One month liquidity index=10(.981)+90(.85.92)=0.93Liquidity .pdf

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One month liquidity index=10%*(.98/1)+90%*(.85/.92)=0.93 Liquidity index value lies between 0 to 1. Smaller the liquidity index signifies higher liquidity risk. Solution One month liquidity index=10%*(.98/1)+90%*(.85/.92)=0.93 Liquidity index value lies between 0 to 1. Smaller the liquidity index signifies higher liquidity risk..

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import java.awt.; Using AWT container and component classes im.pdf

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If possible let T+e1+e2+e2 be non planar Then either one of e1, e2 or e3 or T must be in another plane T is a tree of order atleast 4. i.e maximum number of children per root is more than or equal to 4. Hence vertices containing e1, e2, e3 have to be planar. T being a tree with root is also in the same plane Hence proved Solution If possible let T+e1+e2+e2 be non planar Then either one of e1, e2 or e3 or T must be in another plane T is a tree of order atleast 4. i.e maximum number of children per root is more than or equal to 4. Hence vertices containing e1, e2, e3 have to be planar. T being a tree with root is also in the same plane Hence proved.

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How should the judicial system treat juvenile criminal offenders? Juveniles must be prevented from having frequent contact with the criminal justice system and intervening to support juveniles desist from crime. They should be assisted to grow out of crime are the ways in which the judicial system should treat juvenile criminal offenders. How does adolescent brain science influence juvenile justice cases? Juveniles usually offense against property than offense against the person. The juvenile offenses are mostly attention seeking and need driven. Hence the adolescent brain science treats the juvenile justice cases as a crime where they are mature and do not think about the immature brain. What are some of the consequences of relying on brain science to defend adolescents? Some defendants deserve special consideration because they have brains that are immature or impaired, due to which a specific age group cannot be brought into consideration, some individuals brain will be completely matured by the age of 16 hence can be considered as adult. It will also provide way for many individuals to escape from the justice for their offense. For example, if a prosecutor used an MRI scan to show that a 16-year-old who committed a capital crime had a very mature brain, \"Would we then insist that we execute that juvenile?\" Solution How should the judicial system treat juvenile criminal offenders? Juveniles must be prevented from having frequent contact with the criminal justice system and intervening to support juveniles desist from crime. They should be assisted to grow out of crime are the ways in which the judicial system should treat juvenile criminal offenders. How does adolescent brain science influence juvenile justice cases? Juveniles usually offense against property than offense against the person. The juvenile offenses are mostly attention seeking and need driven. Hence the adolescent brain science treats the juvenile justice cases as a crime where they are mature and do not think about the immature brain. What are some of the consequences of relying on brain science to defend adolescents? Some defendants deserve special consideration because they have brains that are immature or impaired, due to which a specific age group cannot be brought into consideration, some individuals brain will be completely matured by the age of 16 hence can be considered as adult. It will also provide way for many individuals to escape from the justice for their offense. For example, if a prosecutor used an MRI scan to show that a 16-year-old who committed a capital crime had a very mature brain, \"Would we then insist that we execute that juvenile?\".

How should the judicial system treat juvenile criminal offendersJ.pdf

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coordinated movements of gastrulation begin. This dramatic process transforms the simple hollow ball of cells into a multilayered structure with a central gut tube and bilateral symmetry.many of the cells on the outside of the embryo are moved inside it. Subsequent development depends on the interactions of the inner, outer, and middle layers of cells thus formed: the endoderm on the inside, consisting of the cells that have moved into the interior to form the primitive gut; the ectoderm on the outside, consisting of cells that have remained external; and the mesoderm between them, consisting of cells that detach from the epithelium to form a more loosely organized embryonic connective tissue. eventual pattern of muscles-in the limbs, for example-is determined by the routes that the migrant cells follow and the selection of sites that they colonize. The embryonic connective tissues form the framework through which the myoblasts travel and provide signals that guide their distribution. No matter which somite they come from, myoblasts that migrate into a forelimb bud will form the pattern of muscles appropriate to a forelimb, and those that migrate into a hindlimb bud will form the pattern appropriate to a hindlimb. Limb formation results from series of epithelial-mesenchymal inductions between the mesenchymal cells of the lateral plate mesoderm and the overlying ectodermal cells. Cells from the lateral plate mesoderm and the myotome migrate to the limb field and proliferate to create the limb bud. The lateral plate cells produce the cartilaginous and skeletal portions of the limb while the myotome cells produce the muscle components. The lateral plate mesodermal cells secrete a fibroblast growth factor (FGF7 andFGF10, presumably) to induce the overlying ectoderm to form an important organizing structure called the apical ectodermal ridge(AER).The AER reciprocatively secretes FGF8 and FGF4 which maintains the FGF10 signal and induces proliferation in the mesoderm.[citation needed] The position of FGF10 expression is regulated by Wnt8c in the hindlimb and Wnt2b in the forelimb. The forelimb and the hindlimb are specified by their position along the anterior/posterior axis and possibly by two T-box containing transcription factors: Tbx5 and Tbx4, respectively. DKK1 activation in osteoblasts is the underlying cause of glucocorticoid- and estrogen deficiency–mediated osteoporosis, and at least partially underlies the teratogenic effects of thalidomide on limb development. both thalidomide-induced PCD and limb deformities could be partially inhibited by blocking Dkk1 or activating Wnt signaling downstream of the ligand-receptor interaction Solution coordinated movements of gastrulation begin. This dramatic process transforms the simple hollow ball of cells into a multilayered structure with a central gut tube and bilateral symmetry.many of the cells on the outside of the embryo are moved inside it. Subsequent development depends on the interactions .

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Cu(II)Oxide has solid with mono clinic structure .m.p is 12010 C and b.p is 2000 0C. It is insoluble in water.it appears as block powder. Cu (II) nitrate has solid with orthorhobic structure . it appears in blue color. m.p 2560C an d b.p is 1700 C.it is very soluble in water. Cu (II) hydroxide is a blue green solid and m.p is 800 C. it is soluble in NH4OH. Cu(II) sulfate is blue color solid with m.p is 1100 C . It has triclinic crystal structure. it is soluble in water and it has octahedral geometry. Solution Cu(II)Oxide has solid with mono clinic structure .m.p is 12010 C and b.p is 2000 0C. It is insoluble in water.it appears as block powder. Cu (II) nitrate has solid with orthorhobic structure . it appears in blue color. m.p 2560C an d b.p is 1700 C.it is very soluble in water. Cu (II) hydroxide is a blue green solid and m.p is 800 C. it is soluble in NH4OH. Cu(II) sulfate is blue color solid with m.p is 1100 C . It has triclinic crystal structure. it is soluble in water and it has octahedral geometry..

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Assign a major brain structure to each of the cehalon below diencephalon The diencephalon is a piece of the prosencephalon (forebrain), which creates from the first essential cerebral vesicle. The prosencephalon separates into a caudal diencephalon and rostral telencephalon. The cerebral sides of the equator create from the sides of the telencephalon, each containing a parallel ventricle. The diencephalon comprises of structures that are sidelong to the third ventricle, and incorporates the thalamus, the hypothalamus, the epithalamus and the subthalamus.The diencephalon comprises of the accompanying structures: Myelencephalon The myelencephalon or afterbrain is the most back area of the embryonic hindbrain, from which the medulla oblongata creates it comprises of the accompanying structures: Tectum, Cerebral peduncle, Pretectum, Cerebral water system. metacephalon The metencephalon is the embryonic part of the hindbrain that separates into the pons and the cerebellum. It contains diencephalon The diencephalon is a piece of the prosencephalon (forebrain), which creates from the first essential cerebral vesicle. The prosencephalon separates into a caudal diencephalon and rostral telencephalon. The cerebral sides of the equator create from the sides of the telencephalon, each containing a parallel ventricle. The diencephalon comprises of structures that are sidelong to the third ventricle, and incorporates the thalamus, the hypothalamus, the epithalamus and the subthalamus.The diencephalon comprises of the accompanying structures:ThalamusHypothalamus including the NeurohypophysisEpithalamus which comprises ofForemost and Posterior Paraventricular coresAverage and parallel Habenular coresStria medullaris thalamiBack commissurePineal bodySubthalamus Myelencephalon The myelencephalon or afterbrain is the most back area of the embryonic hindbrain, from which the medulla oblongata creates it comprises of the accompanying structures: Tectum, Cerebral peduncle, Pretectum, Cerebral water system. metacephalon The metencephalon is the embryonic part of the hindbrain that separates into the pons and the cerebellum. It containsa segment of the fourth ventricle andthe trigeminal nerve (CN V),abducens nerve (CN VI),facial nerve (CN VII), anda bit of the vestibulocochlear nerve (CN VIII). Solution Assign a major brain structure to each of the cehalon below diencephalon The diencephalon is a piece of the prosencephalon (forebrain), which creates from the first essential cerebral vesicle. The prosencephalon separates into a caudal diencephalon and rostral telencephalon. The cerebral sides of the equator create from the sides of the telencephalon, each containing a parallel ventricle. The diencephalon comprises of structures that are sidelong to the third ventricle, and incorporates the thalamus, the hypothalamus, the epithalamus and the subthalamus.The diencephalon comprises of the accompanying structures: Myelencephalon The myelencephalon or afterbrain is the m.

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B. Assets are listed in descending order of liquidity. The most liquid assets are presented first, followed by less liquid assets. Current assets = net working capital - current liabilities.(so A is false) Share holders equity = total assets - total liabilities (so C D and E are incorrect) Solution B. Assets are listed in descending order of liquidity. The most liquid assets are presented first, followed by less liquid assets. Current assets = net working capital - current liabilities.(so A is false) Share holders equity = total assets - total liabilities (so C D and E are incorrect).

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#include #include #include double total(int array[], int n) { int i, sum =0; for (i = 0; i < n; i++) { sum = sum + array[i]; } return sum; } int main() { srand(time(NULL)); int i, n, rand_number; printf(\"Enter an integer: \"); scanf(\"%d\",&n); int array[n]; for (i = 0; i < n; i++) { rand_number = rand() % 1000; array[i] = rand_number; } int max = array[0]; int min = array[0]; int min_pos; int max_pos; for (i = 0; i < n; i++) { if(array[i] > max) { max = array[i]; max_pos = i; } if(array[i] < min) { min = array[i]; min_pos = i; } } int sum = total(array,n); double average = sum/n; printf(\"min: %d pos: %d\ \",min, min_pos); printf(\"max: %d pos: %d\ \",max, max_pos); printf(\"sum: %d \ \",sum); printf(\"average: %lf \ \",average); printf(\"Pos\\t|\\tval\ \"); printf(\"-----------------------\ \"); for (i = 0; i < n; ++i) { printf(\"%d\\t|\\t%d\ \",i,array[i]); } printf(\"\ \"); return 0; } /* output: Enter an integer: 30 min: 5 pos: 28 max: 998 pos: 0 sum: 14093 average: 469.000000 Pos | val ----------------------- 0 | 998 1 | 403 2 | 523 3 | 787 4 | 37 5 | 14 6 | 815 7 | 232 8 | 594 9 | 555 10 | 531 11 | 52 12 | 550 13 | 530 14 | 265 15 | 839 16 | 247 17 | 902 18 | 99 19 | 120 20 | 648 21 | 101 22 | 6 23 | 983 24 | 759 25 | 664 26 | 371 27 | 705 28 | 5 29 | 758 */ Solution #include #include #include double total(int array[], int n) { int i, sum =0; for (i = 0; i < n; i++) { sum = sum + array[i]; } return sum; } int main() { srand(time(NULL)); int i, n, rand_number; printf(\"Enter an integer: \"); scanf(\"%d\",&n); int array[n]; for (i = 0; i < n; i++) { rand_number = rand() % 1000; array[i] = rand_number; } int max = array[0]; int min = array[0]; int min_pos; int max_pos; for (i = 0; i < n; i++) { if(array[i] > max) { max = array[i]; max_pos = i; } if(array[i] < min) { min = array[i]; min_pos = i; } } int sum = total(array,n); double average = sum/n; printf(\"min: %d pos: %d\ \",min, min_pos); printf(\"max: %d pos: %d\ \",max, max_pos); printf(\"sum: %d \ \",sum); printf(\"average: %lf \ \",average); printf(\"Pos\\t|\\tval\ \"); printf(\"-----------------------\ \"); for (i = 0; i < n; ++i) { printf(\"%d\\t|\\t%d\ \",i,array[i]); } printf(\"\ \"); return 0; } /* output: Enter an integer: 30 min: 5 pos: 28 max: 998 pos: 0 sum: 14093 average: 469.000000 Pos | val ----------------------- 0 | 998 1 | 403 2 | 523 3 | 787 4 | 37 5 | 14 6 | 815 7 | 232 8 | 594 9 | 555 10 | 531 11 | 52 12 | 550 13 | 530 14 | 265 15 | 839 16 | 247 17 | 902 18 | 99 19 | 120 20 | 648 21 | 101 22 | 6 23 | 983 24 | 759 25 | 664 26 | 371 27 | 705 28 | 5 29 | 758 */.

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Let x be the natual abundance of lower mass Since B has only two isotopes so the abundance of heavear mass is 1 - x The average atomic mass = [(mass of lower isotope * its abundance) + ( mass of heavier mass* itys abundance)] / no.of isotopes 10.845 = [(10.013* x ) + ( 11.009*(1-x) ] / 2 ---> x = 0.508 % abundance = 0.508*100 = 50.8 % Solution Let x be the natual abundance of lower mass Since B has only two isotopes so the abundance of heavear mass is 1 - x The average atomic mass = [(mass of lower isotope * its abundance) + ( mass of heavier mass* itys abundance)] / no.of isotopes 10.845 = [(10.013* x ) + ( 11.009*(1-x) ] / 2 ---> x = 0.508 % abundance = 0.508*100 = 50.8 %.

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1) general media like TSA allows growth of any bacteria, while selective/ differential media are used to select for specific species/ and differentiate among closely related species of bacteria. 2)hemolysis helps to detect bacteria that can produce enzymes to lyse blood cells. Different species of Streptococcus shows a different level of enzyme production, partial decomposition gives green color and complete decomposition of blood gives red color. 3) The methylene blue indicator is used to verify the anaerobic condition in the jar. The blue color must become colourless after incubation to confirm the absence of oxygen. The strip color was observed Blue thus indicating presence of oxygen. The organisms growing on the plate are aerobes and aerobes should not grow in anaerobe condition. 4) enteric bacteria are commonly found in feces and the UTI is caused by the infection of aerobic fecal flora. Solution 1) general media like TSA allows growth of any bacteria, while selective/ differential media are used to select for specific species/ and differentiate among closely related species of bacteria. 2)hemolysis helps to detect bacteria that can produce enzymes to lyse blood cells. Different species of Streptococcus shows a different level of enzyme production, partial decomposition gives green color and complete decomposition of blood gives red color. 3) The methylene blue indicator is used to verify the anaerobic condition in the jar. The blue color must become colourless after incubation to confirm the absence of oxygen. The strip color was observed Blue thus indicating presence of oxygen. The organisms growing on the plate are aerobes and aerobes should not grow in anaerobe condition. 4) enteric bacteria are commonly found in feces and the UTI is caused by the infection of aerobic fecal flora..

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