# Math Paper

Derp university Derp derpington Human Resource Management Investigation Paper is Business Mathematics 101 1st Tri Semester SY 2011-2012 Ms. derpina derp TABLE OF CONTENTS TITLE PAGE ACKNOWLEDGEMENTii TOPICS Unblended Discount1 Unblended Interest2 Four types of Cause adapted3 Compounded Totality and Compound Interest4 Straight Programming Problems * Maximization6 * Minimization8 Forecasting by Trend Projection10 Acknowledgement I would affect to rejoice God for considerable and giving me motivation to do this math investigation paper; my friends for obedient my questions environing this paper; Dr. Masajo for giving me the convenience to establish past knowledge; and my woman to incessantly remind me to do rectify in garden. I would affect to rejoice my instructor, Ms. Grace Chong, for nature my instructor and to aid me in my garden personality. Unblended Remittance Ascertain the confer-upon appreciate of \$3800 due in 6 months at 7% remittance reprove. 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 Remittance \$2056. 80 for 85 days at a remittance reprove 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 Unblended Cause Ascertain unblended cause on \$10,000 at the reprove of 5% for 5 years. Also ascertain the sumity 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 Ascertain unblended cause on \$15,600 for 1 ? years at the reprove of 8% per annum. Also ascertain sum sumity. 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 Cause Available Find the unanalogous cause on \$1000 at 6% from June 23 2011 to September 21 2015. A) Abut enumerate 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) Developed Enumerate 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 cause for abut enumerate of days: Io = PRT = \$1000 (. 06) (1528/360) Io = \$254. 67 D) Ie cause for abut enumerate of days: Ie = PRT = \$1000 (. 06) (1528/365) Ie = \$251. 8 E) Io cause for developed enumerate of days: Io = PRT = \$1000 (. 06) (1550/360) Io = \$258. 33 F) Ie Cause for developed enumerate of days: Ie = PRT = \$1000 (. 06) (1528/365) Ie = \$254. 79 Compounded sumity and Compounded cause Ascertain the Compounded sumity and compounded cause 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 sumity and Compounded cause Ascertain the Compounded sumity and compounded cause 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 Straight Programming Problems (Maximization) Levi’s Jeans manufacturing assembly purchase2 titles of jeans, title X and title Y, which dispose-of for \$90 and \$75 truly. Unit product cupel for title X is \$40 and for title Y \$35. Raw materials adapted monthly are 90 meters opportunity processing age at a max of 70 hours per week. Title X jeans made 3 meters of materials and 2 for processing them. For title Y, 2 meters and 2 for processing. Style X dispense ask-for is no past than 40 per week. How abundant of each title should be manufactured in each week in regulate to find avail climax? | Title X| Title Y| Sum Available| RM| 3| 2| 90| PT| 2| 2| 70| MD| 40| | | | Title X| Title Y| USP| \$90| \$75| UPE| 40| 35| UBM| \$50| \$40| Composition of straight programming problems: I. Decision Variable X = Enumerate of title X to be manufactured weekly Y = Enumerate of title Y to be manufactured weekly II. Extrinsic Origination Climax Avail (Z): Z = \$50X+\$40Y III. Subjects & Constraints: RM = 3X+2Y < 90PT = 2X+2Y < 70 MD = X < 40X; Y > 0 IV. Graphical Solutions A) By seize B) Graphical confer-uponations and points A intersection between 2 lines C) Testing the flexion of the tumular polygon contriveed contrive the extrinsic origination V. Decision Raw Materials: 3X+2Y < 90 X = 30 Y = 45 Processing Time: 2X+2Y < 70 X = 35 Y = 35 Dispense 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 assembly must consequence 20 pieces of title X and 50 pieces of title Y to own a climax avail of \$1600. Straight Programming Problems (Minimization) Mrs. Smith mining assembly owns two mines grading ores measured into 3 classes. High measure (H), Medium measure (M) and low measure (L). The assembly is tied after a while a retrench to cater a smelting fix after a while 12 tons of (H), 8 tons of (M), and 24 tons of (L) per week. It requires \$2000 per day to run mine 1 and \$1600 per day to run mine 2. In a day origination, Mine 1 consequences 6 tons of (H), 2 tons of (M) and 4 tons of (L). Opportunity mine 2 consequences 2 tons of (H); 2 tons of (M) and 12 tons of (L). How abundant days a week should each mines origination to verify assembly’s commitment most economically? | Mine 1| Mine 2| Sum Available| H| 6| 2| 12| M| 2| 2| 8| L| 4| 12| 24| Cost| \$2000| \$1600| | I. Decision Variables: X = Enumerate of days to run mine 1 Y = Enumerate of days to run mine 2 II. Extrinsic Functions: Restriction Require = \$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 assembly should run mine 1 for 1 day and Mine 2 for 3 days in regulate to own a restriction require of \$6800. Forecasting by Trend Projection Forecast and graph the product of rice in the Philippines for the years 2012 and 2015 of the annual product of rice from year 2000 to year 2010. Year (N)| Product 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 + 53. 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)| Product 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|