Ikrandraco - Facts and Figures

Name Ikdrandraco (Ikran dragon, after the flying creatures from Avatar); pronounced EE-krahn-DRAY-coe Habitat Rivers and lakes of Asia Historical Period Early Cretaceous (120 million years ago) Size and Weight About 30 inches long and a few pounds Diet Fish Distinguishing Characteristics Moderate size; distinctive bill structure; possible throat patch for holding fish About Ikrandraco Ikrandraco is an odd choice to honor the Ikran, or mountain banshees, of Avatar: this early Cretaceous pterosaur was only about two and a half feet long and a few pounds, whereas the Ikran from the hit movie are majestic, horse-sized, flying creatures that the Navi ride into battle against their human antagonists. Once you get past its name, though, Ikrandraco avatar may have been a truly unique pterosaur: some paleontologists claim that it had a pouch on the underside of its distinctively shaped bill in which it stored recently caught fish, which would make it similar to the modern pelican. However, not everyone is convinced by this putative anatomical feature of Ikrandraco (made of soft tissue, a throat pouch would have no chance of surviving in the fossil record), nor by the hypothesis that this pterosaur skimmed over the surface of lakes and trapped wiggling prey in its submerged lower jaw. The fact is that it can be difficult to infer the everyday behavior of a 120-million-year-old reptile by analogy with modern birds, and the possibility remains that Ikrandraco fed in more conventional fashion, like other pterosaurs of the early Cretaceous period, simply diving into the water and swallowing its fill of fish.

Brief Summary Thematic Analysis of A Midsummer Nights...

â€Å"Lord, what fools these mortals be!† This line, uttered by the fairy king’s servant and trickster Robin Goodfellow, is very telling of how ridiculous the central four characters in William Shakespeare’s A Midsummer Night’s Dream are in their thoughts and actions. The true motivation behind their actions, though, is not found in witty quips by knavish fairies, but rather in the symbolic nature of the play’s setting. The varied settings in the play, from Duke Theseus’s regal estate to Fairy Queen Titania’s forest bower, serve to set the mood of every scene, and to accentuate the characters actions throughout the play. By observing the rich yet subtle backdrops of A Midsummer Night’s Dream, it is possible to glean greater understanding of†¦show more content†¦Once everyone is back in the confines of the city walls, order returns to the character’s actions – this is seen in their inability to justify or s ubstantiate their supposed dream, which they cannot see being possible now that they are thinking with normal reasoning. However, there are two events that in particular show that disorder is never truly gone - first, Thesus’ refusal to follow both the demands of Egeus and the Athenean law regarding Hermia’s marriage, and second, Robin’s final soliloquy, which encourages the audience to believe that the whole play was just an irrational dream driven by the streak of disorder inside all of us. It is on that thought, then, that I wish to concentrate. Shakespeare shows that both extremes – complete order, as represented by Theseus’ estate and the greater city of Athens, and complete disorder, as represented by the wild forest and the world of the fairies, both have problems in pure form. When Egeus demands that Theseus uphold the absolute, complete, and unyielding order of ancient Athenean law, while not bothering to think of his daughter’s tr ue feelings, Shakespeare shows that by-the-book proper behavior and law is often ridiculously unreasonable. Conversely, when the raw disorder of the fairy world is channeledShow MoreRelatedDeveloping Management Skills404131 Words   |  1617 PagesWhat Are Management Skills? 9 Improving Management Skills 12 An Approach to Skill Development 13 Leadership and Management 16 Contents of the Book 18 Organization of the Book 19 Practice and Application 21 Diversity and Individual Differences 21 Summary 23 SUPPLEMENTARY MATERIAL 24 Diagnostic Survey and Exercises 24 Personal Assessment of Management Skills (PAMS) 24 What Does It Take to Be an Effective Manager? 28 SSS Software In-Basket Exercise 30 SCORING KEY AND COMPARISON DATA 42 Personal Assessment

Math Paper Free Essays

Derp university Derp derpington Human Resource Management Research Paper is Business Mathematics 101 1st Tri Semester SY 2011-2012 Ms. derpina derp TABLE OF CONTENTS TITLE PAGE ACKNOWLEDGEMENTii TOPICS Simple Discount1 Simple Interest2 Four types of Interest available3 Compounded Amount and Compound Interest4 Linear Programming Problems * Maximization6 * Minimization8 Forecasting by Trend Projection10 Acknowledgement I would like to thank God for guiding and giving me motivation to do this math research paper; my friends for answering my questions about this paper; Dr. Masajo for giving me the opportunity to gain more knowledge; and my mother to constantly remind me to do better in college. We will write a custom essay sample on Math Paper or any similar topic only for you Order Now I would like to thank my mentor, Ms. Grace Chong, for being my mentor and to aid me in my college life. Simple Discount Find the present value of $3800 due in 6 months at 7% discount rate. A) F = $3800 d = 7% = . 07 t = 6 / 12 = 1/2 Formula: D = Fdt Solution: D = $3800 (. 07) (1/2) D = $133 P = F – D P = $3800 – $133 P = $3667 Discount $2056. 80 for 85 days at a discount rate of 6 ? % B) F = 2056. 80 d = 6 ? % = . 065 t = 85 / 360 = 17 / 72 years Formula: D = Fdt Solution: D = $2056. 80 (. 065) (17/72) D = $31. 57 P = F – D P = $2056. 80 – $31. 57 P = $2025. 13 Simple Interest Find simple interest on $10,000 at the rate of 5% for 5 years. Also find the amount for 5 years. A) P = $10,000 R = 5% = . 05 T = 5 years = n = 5 I = PRT I = $10,000 (. 05) (5) I = $2500 A = P + I A = $10,000 + $2500 A = $12,500 Find simple interest on $15,600 for 1 ? years at the rate of 8% per annum. Also find total amount. B) P = $15,600 R = 8% = . 08 T = 1 ? = n = 1 ? I = PRT I = $15,600 (. 08) (1? ) I = $1872 A = P + I A = $15,600 + $1872 A = $17472 4 Types of Interest Available Find the different interest on $1000 at 6% from June 23 2011 to September 21 2015. A) Approximate number of days: Year: 2015 – 2011 = 4 Month: 8 – 6 = 2 Days: 51 – 23 = 28 4 x 360 = 1440 2 x 30 = 60 28 = 28 = 1528 Days B) Actual Number of days: 4 years x 365 days = 1463 days January 30 – June 23 = 173 days January 30 – September 21 = 263 days 1463 Days – 173 days = 1287 days 1287 Days + 263 days = 1550 days = 1550 days C) Io interest for approximate number of days: Io = PRT = $1000 (. 06) (1528/360) Io = $254. 67 D) Ie interest for approximate number of days: Ie = PRT = $1000 (. 06) (1528/365) Ie = $251. 8 E) Io interest for actual number of days: Io = PRT = $1000 (. 06) (1550/360) Io = $258. 33 F) Ie Interest for actual number of days: Ie = PRT = $1000 (. 06) (1528/365) Ie = $254. 79 Compounded amount and Compounded interest Find the Compounded amount and compounded interest of $1000 at 7% for 3 years A) B) Compounded Annually P = $1000 R = 7% = . 07 T = 3 years = N = 3 x 1 = 3 A = P (1+i) ^ n A = $1000 (1+0. 7) ^ 3 A = $1225. 043 I = A – P I = $1225. 043 – $1000 I = $225. 043 C) Compounded Semi – Annually P = $1000 R = 7 / 2 % = 3. 5 = . 035 T = 3 years = N = 3 x 2 = 6 A = P (1+i) ^ n A = $1000 (1+0. 5) ^ 6 A = $1229. 36 I = A – P I = $1229. 36 – $1000 I = $229. 36 D) Compounded Quarterly P = $1000 R = 7 / 4% = 1. 75 = . 0175 T = 3 years = N = 3 x 4 = 12 A = P (1+i) ^ n A = $1000 (1+0. 175) ^ 12 A = $1231. 44 I = A – P I = $1231. 44 – $1000 I = $231. 44 E) Compounded Monthly P = $1000 R = 7 / 12% = . 5833 = . 00583 T = 3 years = N = 3 x 12 = 36 A = P (1+i) ^ n A = $1000 (1+. 00583) ^ 36 A = $1232. 78 I = A – P I = $1232. 78 – $1000 I = $232. 78 Compounded amount and Compounded interest Find the Compounded amount and compounded interest of $1500 at 5% for 3 years A) B) Compounded Annually P = $1500 R = 5% = . 05 T = 3 years = N = 3 x 1 = 3 A = P (1+i) ^ n A = $1500 (1+. 05) ^ 3 A = $1736. 4375 I = A – P I = $1736. 4375 – $1500 I = $236. 4375 C) Compounded Semi – Annually P = $1500 R = 5 / 2 % = 2. 5 = . 025 T = 3 years = N = 3 x 2 = 6 A = P (1+i) ^ n A = $1500 (1+. 025) ^ 6 A = $1739. 540127 I = A – P I = $1739. 540127 – $1500 I = $739. 540127 D) Compounded Quarterly P = $1500 R = 5 / 4% = 1. 25 = . 0125 T = 3 years = N = 3 x 4 = 12 A = P (1+i) ^ n A = $1500 (1+. 0125) ^ 12 A = $1741. 131777 I = A – P I = $1741. 131777 – $1500 I = $741. 131777 E) Compounded Monthly P = $1500 R = 5 / 12% = . 41666 = . 00416 T = 3 years = N = 3 x 12 = 36 A = P (1+i) ^ n A = $1500 (1+. 00416) ^ 36 A = $1741. 792 I = A – P I = $1741. 792 – $1500 I = $741. 792 Linear Programming Problems (Maximization) Levi’s Jeans manufacturing company purchase2 styles of jeans, style X and style Y, which sell for $90 and $75 appropriately. Unit production test for style X is $40 and for style Y $35. Raw materials available monthly are 90 meters while processing time at a max of 70 hours per week. Style X jeans made 3 meters of materials and 2 for processing them. For style Y, 2 meters and 2 for processing. Style X market demand is no more than 40 per week. How many of each style should be produced in each week in order to make profit maximum? | Style X| Style Y| Total Available| RM| 3| 2| 90| PT| 2| 2| 70| MD| 40| | | | Style X| Style Y| USP| $90| $75| UPE| 40| 35| UBM| $50| $40| Composition of linear programming problems: I. Decision Variable X = Number of style X to be produced weekly Y = Number of style Y to be produced weekly II. Objective Function Maximum Profit (Z): Z = $50X+$40Y III. Subjects Constraints: RM = 3X+2Y 90PT = 2X+2Y 70 MD = X 40X; Y 0 IV. Graphical Solutions A) By intercept B) Graphical presentations and points A intersection between 2 lines C) Testing the curve of the convex polygon formed form the objective function V. Decision Raw Materials: 3X+2Y 90 X = 30 Y = 45 Processing Time: 2X+2Y 70 X = 35 Y = 35 Market Demand: X = 40 A) Z = $50X + $40Y = $50(0) + $40(35) =$1400 B) Z = $50X + $40Y = $50(20) + $40(75) =$1600 C) Z = $50X + $40Y = $50(30) + $40(0) =$1500 Choose B. Decision: The Levi’s manufacturing company must produce 20 pieces of style X and 50 pieces of style Y to have a maximum profit of $1600. Linear Programming Problems (Minimization) Mrs. Smith mining company owns two mines grading ores graded into 3 classes. High grade (H), Medium grade (M) and low grade (L). The company is tied with a contract to provide a smelting plant with 12 tons of (H), 8 tons of (M), and 24 tons of (L) per week. It costs $2000 per day to run mine 1 and $1600 per day to run mine 2. In a day operation, Mine 1 produces 6 tons of (H), 2 tons of (M) and 4 tons of (L). While mine 2 produces 2 tons of (H); 2 tons of (M) and 12 tons of (L). How many days a week should each mines operation to fulfil company’s commitment most economically? | Mine 1| Mine 2| Total Available| H| 6| 2| 12| M| 2| 2| 8| L| 4| 12| 24| Cost| $2000| $1600| | I. Decision Variables: X = Number of days to run mine 1 Y = Number of days to run mine 2 II. Objective Functions: Minimum Cost = $2000X + $1600Y III. Subjects to Constraints: H = 6X + 2Y 12 M = 2X + 2Y 8 L = 4X + 12Y 24 X; Y 0 IV. Graphical Solutions H = 6X + 2Y 12M = 2X + 2Y 8L = 4X + 12Y 24 X = 2 Y = 6X = 4 Y = 4X = 6 Y = 2 P1 (0,6) Min C = $2000(0) + $1600(6) = $9600 P2 (1,3) Min C = $2000(1) + $1600(3) = $6800 P3 (3,1) Min C = $2000(3) + $1600(1) = $7600 P4 (6,0) Min C = $2000(6) + $1600(0) = $12000 Choose P2 V. Decision: Mrs. Smith’s mining company should run mine 1 for 1 day and Mine 2 for 3 days in order to have a minimum cost of $6800. Forecasting by Trend Projection Forecast and graph the production of rice in the Philippines for the years 2012 and 2015 of the annual production of rice from year 2000 to year 2010. Year (N)| Production of Rice (Y)| X| XY| Y’| X^2| 2000| 20| 0| 0| | 0| 2001| 22| 1| 22| | 1| 2002| 18| 2| 36| | 4| 2003| 19| 3| 57| | 9| 2004| 21| 4| 84| | 16| 2005| 24| 5| 120| | 25| 2006| 22| 6| 132| | 36| 2007| 26| 7| 182| | 49| 2008| 28| 8| 224| | 64| 2009| 25| 9| 225| | 81| 010| 30| 10| 300| | 100| | ? (Y) = 255| ? (X) = 55| ? (XY)=1382| | ? (X^2) = 385| 2 Normal Equations: ?(Y) = NA + B? (X)Equation 1 ?(XY) = A? (X) + B? (X^2)Equation 2 Solve for B) 255 = 11A + 55B (-5) 1382 = 55A + 385B -1275 = -55A – 275B 1382 = 55A + 385B 107 /100 = 110B /100 B = . 97272727 Solve for A) 255 = 11A + 55B 11A + 55B = 255 11A +55(. 97272727) = 255 11A + 5 3. 5 = 255 11 A = 255 – 53. 5 11A /11 = 201. 5 /11 A = 18. 31818182 A = 18. 32 B = 0. 97 Formula Y’ = A+Bx Year 2000 = 18. 32 + 0. 97(0) Y’ = 18. 32 Year 2001 = 18. 32 + 0. 97(1) Y’ = 19. 29 Year 2002 = 18. 32 + 0. 92(2) Y’ = 20. 6 Year 2003 = 21. 23 Year 2004 = 22. 2 Year 2005 = 23. 17 Year 2006 = 24. 14 Year 2007 = 25. 11 Year 2008 = 26. 08 Year 2009 = 27. 05 Year 2010 = 28. 02 In the table: Year (N)| Production of Rice (Y)| X| XY| Y’| X^2| 2000| 20| 0| 0| 18. 32| 0| 2001| 22| 1| 22| 19. 29| 1| 2002| 18| 2| 36| 20. 26| 4| 2003| 19| 3| 57| 21. 23| 9| 2004| 21| 4| 84| 22. 2| 16| 2005| 24| 5| 120| 23. 17| 25| 2006| 22| 6| 132| 24. 14| 36| 2007| 26| 7| 182| 25. 11| 49| 2008| 28| 8| 224| 26. 08| 64| 2009| 25| 9| 225| 27. 05| 81| 2010| 30| 10| 300| 28. 02| 100| | ? (Y) = 255| ? (X) = 55| ? (XY)=1382| | ? (X^2) = 385| How to cite Math Paper, Papers