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Time to Pay Off Debt. Victor owes $20,000 on his credit

Algebra and Trigonometry | 3rd Edition | ISBN: 9780470648032 | Authors: Cynthia Y. Young ISBN: 9780470648032 218

Solution for problem 44 Chapter 5

Algebra and Trigonometry | 3rd Edition

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Algebra and Trigonometry | 3rd Edition | ISBN: 9780470648032 | Authors: Cynthia Y. Young

Algebra and Trigonometry | 3rd Edition

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Problem 44

Time to Pay Off Debt. Victor owes $20,000 on his credit card. The annual interest rate is 17%. a. Approximately how many years will it take him to pay off this credit card if he makes a monthly payment of $300? b. Approximately how many years will it take him to pay off this credit card if he makes a monthly payment of $400?

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ANEQ 328 Foundations In Animal Genetics Week 13 Class Notes (4/12/16-4/14/16) Must Know Statistics: Especially To Understand Quantitative Genetics  Important Background Information o Population  A breeding population of animals. o Value  A numeric value applied to an individual as opposed to a population.  Phenotypic Trait o Trait  A genetically determined characteristics.  Trait doesn’t always equal phenotype. o Simple Trait  When a gene is controlled by 1 or a few loci.  Can be qualitative or categorical.  Ex. The coat color of a horse is only controlled by a few loci. o Complex Trait  When a gene is controlled by many loci.  Can be quantitative or polygenic.  Ex. Milk production in cows in controlled by many loci. o Mean  The average of a set of numbers.  To calculate: Just add up all the numbers, then divide by how many numbers there are.  Normal Distribution (Bell Curve) o The statistical distribution that appears graphically as a symmetric, bell- shaped curve. Frequency Category ANEQ 328 Foundations In Animal Genetics Week 13 Class Notes (4/12/16-4/14/16)  About 66% of the animal population will fall within the Standard Deviation of the mean.  About 95% of the animal population will fall with 2 Standard Deviations of the mean.  99% of the animal population will fall within 3 Standard Deviations of the mean.  Mean () o The average of a set of numbers.  To calculate: Just add up all the numbers, then divide by how many numbers there are.  Units: o Using the following data to calculate the mean of the weaning and yearling weights. Calf ID Weaning Weight (lbs; X) Yearling Weight (lbs; Y) CSU 2001 500 1100 CSU 2002 475 1050 CSU 2003 525 1150  Weaning Weight Mean Calculation (500 + 475 + 425) 1400 = = 466.67 3 3  Yearling Weight Mean Calculation (1100 + 1050 + 1150) 3300 = = 1100 3 3  Variance ( ) o Differences among individuals within a population. 2 ∑(− ) 2 (1 − ) +2 ) 3(− ) ……  Equation: = −1 or = −1 o Using the following data to calculate the variance of the weaning and yearling weights. Calf ID Weaning Weight (lbs; X) Yearling Weight (lbs; Y) CSU 2001 500 1100 CSU 2002 475 1050 CSU 2003 525 1150 o We know that the number in our population is 3 calves, so n=3. ANEQ 328 Foundations In Animal Genetics Week 13 Class Notes (4/12/16-4/14/16)  Weaning Weight Variance Calculation (500 − 500) + (475 − 500) + (525 − 500) ()2 + −25 )2+ 25 ) 0 + 625 + 625 1250 = = = = 625 = 2 3 − 1 2 2 2  Yearling Weight Variance Calculation (1100 − 1100) + (1050 − 1100) + (1150 − 110)+ −50)2+ 50)2 0 + 2500 + 25005000 2 2 3 − 1 = 2 = 2 = 2 = 2500=  Standard Deviation () o The average deviation from the mean.  The square root of the variance.  Equation: √ 2 o Using the following data to calculate the standard variation of the weaning and yearling weights. Calf ID Weaning Weight (lbs; X) Yearling Weight (lbs; Y) CSU 2001 500 1100 CSU 2002 475 1050 CSU 2003 525 1150 o Based on previous calculations we know that 2 = 625 and 2= 2500 .2  Weaning Weight Standard Deviation Calculation 2 = 625 = 62√ = 25=  Yearling Weight Standard Deviation Calculation 2 2 = 2500 = √2500 = 50 =  Standard Error (SE) o Equation: or √ √ o Using the following data to calculate the standard error of the weaning and yearling weights. ANEQ 328 Foundations In Animal Genetics Week 13 Class Notes (4/12/16-4/14/16) Calf ID Weaning Weight (lbs; X) Yearling Weight (lbs; Y) CSU 2001 500 1100 CSU 2002 475 1050 CSU 2003 525 1150 o Based on previous calculations we know that = 25 and = 50 and the number in our population is 3 calves, so n=3.  Weaning Weight Standard Error Calculation 25 25 = = 15 = .500 ± 15 √ 3  Yearling Weight Standard Error Calculation 50 50 = 3 = 30 = .1100 ± 30 √  Covariance (Cov(x,y)) o How two traits or values vary together in a population.  Equation: (1−)(1−)+ 2 −(2−)+ 3−(3−). −1  Has no units, just a positive or negative number. o Using the following data to calculate the covariance of the weaning and yearling weights. Calf ID Weaning Weight (lbs; X) Yearling Weight (lbs; Y) CSU 2001 500 1100 CSU 2002 475 1050 CSU 2003 525 1150 o We know that the number in our population is 3 calves, so n=3.  Weaning Weight and Yearling Weight Covariance Calculation (500 − 500 1100 − 1100 + 475 − 500 1050 − 1100 + 525 − 500 1150 − 1100 ) 2500 = = 1250 3 − 1 2 o Covariation  1.) Positive or Negative Number  2.) Correlation shows the strength of a relationship.  3.) Regression show how much (amount). ANEQ 328 Foundations In Animal Genetics Week 13 Class Notes (4/12/16-4/14/16)  Correlation (r) o A measure of strength of the relationship between two variables. (Cov(x,y))  Equation: =  If the correlation is between 0 and 1 it demonstrates a strong relationship.  If the correlation is between 0 and -1 it demonstrates a weak relationship. o Using the following data to calculate the correlation of the weaning and yearling weights. Calf ID Weaning Weight (lbs; X) Yearling Weight (lbs; Y) CSU 2001 500 1100 CSU 2002 475 1050 CSU 2003 525 1150 o Based on previous calculations we know that Cov(x,y( ))= 1250lbs and = 25 and 50.  Weaning Weight and Yearling Weight Standard Correlation Calculation (Cov x,y ) = 1250lbs 1250 = = 1 (= 25)(= 50) (25)(50)  Regression (b) o The expected or average change in one variable (y) per unit change in another (x). (Cov x,y )  Equation: ∗= 2 o Using the following data to calculate the regression of the weaning and yearling weights. Calf ID Weaning Weight (lbs; X) Yearling Weight (lbs; Y) CSU 2001 500 1100 CSU 2002 475 1050 CSU 2003 525 1150 o Based on previous calculations we know that Cov(x,y( ))= 1250lbs and 2 = 625.  Weaning Weight and Yearling Weight Standard Regression Calculation (Cov x,y))= 1250lbs = 1250 = 2 2= 625 625  For every 1lb increase in weaning weight, yearling weight increases an average of 2 lbs. ANEQ 328 Foundations In Animal Genetics Week 13 Class Notes (4/12/16-4/14/16) Introduction to the Genetic Model for Quantitative Traits  Important Background Information o Population  A breeding population of animals. o Value  A numeric value applied to an individual as opposed to a population.  Phenotypic Trait o Trait  A genetically determined characteristics.  Trait doesn’t always equal phenotype. o Simple Trait  When a gene is controlled by 1 or a few loci.  Can be qualitative or categorical.  Ex. The coat color of a horse is only controlled by a few loci. o Complex Trait  When a gene is controlled by many loci.  Can be quantitative or polygenic.  Ex. Milk production in cows in controlled by many loci.  Genetic Model for Quantitative Traits o Equation: P=+G+E  P= Phenotypic Value  The performance of an individual animal for a specific trait.  = Population mean  The average phenotypic value for the specific trait for all animals in the population.  G= Genotypic value  The genotypic values of the individual for the specific trait.  E= Environmental Effect  The environmental effects on the individual’s performance for the trait.  Genetic Model for Quantitative Traits Including Breeding Value and Gene Combination Value o Equation: P=+BV+GCV+E  P= Phenotypic Value  The performance of an individual animal for a specific trait.  = Population mean  The average phenotypic value for the specific trait for all animals in the population.  BV= Breeding Value  The value of the individual as a parent, (the sum of the independent genotypes). ANEQ 328 Foundations In Animal Genetics Week 13 Class Notes (4/12/16-4/14/16)  Ex. A= +5, a= -5  AA= Value of 10  aa= Value of -10  Aa= Value of 0  GCV= Gene Combination Value  An individual’s genotypic value of gene interaction.  E= Environmental Effect  The environmental effects on the individual’s performance for the trait.  Progeny Difference (PD) o The expected difference between the mean performance of the individual’s progeny and the mean performance of all progeny.  Equation: PD= ½BV  BV= Breeding Value  The value of the individual as a parent, (the sum of the independent genotypes).  Genetic Prediction and Its Data o The bigger the data the more accurate the EPD’s are.  EPD’s get better as time goes on.  By adding multiple traits, animal models, and pedigree’s to an EPD it makes it more accurate the younger the animal is.  Genetic Model For Repeated Quantitative Traits o Producing Ability (PA)  The performance potential of an individual for a repeated trait.  Equation: PA= +BV+GCV+E +E p t  = Population mean  The average phenotypic value for the specific trait for all animals in the population.  BV= Breeding Value  The value of the individual as a parent, (the sum of the independent genotypes). Ex. A= +5, a= -5 o AA= Value of 10 ANEQ 328 Foundations In Animal Genetics Week 13 Class Notes (4/12/16-4/14/16) o aa= Value of -10 o Aa= Value of 0  GCV= Gene Combination Value  An individual’s genotypic value of gene interaction.  E = Permanent Environmental Effect p  An environmental effect that permanently influences an individual’s performance for a repeated trait. Ex. Increase or decrease in performance.  E t Temporary Environmental Effect  An environmental effect that influences a single performance record of an individual but does not permanently affect an individual’s performance potential for a repeated trait. Ex. Dry summer results in low numbers being collected for the particular trait.

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Chapter 5, Problem 44 is Solved
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Textbook: Algebra and Trigonometry
Edition: 3
Author: Cynthia Y. Young
ISBN: 9780470648032

Algebra and Trigonometry was written by and is associated to the ISBN: 9780470648032. This full solution covers the following key subjects: . This expansive textbook survival guide covers 13 chapters, and 10127 solutions. Since the solution to 44 from 5 chapter was answered, more than 255 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 44 from chapter: 5 was answered by , our top Math solution expert on 01/04/18, 09:28PM. This textbook survival guide was created for the textbook: Algebra and Trigonometry, edition: 3. The answer to “Time to Pay Off Debt. Victor owes $20,000 on his credit card. The annual interest rate is 17%. a. Approximately how many years will it take him to pay off this credit card if he makes a monthly payment of $300? b. Approximately how many years will it take him to pay off this credit card if he makes a monthly payment of $400?” is broken down into a number of easy to follow steps, and 64 words.

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