Q1) Create a frequency table for each of the following variables..pdf

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The document provides instructions for analyzing several variables from a dataset. It asks to: 1) Create frequency tables for race and political identity variables. 2) Create histograms for age and police contact variables. 3) Calculate measures of central tendency and dispersion for self-control and trust in police variables. 4) Transform school commitment raw scores to z-scores.

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Q1) Create a frequency table for each of the following variables.
a) Race
Let Hispanic be 1
Let White be 2
Let African American be 3
Let Asian be 4
b) Political Identity
Let Liberal be 1
Let Moderate be 2
Let Very Liberal be 3
Let Conservative be 4
Let Very Conservative 5
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Q2) Create a histogram showing the distribution of each of the following variables.
a) Age
b) Police Contact
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Q3) For each of the following variables, calculate the MEAN, MEDIAN, MODE, and Standard
Deviation.
a) Self-Control
b) Trust in Police
--------------------------------------------------------------------------------------------------------------------
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Q4) For each of the following variable, transform each RAW SCORE into a Z SCORE. a)
School Commitment

Race cumulativeloweruppermidpointwidthfrequencypercent
frequencypercent0.0<1.00.51.000.000.01.0<2.01.51.0426.7426.72.0<3.02.51.0746.71173.33.0<4
.03.51.0320.01493.34.0<5.04.51.016.715100.0
15100.0
Solution
Q1) Create a frequency table for each of the following variables.
a) Race
Let Hispanic be 1
Let White be 2
Let African American be 3
Let Asian be 4

b) Political Identity
Let Liberal be 1
Let Moderate be 2
Let Very Liberal be 3
Let Conservative be 4
Let Very Conservative 5
--------------------------------------------------------------------------------------------------------------------
-----------
Q2) Create a histogram showing the distribution of each of the following variables.
a) Age
b) Police Contact
--------------------------------------------------------------------------------------------------------------------

-----------
Q3) For each of the following variables, calculate the MEAN, MEDIAN, MODE, and Standard
Deviation.
a) Self-Control
b) Trust in Police
--------------------------------------------------------------------------------------------------------------------
-----------
Q4) For each of the following variable, transform each RAW SCORE into a Z SCORE. a)
School Commitment
Race cumulativeloweruppermidpointwidthfrequencypercent
frequencypercent0.0<1.00.51.000.000.01.0<2.01.51.0426.7426.72.0<3.02.51.0746.71173.33.0<4
.03.51.0320.01493.34.0<5.04.51.016.715100.0

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Ans: Triploidy is a desirable trait in all angiosperms except orchidaceae and Triploidy could be a desirable trait in agricultural species where the economic part especially fruits. Triploid hybrids have one of the most important traits, seedlessness, which is the characteristic for the fresh-fruit market. Triploid embryos are found in small seeds that do not germinate. Triploid plants raised from endosperm are generally sterile. Endosperm-raised triploid plants are of commercial value, e.g. timber-yielding plants, edible fruit plants or ornamentals propagated vegetatively and multiplied mainly through micropropagation. Triploid plants often show vigorous vegetative growth and in some species produce excellent seedless fruit. All cultivated bananas are triplods, and must be propagated vegetatively as they bear no seed. Some sugar beet varieties are triploid and these tend to be higher yielding than diploid varieties. In some species it is not desirable as they produce little or no seed, each generation of a triploid variety has to be made afresh by the hybridization of a diploid and a tetraploid parent. Eg: Commercial use of triploid fruits can be found in watermelons and are produced artificially by first developing tetraploids whicg are then crossed with diploid watermelon. In order to set fruit, the triploid watermelon crossed with desirable diploid pollen donar. It is an undesirable trait In some species it is not desirable as they produce little or no seed (may be sterile because they cannot form gametes with a balanced assortment of chromosomes), each generation of a triploid variety has to be made afresh by the hybridization of a diploid and a tetraploid parent. Eg: Diana\' (1963, USA) has large single, pure white petals with crimped edges. It is a triploid, so produces little or no viable seed, and thus continues to flower over a long period. It was followed in the 1980s by more triploid cultivars, in particular \'Aphrodite\' polyploidy leads to significant reduction in the amount of genetic diversity in some species. As genetically identical clones, a pest or disease that affects one individual is likely capable of affecting each of its clones. Eg: vast majority of commercially produced bananas are cloned from a single source, the Cavendish cultivar, and those plants are currently threatened worldwide by a newly discovered fungal disease to which they are highly susceptible. It is an undesirable trait in crop species where the economic part is seed Example : Orchidacea The Phalaenopsis species bear a very brightly colored and patterned flowers were reintroduced into cultivation. Many of these novelty species however produce inflorescences with only a very few, tiny, starry-shaped flowers. These diploid species have been crossed with the modern autopolyploid white types to improve their size, form, and floriferousness. In many instances, these characteristics were improved but flower color was substantially diluted. In addition, many .

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Phosphorous reacts readily with hydrogen to form hydrides; with halogens to form hallides; With oxygen to form oxides; With sulfur to form sulfides These are few of the important reactions of Phosphorous Solution Phosphorous reacts readily with hydrogen to form hydrides; with halogens to form hallides; With oxygen to form oxides; With sulfur to form sulfides These are few of the important reactions of Phosphorous.

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Molarity = moles / Litres of solution. Work out how many moles of each reagent you add to the initial solution. Then use the total volume of the initial solution to determine the molarity. Use this equation to solve. Solution Molarity = moles / Litres of solution. Work out how many moles of each reagent you add to the initial solution. Then use the total volume of the initial solution to determine the molarity. Use this equation to solve..

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How many ml of .20 M NaOh are required to neutralize 50.0 ml of .41 M HBr? M1 = molarity of HBr = 0.41 M : V1 = volume of HBr = 50.0mL M2 = molarity of NaOH = 0.20 : V2 = volume of NaOH By titration we have M1V1=M2V2 V2=(M1V1)/M2= (0.41*50)/0.20 = 102.5mL NaOH What is the pH of a solution to which 35.0 ml of .20 M HCl and 50.0 ml of .42M NaOH have been combined? Moles = molarity * volume Moles of HCl = 0.50M*(35/1000)L = 0.0175 mol HCl Moles of NaOH = 0.42 M * (50/1000)L=0.021 mol NaOH Moles of NaOH left in the solution = 0.021 - 0.0175 = 0.0035 mol NaOH Molarity of NaOH = moles of NaOH left/ total volume of the solution =0.0034/((50+35)/1000)L = 0.04 M pOH = -log[OH-] = -log(0.04)=1.40 pH + pOH = 14 pH = 14-pOH = 14-1.40 = 12.6 Hope it has help you.... !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Solution How many ml of .20 M NaOh are required to neutralize 50.0 ml of .41 M HBr? M1 = molarity of HBr = 0.41 M : V1 = volume of HBr = 50.0mL M2 = molarity of NaOH = 0.20 : V2 = volume of NaOH By titration we have M1V1=M2V2 V2=(M1V1)/M2= (0.41*50)/0.20 = 102.5mL NaOH What is the pH of a solution to which 35.0 ml of .20 M HCl and 50.0 ml of .42M NaOH have been combined? Moles = molarity * volume Moles of HCl = 0.50M*(35/1000)L = 0.0175 mol HCl Moles of NaOH = 0.42 M * (50/1000)L=0.021 mol NaOH Moles of NaOH left in the solution = 0.021 - 0.0175 = 0.0035 mol NaOH Molarity of NaOH = moles of NaOH left/ total volume of the solution =0.0034/((50+35)/1000)L = 0.04 M pOH = -log[OH-] = -log(0.04)=1.40 pH + pOH = 14 pH = 14-pOH = 14-1.40 = 12.6 Hope it has help you.... !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!.

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In H3O+ ion there are four electronpairs on the valency shell of the central atom . Out of this 4, oneis lone pair of electrons and other three is bond pair electrons.So it is in trigonal pyramidal geometry. Solution In H3O+ ion there are four electronpairs on the valency shell of the central atom . Out of this 4, oneis lone pair of electrons and other three is bond pair electrons.So it is in trigonal pyramidal geometry..

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According to periodic properties Theatomic radius is increases from top to bottom and decreases fromleft to right . Cs+is at IA group lastbutone in that group .So the radius of that ion is having more. And the remaining elements S andSe2+ comes under same group.They are having more radiusthan F . Among the S and Se2+ , S ia having small radius than Se2+.becausein groups size of the ion is increases. The priority Order is Cs+ > Se2+> S > F Fluorine got number 1 . and Cs+got number 4 Solution According to periodic properties Theatomic radius is increases from top to bottom and decreases fromleft to right . Cs+is at IA group lastbutone in that group .So the radius of that ion is having more. And the remaining elements S andSe2+ comes under same group.They are having more radiusthan F . Among the S and Se2+ , S ia having small radius than Se2+.becausein groups size of the ion is increases. The priority Order is Cs+ > Se2+> S > F Fluorine got number 1 . and Cs+got number 4.

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Why does one would like all of this social control stuff? the most goal is to avoid redundancy in your knowledge. Redundancy will result in numerous anomalies once you modify your knowledge. each truth ought to be keep just one occasion and you ought to grasp wherever to seem for every truth. The social control method brings order to your file. you choose to evolve to bound rules wherever every truth is keep. Nowadays the go-to traditional types ar either the Boyce-Codd traditional form (BCNF), that we\'ll cowl here these days, or the third traditional type (3NF), which can be coated later. (Yes, there are the primary, second, fourth, fifth traditional type. We’ll mention them later.) The Boyce-Codd traditional type A relative schema R is taken into account to be in Boyce–Codd traditional type (BCNF) if, for each one among its dependencies X Y, one among the subsequent conditions holds true: X Y could be a trivial useful dependency (i.e., Y could be a set of X) X could be a superkey for schema R Informally the Boyce-Codd traditional type is expressed as “Each attribute should represent a truth concerning the key, the total key, and zilch however the key.” Solution Why does one would like all of this social control stuff? the most goal is to avoid redundancy in your knowledge. Redundancy will result in numerous anomalies once you modify your knowledge. each truth ought to be keep just one occasion and you ought to grasp wherever to seem for every truth. The social control method brings order to your file. you choose to evolve to bound rules wherever every truth is keep. Nowadays the go-to traditional types ar either the Boyce-Codd traditional form (BCNF), that we\'ll cowl here these days, or the third traditional type (3NF), which can be coated later. (Yes, there are the primary, second, fourth, fifth traditional type. We’ll mention them later.) The Boyce-Codd traditional type A relative schema R is taken into account to be in Boyce–Codd traditional type (BCNF) if, for each one among its dependencies X Y, one among the subsequent conditions holds true: X Y could be a trivial useful dependency (i.e., Y could be a set of X) X could be a superkey for schema R Informally the Boyce-Codd traditional type is expressed as “Each attribute should represent a truth concerning the key, the total key, and zilch however the key.”.

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We know that absorbance , A = log ( Io / It ) Where Io = intensity of incident light It = intensity of transmitted light = 3 % ofIo = 0.03 Io Plug the values we get A = log ( Io /0.03Io ) = log (1/0.03) = 1.52 Solution We know that absorbance , A = log ( Io / It ) Where Io = intensity of incident light It = intensity of transmitted light = 3 % ofIo = 0.03 Io Plug the values we get A = log ( Io /0.03Io ) = log (1/0.03) = 1.52.

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The race industry and its elite enablers take it as self-evident that high black incarceration rates result from discrimination. At a presidential primary debate this Martin Luther King Day, for instance, Senator Barack Obama charged that blacks and whites “are arrested at very different rates, are convicted at very different rates, [and] receive very different sentences . . . for the same crime.” Not to be outdone, Senator Hillary Clinton promptly denounced the “disgrace of a criminal-justice system that incarcerates so many more African-Americans proportionately than whites.” Racial activists usually remain assiduously silent about that problem. But in 2005, the black homicide rate was over seven times higher than that of whites and Hispanics combined, according to the federal Bureau of Justice Statistics. From 1976 to 2005, blacks committed over 52 percent of all murders in America. In 2006, the black arrest rate for most crimes was two to nearly three times blacks’ representation in the population. Blacks constituted 39.3 percent of all violent-crime arrests, including 56.3 percent of all robbery and 34.5 percent of all aggravated- assault arrests, and 29.4 percent of all property-crime arrests. The advocates acknowledge such crime data only indirectly: by charging bias on the part of the system’s decision makers. As Obama suggested in the Martin Luther King debate, police, prosecutors, and judges treat blacks and whites differently “for the same crime.” The media love to target the federal crack penalties because crack defendants are likely to be black. In 2006, 81 percent of federal crack defendants were black, while only 27 percent of federal powder-cocaine defendants were. Since federal crack rules are more severe than those for powder, and crack offenders are disproportionately black, those rules must explain why so many blacks are in prison, the conventional wisdom holds. The press has covered this development voraciously, serving up a massive dose of crack revisionism aimed at proving the racist origins of the war on crack. Crack was never a big deal, the revisionist story line goes. But when Boston Celtics draft pick Len Bias died of a crack overdose in 1986, the media went into overdrive covering the crack phenomenon. “Images—or perhaps anecdotes—about the evils of crack, and the street crime it was presumed to stoke” circulated, as the New York Times archly put it in a December 2007 article. A “moral panic” (Michael Tonry’s term) ensued about an imaginary threat from a powerless minority group. Whites feared that addicted blacks would invade their neighborhoods. Sensational stories about “crack babies” surfaced. All this hysteria resulted in the unnecessary federal crack penalties. Those who tar the criminal-justice system as racist often make a broader claim: incarceration doesn’t even lower crime, making the nation’s skyrocketing prison rolls a particularly senseless injustice. Incarceration foes are right about one thing: the.

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