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ROE = 14.3% Payout Ratio = 42% Sustainable Growth Rate = ROE * (1 - Payout Ratio) Sustainable Growth Rate = 14.30% * (1 - 0.42) Sustainable Growth Rate = 14.30% * 0.58 Sustainable Growth Rate = 8.29% Solution ROE = 14.3% Payout Ratio = 42% Sustainable Growth Rate = ROE * (1 - Payout Ratio) Sustainable Growth Rate = 14.30% * (1 - 0.42) Sustainable Growth Rate = 14.30% * 0.58 Sustainable Growth Rate = 8.29%.

ROE = 14.3 Payout Ratio = 42Sustainable Growth Rate = ROE (1.pdf

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Sample code for Discrete Cosine Transform for an imageSolution.pdf

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Present PbNot present HgSolutionPresent PbNot present .pdf

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dear pls put prob 14 and 20... i cant solve witho.pdf

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D) is the answer. note: BF3 is a strong Lewis acid (electron pair acceptor) that reacts with a Lewis base (electron pair donor). CH3NH2 has a lone pair of electrons at N atom and thus acts as a Lewis base to react with BF3. Solution D) is the answer. note: BF3 is a strong Lewis acid (electron pair acceptor) that reacts with a Lewis base (electron pair donor). CH3NH2 has a lone pair of electrons at N atom and thus acts as a Lewis base to react with BF3..

D) is the answer. note BF3 is a strong Lewis aci.pdf

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One major similarity with soils dominated by calcium and sodium ions.pdf

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Interval.SolutionInterval..pdf

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C) II and IV .pdf

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Hi! Since Lithium has 1 single valence electron (it is located in Group/Column 1 on the periodic table) and Fluorine is lacking 1 single valence electron (it is located in Group 7A), the most likely ionic compound to form here would be: LiF, Lithium Fluoride I hope this helps! Please rate if it does :) Solution Hi! Since Lithium has 1 single valence electron (it is located in Group/Column 1 on the periodic table) and Fluorine is lacking 1 single valence electron (it is located in Group 7A), the most likely ionic compound to form here would be: LiF, Lithium Fluoride I hope this helps! Please rate if it does :).

Hi!Since Lithium has 1 single valence electron (it is located in G.pdf

angelcolluctions

I think that 0.434SolutionI think that 0.434.pdf

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B)diffusion a)effusion .pdf

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Determine decimal fraction of C-14 remaining: 43.3 / 58.2 = 0.744 Determine how many half-lives have elapsed (1/2)n = 0.744 n log (1/2) = log (0.744) n = 0.42 Determine length of time elapsed: 5715 yr X 0.42 = 2438.16 Solution Determine decimal fraction of C-14 remaining: 43.3 / 58.2 = 0.744 Determine how many half-lives have elapsed (1/2)n = 0.744 n log (1/2) = log (0.744) n = 0.42 Determine length of time elapsed: 5715 yr X 0.42 = 2438.16.

$Determine decimal fraction of C-14 remaining 43.3 58.2 = 0.74.pdf$ $Determine decimal fraction of C-14 remaining 43.3 58.2 = 0.74.pdf$

Determine decimal fraction of C-14 remaining 43.3 58.2 = 0.74.pdf

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c) All colonies sub-cultured from blood culture broth media must be identified. Multiple bottles improves the likelihood that diphtheroids and coagulase negative Staphylococcus will be treated. During sepsis, multiple organsims may be present in the infection. Time separation and culture of blood in different environement allows different types of bacteria to be easily separated and identified. Solution c) All colonies sub-cultured from blood culture broth media must be identified. Multiple bottles improves the likelihood that diphtheroids and coagulase negative Staphylococcus will be treated. During sepsis, multiple organsims may be present in the infection. Time separation and culture of blood in different environement allows different types of bacteria to be easily separated and identified..

c) All colonies sub-cultured from blood culture broth media must be .pdf

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Answer:Four phenotypes are possible in a flock of ducks that contains all the possible combinations of these three alleles.As in the ABO blood system 4 phenotypes are possible in this case.GenotypePhenotypeHH,Hi(H)II,Ii(I)HI(HI)ii(i) Solution Answer:Four phenotypes are possible in a flock of ducks that contains all the possible combinations of these three alleles.As in the ABO blood system 4 phenotypes are possible in this case.GenotypePhenotypeHH,Hi(H)II,Ii(I)HI(HI)ii(i).

AnswerFour phenotypes are possible in a flock of ducks that contain.pdf

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Answer: 1. b. Agrobacterium The ability of Agrobacterium to transfer genes to plants and fungi is used in biotechnology, in particular, genetic engineering for plant improvement. A modified Ti or Ri plasmid can be used. 2. c. Artificial active Artificially acquired active immunity can be induced by a vaccine, a substance that contains the antigen. 3. c. Attenuated An attenuated vaccine is a vaccine created by reducing the virulence of a pathogen, but still keeping it viable (or \"live\"). Attenuation takes an infectious agent and alters it so that it becomes harmless or less virulent. 4. a. Sign Symptom is nothing but the abnormal phenotype expressed in living organisms. Symdrome means the all abnormalities associated with the disease. 5. b. The eye Normal flora can be found in many sites of the human bodyincluding the skin (especially the moist areas, such as the groin and between the toes), respiratory tract (particularly the nose), urinary tract, and the digestive tract (primarily the mouth and the colon). Solution Answer: 1. b. Agrobacterium The ability of Agrobacterium to transfer genes to plants and fungi is used in biotechnology, in particular, genetic engineering for plant improvement. A modified Ti or Ri plasmid can be used. 2. c. Artificial active Artificially acquired active immunity can be induced by a vaccine, a substance that contains the antigen. 3. c. Attenuated An attenuated vaccine is a vaccine created by reducing the virulence of a pathogen, but still keeping it viable (or \"live\"). Attenuation takes an infectious agent and alters it so that it becomes harmless or less virulent. 4. a. Sign Symptom is nothing but the abnormal phenotype expressed in living organisms. Symdrome means the all abnormalities associated with the disease. 5. b. The eye Normal flora can be found in many sites of the human bodyincluding the skin (especially the moist areas, such as the groin and between the toes), respiratory tract (particularly the nose), urinary tract, and the digestive tract (primarily the mouth and the colon)..

Answer1. b. AgrobacteriumThe ability of Agrobacterium to transf.pdf

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H2SO4 is stronger than HSO4- since HSO4 already has one negative charge - gaining a second is unfavorable compared to H2SO4 which would gain only a -1 charge after deprotonation. HNO3 is stronger than HNO2 since the greater number of O atoms exerts a stronger electron- withdrawing effect on the O-H electrons. Additionally, the NO3- ion has more resonance structures and thus greater resonance stability than the NO2- ion. H2S is a stronger acid than H2O since hydrogen forms weaker bonds to elements lower in the periodic table (larger elements). This is why HI is stronger than HBr, for example. H2S is a stronger acid than PH3 since S is more electronnegative (further to the right in the periodic table) than P. Thus, S prefers to have a negative charge compared to P. For part 2: We know that Kw = [H3O*][OH-] = 10^- 14 M. In pure water, these ions are at equal concentrations, so [H+] = [OH-] = 10^-7 M. Now, we have 1.00 mL = 0.001 L. So, we have 10-7 mol/L * 0.001 L = 10^-10 moles of each ion. Now, 1 mol = 6.02 * 10^23 ions, so we have: 6.02*10^23 * 10^-10 = 6.02 * 10^13 [H3O*] ions and 6.02 * 10^13 [OH-] ions. Solution H2SO4 is stronger than HSO4- since HSO4 already has one negative charge - gaining a second is unfavorable compared to H2SO4 which would gain only a -1 charge after deprotonation. HNO3 is stronger than HNO2 since the greater number of O atoms exerts a stronger electron- withdrawing effect on the O-H electrons. Additionally, the NO3- ion has more resonance structures and thus greater resonance stability than the NO2- ion. H2S is a stronger acid than H2O since hydrogen forms weaker bonds to elements lower in the periodic table (larger elements). This is why HI is stronger than HBr, for example. H2S is a stronger acid than PH3 since S is more electronnegative (further to the right in the periodic table) than P. Thus, S prefers to have a negative charge compared to P. For part 2: We know that Kw = [H3O*][OH-] = 10^- 14 M. In pure water, these ions are at equal concentrations, so [H+] = [OH-] = 10^-7 M. Now, we have 1.00 mL = 0.001 L. So, we have 10-7 mol/L * 0.001 L = 10^-10 moles of each ion. Now, 1 mol = 6.02 * 10^23 ions, so we have: 6.02*10^23 * 10^-10 = 6.02 * 10^13 [H3O*] ions and 6.02 * 10^13 [OH-] ions..

H2SO4 is stronger than HSO4- since HSO4 already has one negative cha.pdf

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ANS: #include #include int main() { int x,y=31; printf(\"input %d\",l_bit); binar_ar(y); for (int i = 7; i >= 0; i--) printf(\"%d/t%d\",get_bit(y,i),l_bit); x=1; printf(\"%d th bit %d\",x,l_bit); y = set_bit(y, x,true); for (int i =15; i >= 0; i--) printf(\"%d/t%d\",get_bit(y,i),l_bit); x = 1; printf(\"%d th bit %d\",x,l_bit); y = set_bit(y, x, false); for (int i=31; i >= 0; i--) printf(\"%d/t%d\",get_bit(y,i),l_bit); } void binar_ar(int n) { for(int i = 31; i > 0; i = i/2) { if(n & i) printf(\"1\"); else printf(\"0\"); } printf(\"%d\",l_bit); } bool get_bit(int n, int x) { return ( (n & (1 << x) ) > 0); } int set_bit(int n, int x, bool b) { if(b) return (n | (1 << x)) ; else { int ma = ~(1 << x); return n & ma; } } Solution ANS: #include #include int main() { int x,y=31; printf(\"input %d\",l_bit); binar_ar(y); for (int i = 7; i >= 0; i--) printf(\"%d/t%d\",get_bit(y,i),l_bit); x=1; printf(\"%d th bit %d\",x,l_bit); y = set_bit(y, x,true); for (int i =15; i >= 0; i--) printf(\"%d/t%d\",get_bit(y,i),l_bit); x = 1; printf(\"%d th bit %d\",x,l_bit); y = set_bit(y, x, false); for (int i=31; i >= 0; i--) printf(\"%d/t%d\",get_bit(y,i),l_bit); } void binar_ar(int n) { for(int i = 31; i > 0; i = i/2) { if(n & i) printf(\"1\"); else printf(\"0\"); } printf(\"%d\",l_bit); } bool get_bit(int n, int x) { return ( (n & (1 << x) ) > 0); } int set_bit(int n, int x, bool b) { if(b) return (n | (1 << x)) ; else { int ma = ~(1 << x); return n & ma; } }.

ANS#includestdio.h#includeconio.h int main() { int.pdf

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a. An interval or ratio scale.Solutiona. An interval or ratio .pdf

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A) Germ cell specification in Drosophila occurs by maternally contributed determinants. In Drosophilla, germ cells can easily be identified very early in embryogenesis, when their differentiation as germ cells is assured by the localisation of maternally inherited determinants before, or immediately following, fertilisation. During oogenesis in Drosophila melanogaster, RNAs and proteins are synthesised by the nurse cells. These products are transported through cytoplasmic bridges to the oocyte where They become localised to the posterior of the ooplasm both by molecular anchoring at the posterior of the oocyte, and by posterior-specific translational and transcriptional regulation. This posterior ooplasm is the germ plasm, or germ line determinant. During early embryogenesis, cells which inherit the germ plasm become the primordial germ cells. One of the important gene that contributes to germ cell specification in Drosophila is Nanos. b) Neural crest cell specification is due to the cell signalling and depends on BMP signalling in the prospective epidermis and Wnt signalling from the underlying mesoderm. C) Anterior -posterior specification in Drosophila is initiated by maternal effect genes that produce messenger RNA that are placed in different regions of egg . These mRNA encode transcriptional and translational regulatory protiens that diffuse through the syncytial blastoderm and activates or repress the expression of certain zygotic genes. One important maternal effect gene in this respect is bicoid gene. Solution A) Germ cell specification in Drosophila occurs by maternally contributed determinants. In Drosophilla, germ cells can easily be identified very early in embryogenesis, when their differentiation as germ cells is assured by the localisation of maternally inherited determinants before, or immediately following, fertilisation. During oogenesis in Drosophila melanogaster, RNAs and proteins are synthesised by the nurse cells. These products are transported through cytoplasmic bridges to the oocyte where They become localised to the posterior of the ooplasm both by molecular anchoring at the posterior of the oocyte, and by posterior-specific translational and transcriptional regulation. This posterior ooplasm is the germ plasm, or germ line determinant. During early embryogenesis, cells which inherit the germ plasm become the primordial germ cells. One of the important gene that contributes to germ cell specification in Drosophila is Nanos. b) Neural crest cell specification is due to the cell signalling and depends on BMP signalling in the prospective epidermis and Wnt signalling from the underlying mesoderm. C) Anterior -posterior specification in Drosophila is initiated by maternal effect genes that produce messenger RNA that are placed in different regions of egg . These mRNA encode transcriptional and translational regulatory protiens that diffuse through the syncytial blastoderm and activates or repress the expression of certa.

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