I in order are 3, 2, 1, 4 NADH is oxidized to NAD+ by reducing FMN to FMNH2 while oxidising Fe+2 to Fe+3 for the electron transport. Finally the CoQ is reduced to CoQH2 after which the electron enters in Complex II.
ANSWER The events occuring in Complex I in order are 3, 2, 1, 4NA.pdf
ANSWER The events occuring in Complex I in order are 3, 2, 1, 4NA.pdf
aesalem06
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Nope. When an atom is ionized it only affects the.pdf
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HOCl (aq) + HCI(aq) - H2O(l) + Cl2 (g) 6Cl2(g) +.pdf
HOCl (aq) + HCI(aq) - H2O(l) + Cl2 (g) 6Cl2(g) +.pdf
In Atmospehere Pressure = 105760 = 0.138 Atmo.pdf
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H2SO4 is a strong electrolyte. It dissociates alm.pdf
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DE is positive in exothermic reactions .pdf
DE is positive in exothermic reactions .pdf
The yield to maturity (YTM) is the measure of bonds rate of return t.pdf
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TDS (Total Dissolved Solids ) is a measure of the combined content o.pdf
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magnesium oxide + water = magnesium hydroxideMgO (s) + H2O (l) =.pdf
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Meiosis, inheritance and variationThe uniqueness and similarities.pdf
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imple cellular phones that do very little over voice and text line o.pdf
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Confidence IntervalCI = xSolutionConfidence IntervalCI = x.pdf
Confidence IntervalCI = xSolutionConfidence IntervalCI = x.pdf
A)2moles of calcium reacts with 1mole of O2 to fo.pdf
A)2moles of calcium reacts with 1mole of O2 to fo.pdf
AnswerTwo Yellow colored and two ebony colored fruit flies are pr.pdf
AnswerTwo Yellow colored and two ebony colored fruit flies are pr.pdf
answer B) Cl - FF creates strong dipole momentfor a, trigonal pl.pdf
answer B) Cl - FF creates strong dipole momentfor a, trigonal pl.pdf
ANSWER The events occuring in Complex I in order are 3, 2, 1, 4NA.pdf
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Program code examples (known also as worked examples) play a crucial role in learning how to program. Instructors use examples extensively to demonstrate the semantics of the programming language being taught and to highlight the fundamental coding patterns. Programming textbooks allocate considerable space to present and explain code examples. To make the process of studying code examples more interactive, CS education researchers developed a range of tools to engage students in the study of code examples. These tools include codecasts (codemotion,codecast,elicasts), interactive example explorers (WebEx, PCEX), and tutoring systems (DeepTutor). An important component in all types of worked examples is code explanations associated with specific code lines or code chunks of an example. The explanations connect examples with general programming knowledge explaining the role and function of code fragments or their behavior. In textbooks, these explanations are usually presented as comments in the code or as explanations on the margins. The example explorer tools allow students to examine these explanations interactively. Tutoring systems, which engage students in explaining the code, use these model explanations to check student responses and provide scaffolding. In all these cases, to make a worked example re-usable beyond its presentation in a lecture, the explanations have to be authored by instructors or domain experts i.e., produced and integrated into a specific system. As the experience of the last 10 years demonstrated, these explanations are hard to obtain. Those already collected are usually “locked” in a specific example-focused system and can’t be reused. The purpose of this working group is to support broader re-used of worked examples augmented with explanations. Our current plan is to develop а standard approach to represent explained examples. This approach will enable an example created for any of the existing systems to be explored in a standard format and imported into any other example-focused system. We plan to follow a successful experience of the PEML working group focused on re-using programming exercises.
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The Story of Village Palampur Class 9 Free Study Material PDF
The Story of Village Palampur Class 9 Free Study Material PDF
Vivekanand Anglo Vedic Academy
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Graduate Outcomes Presentation Slides - English (v3).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
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