320-HW9
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320-HW9

Course Number: BRAE 3934, Fall 2009

College/University: Cal Poly

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BRAE 340 Irrig. Water Management HW 9 - Drainage & Salinity Name: Solution Key (only odd numbered problems will be graded) [C] 1. Over the course of an irrigation season, an irrigation system applies a gross amount of 35.5 inches. Pre-infiltration losses are 2.3 inches; leaching for salt removal is 2.8 inches; deep percolation due to poor timing is 4.3 inches; deep percolation due to non-uniformity is 5.1...

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340 BRAE Irrig. Water Management HW 9 - Drainage & Salinity Name: Solution Key (only odd numbered problems will be graded) [C] 1. Over the course of an irrigation season, an irrigation system applies a gross amount of 35.5 inches. Pre-infiltration losses are 2.3 inches; leaching for salt removal is 2.8 inches; deep percolation due to poor timing is 4.3 inches; deep percolation due to non-uniformity is 5.1 inches; and the remainder is used to satisfy crop ET. Rainfall is negligible. What is the total amount of water that must be drained away to prevent build up of the water table? A) 10.5 inches B) 8.7 inches C) 12.2 inches D) 13.8 inches E) 15.5 inches Solution: Drainage must remove all deep percolation, including beneficial leaching for salt removal Deep Percolation = 2.8 in + 4.3 in + 5.1 in = 12.2 in [A] 2. Over the course of an irrigation season, an irrigation system applies a gross amount of 40 inches. Pre-infiltration losses are 3 inches; total deep percolation is 9.0 inches, of which 4 is beneficial for salt removal; and the remainder is used to satisfy crop ET. Rainfall = 0. What is the total amount of water that must be drained away to prevent build up of the water table? A) 9.0 inches B) 7.4 inches C) 5.1 inches D) 10.0 inches E) 6.4 inches Solution: Drainage must remove all deep percolation, including beneficial leaching for salt removal Deep Percolation = 9.0 in = total DP [B] 3. Peak crop ET = 0.25 in/day; LR = 4 %; pre-infiltration loss = 0. With good management, AE = 75 %. Rainfall = 0. A drainage system should be able to handle what drainage coefficient? A) 0.108 in/day B) 0.097 in/day C) 0.124 in/day D) 0.084 in/day E) 0.070 in/day Solution: Gross = net / [AE/100 x (1-LR)] = 0.25 in/day / [(75/100) x (1-(4/100))] = 0.347 in/day Deep Percolation = gross - ETc = 0.347 in/day - 0.25 in/day = 0.097 in/day All deep perc (including leaching) must be drained away. So, drainage coefficient = 0.097 in/day [A] 4. Peak crop ET = 0.29 in/day; LR = 4 %; pre-infiltration loss = 0. With good management, AE = 76 %. Rainfall = 0. A drainage system should be able to handle what drainage coefficient? A) 0.107 in/day B) 0.081 in/day C) 0.118 in/day D) 0.093 in/day E) 0.059 in/day Solution: Gross = net / [AE/100 x (1-LR)] = 0.29 in/day / [(76/100) x (1-(4/100))] = 0.397 in/day deep perc. = gross - ET = 0.397 in/day - 0.29 in/day = 0.107 in/day All deep perc (including leaching) must be drained away. So, drainage coefficient = 0.107 in/day Prob. 5-13: Assume the depth to the impermeable barrier and the root zone depth which must remain free of saturation are known and fixed. If the item listed is INCREASED, determine whether the required drain spacing will [I] Increase or [D] Decrease. I) Increase D) Decrease [D] [D] [D] [I] [D] [I] [I] [I] [I] 5. 6. 7. 8. 9. 10. 11. 12. 13. LR - Leaching Requirement (dimensionless fraction) a - height of drains above impermeable barrier (feet) [careful - try a numerical example to be sure] ETc - Annual crop ET (inches) d - depth of drains below ground surface (feet) q - drainage coefficient (inches/day) m - height of water table (at its highest point) above drains (feet) DU - Distribution Uniformity (%) K - saturated hydraulic conductivity of the soil (inches/day) PI Loss - Pre-infiltration losses (%) Explanation As K or m increase, the drain spacing formula will Increase S As a increases, d and m must DECREASE (c and a+d are fixed), Decreasing S As q increases, the drain spacing formula will Decrease S Increasing ETc increases the salt brought onto the field, increasing the need for leaching. So the drainage coefficient will go up, Decreasing S Increasing DU will decrease deep percolation, decreasing q and Increasing S Increasing LR means more leaching, increasing q and Decreasing S [D] 14. Peak crop ET = 0.26 in/day; LR = 5 %; pre-infiltration loss = 0. With good management, AE = 76 %. Rainfall = 0. What drain spacing is required if we are to maintain a 3 foot root zone above the water table? Drains will be 7 feet deep. The impermeable barrier is 19 feet below the surface. The saturated hydraulic conductivity = 17.39 in/day. A) 372 ft B) 316 ft C) 438 ft D) 279 ft E) 407 ft Solution: Gross = net / [AE/100 x (1-LR)] = 0.26 in/day / [(76/100) x (1-(5/100))] = 0.360 in/day deep perc. = gross - ET = 0.360 in/day - 0.26 in/day = 0.100 in/day S = [(4 x K x ((m^2) + 2 x a x m)) / (q)]^0.5 S = [(4 x 17.39 in/day x ((4ft)^2) + 2x12 ftx4 ft)) / (0.100 in/day)]^0.5 = 279 ft Soil Surface 0 2 4 6 8 10 12 14 16 18 20 Barrier Gravely Sand K = 17.39 in/day q = 0.100 in/day Drains Soil Surface Water Table Barrier at 19 ft c = 3 ft m = 4 ft d = 7 ft a = 12 ft S = 279 ft Problems 15 and 16: Peak crop ET = 0.32 in/day; LR = 4 %; pre-infiltration loss = 0. With good management, AE = 76 %. Rainfall = 0. We are to maintain a 2 foot root zone above the water table. Drains will be 7 feet deep. The impermeable barrier is 16 feet below the surface. The saturated hydraulic conductivity = 0.11 in/day. [E] 15. What drain spacing is required? A) 27 ft B) 24 ft C) 30 ft D) 19 ft E) 21 ft Solution: Gross = net / [AE/100 x (1-LR)] = 0.32 in/day / [(76/100) x (1-(4/100))] = 0.439 in/day deep perc. = gross - ET = 0.439 in/day - 0.32 in/day = 0.119 in/day S = [(4 K ((m^2) + 2am)) / (q)]^0.5 S = [(4 x 0.11 in/day x ((5ft)^2) + 2x9 ftx5 ft)) / (0.119 in/day)]^0.5 S = 21 ft Soil Surface 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Barrier Clay K = 0.11 in/day q = 0.119 in/day Drains Soil Surface Water Table Barrier at 16 ft c = 2 ft m = 5 ft d = 7 ft a = 9 ft S = 21 ft [C] 16. How much wider can the drains be spaced if the AE were increased 5 percentage points? A) 2.0 feet wider B) 1.6 feet wider C) 3.0 feet wider D) 2.3 feet wider E) 1.1 feet wider Solution: New AE = 76 + 5% = 81% Gross = net / [AE/100 x (1-LR)] = 0.32 in/day / [(81/100) x (1-(4/100))] = 0.412 in/day deep perc. = gross - ET = 0.412 in/day - 0.32 in/day = 0.092 in/day New S = [(4 K ((m^2) + 2am)) / (q)]^0.5 New S = [(4 x 0.11 in/day x ((5ft)^2) + 2x9 ftx5 ft)) / (0.092 in/day)]^0.5 New S = 24 ft Increase in spacing = 24 ft - 21 ft = 3.0 feet wider [C] 17. What salinity (dS/m) is equivalent to a soil moisture tension of 0.3 bars ? A) 0.55 dS/m B) 0.74 dS/m C) 1.00 dS/m D) 1.11 dS/m E) 0.87 dS/m Solution: 1 bar = 100 centibars = 100 cb; 1 dS/m 30 cb 0.3 bars = 30 cb x (1 dS/m)/(30 cb) = 1.00 dS/m [C] 18. What concentration (ppm) is equivalent to a soil moisture tension of 1.1 bars? A) 828 ppm B) 1,219 ppm C) 2,567 ppm D) 2,124 ppm E) 1,590 ppm Solution: 1 bar = 100 centibars = 100 cb; 1 dS/m 30 cb 1.1 bars = 110 cb x (1 dS/m)/(30 cb) = 3.7 dS/m 3.7 dS/m x (700 ppm)/(1 dS/m) = 2,567 ppm [B] 19. If a soil has a moisture tension of 0.14 bars and the salinity of the soil water is 1.42 dS/m, what is the total tension experienced by the plant roots? That is, what is the sum of the moisture tension and the osmotic (due to salinity) tension? A) 65 cb B) 57 cb C) 84 cb D) 52 cb E) 75 cb Solution: total tension = moisture tension + osmotic tension moisture tension = 0.14 bars x 100 cb/bar = 14 cb osmostic tension = 1.42 dS/m x 30 cb/1 dS/m = 43 cb total tension = 14 cb + 43 = cb 57 cb [A] 20. If a soil has a moisture tension of 0.12 bars and the salinity of the soil water is 1.97 dS/m, what is the total tension experienced by the plant roots? That is, what is the sum of the moisture tension and the osmotic (due to salinity) tension? A) 71 cb B) 80 cb C) 99 cb D) 90 cb E) 64 cb Solution: total tension = moisture tension + osmotic tension moisture tension = 0.12 bars x 100 cb/bar = 12 cb osmostic tension = 1.97 dS/m x 30 cb/1 dS/m = 59 cb total tension = 12 cb + 59 cb = 71 cb [D] 21. Soil water with a salt concentration of 1,880 ppm contributes an osmotic tension of how much to the total tension experienced by the plants? A) 55 cb B) 44 cb C) 89 cb D) 81 cb E) 65 cb Solution: 1,880 ppm x 1 dS/m/700 ppm x 30 cb/1 dS/m = 81.0 cb [D] 22. A salt concentration of 683 ppm is equivalent to an electrical conductivity of how many dS/m? A) 1.31 dS/m B) 1.16 dS/m C) 1.55 dS/m D) 0.98 dS/m E) 1.45 dS/m Solution: 700 ppm is approximately 1 dS/m 683 ppm x (1 dS/m)/(700 ppm) = 0.98 dS/m [A] 23. Colorado River salinity is about 1.1 dS/m. What is the salt load in parts per million? A) 770 ppm B) 1,016 ppm C) 1,169 ppm D) 908 ppm E) 1,285 ppm Solution: 1.1 dS/m x 700 ppm / dS/m = 770 ppm [C] 24. ECw = 1.1 dS/m. What is LR for Wheat? A) 0.027 B) 0.031 C) 0.038 D) 0.021 E) 0.042 Solution: Threshold ECe* for Wheat = 6.0 dS/m LR = ECw / [5 ECe* - ECw] = 1.1 / [5 x 6.0 - 1.1] = 0.038 [D] 25. ECw = 1.0 dS/m. What is LR for Tomato? A) 0.121 B) 0.146 C) 0.103 D) 0.087 E) 0.133 Solution: Threshold ECe* for Tomato = 2.5 dS/m LR = ECw / [5 ECe* - ECw] = 1.0 / [5 x 2.5 - 1.0] = 0.087 [C] 26. A cropped soil is irrigated with water of salinity ECw. After drainage to FC, the soil salinity is found to be ECe. The salinity of the soil water is ECsw. As the days pass, and the crop extracts water for ET, what happens to ECe? A) ECe increases until it reaches ECsw B) If MAD = 50%, ECe will double before the next irrigation C) ECe does not change D) ECe increases but will never reach ECsw E) ECe decresaes until it reaches ECw Explanation Crops will extract esentially pure water for ET, leaving the total amount of salt in the soil unchanged. Since ECe measurement requires bringing the soil sample back to saturation, we would have the same amount of salt dissolved in the same amount of water. Thus ECe does not change. [D] 27. A cropped soil is irrigated with water of salinity ECw. After drainage to FC, the soil salinity is found to be ECe. The salinity of the soil water is ECsw. As the days pass, and the crop extracts water for ET, what happens to ECsw? A) ECsw decreases until it reaches ECw B) ECsw does not change C) If MAD = 50%, ECsw will double before the next irrigation D) ECsw increases but we don't know how much E) ECsw increases until it reaches ECe Explanation Crops will extract esentially pure water for ET, leaving the total amount of salt in the soil unchanged. The salt is dissolved in less and less water ech day, as the crop extracts soil water for ET. Therefore, will be based on the same amount of water, and ECsw will increase as time goes on. [T] 28. T/F: Once past the germintion stage, plants tend to respond to average root zone salinity. Explanation See bottom of pg S-13: once established, plants respond to average root zone salinity [T] 29. T/F: ECsw = 1.5 dS/m corresponds to about 45 cb tension. Explanation 1 dS/m 30 cb is the approximate conversion factor [F] 30. T/F: Seeds planted at the highest point on the bed will miss the highest salt concentration areas. Explanation Check figure S-1, pg S-8. The salt will concentrate in the high points [F] 31. T/F: To account for leaching, the water requirement should be multiplied by (1+LR). Explanation The water requirement should be divided by (1-LR) [F] 32. T/F: Even with proper leaching, the spot in the field which receives the least amount of water will have the lowest ECe. Explanation With proper leaching, the spot receiving the least water will have an acceptable ECe. But other spots in the field will have greater leaching and lower ECe values. [F] 33. T/F: The equation LR = ECw/[5 ECe - ECw] is for reclamation (not maintenance) leaching. Explanation This expression is for maintenance leaching only, after the soil has been reclaimed [A] 34. If the tensiometer at a particular depth reads 47 cb and the ECsw = 2.1 dS/m, what is the total tension experienced by the roots at that depth? A) 1.10 bars B) 0.80 bars C) 1.42 bars D) 1.28 bars E) 0.93 bars Solution: 2.1 dS/m x 30 cb/1 dS/m = 63 cb Total tension = moisture tension + osmotic (salinity) tension Total tension = 47 cb + 63 cb = 110 cb 110 cb / 100 cb = 1.10 bar [A] 35. ECw = 1.4 dS/m. What is LR for Corn (sweet)? A) 0.20 B) 0.17 C) 0.22 D) 0.14 E) 0.26 Solution: Threshold ECe* for Corn (sweet) = 1.7 dS/m LR = ECw / [5 ECe* - ECw] = 1.4 / [5 x 1.7 - 1.4] = 0.20 [D] 36. ECw = 1.3 dS/m. What is LR for Onion? A) 0.45 B) 0.37 C) 0.40 D) 0.28 E) 0.33 Solution: Threshold ECe* for Onion = 1.2 dS/m LR = ECw / [5 ECe* - ECw] = 1.3 / [5 x 1.2 - 1.3] = 0.28 [A] 37. The soil salinity is maintained at ECe = 16.8 dS/m. What is the relative yield for Cotton? A) 53 % B) 38 % C) 77 % D) 100 % E) 45 % Solution: 16.8 dS/m is greater than ECe* of 7.7 dS/m by 16.8 - 7.7 = 9.1 dS/m Yield declines by 5.2% per dS/m over the threshold Yield decline = 5.2% per dS/m x 9.1 dS/m = 47.32 % Relative Yield = 100 % - 47.32% = 52.7% 120 Cotton 100 80 60 40 20 ECe* = 7.7 dS/m s = 5.2 % per dS/m Rel. Yield for Cotton ECe = 16.8 dS/m ECe <=> Rel. Yield Rel. Yield = 53 % 0 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 ECe (dS/m) [B] 38. The soil salinity is maintained at ECe = 5.4 dS/m. What is the relative yield for Cotton? A) 32 % B) 100 % C) 21 % D) 58 % E) 77 % Solution: 5.4 dS/m is less than the threshold ECe* = 7.7 dS/m There is no yield decline Relative Yield = 100 % 120 Cotton 100 80 60 40 20 ECe* = 7.7 dS/m s = 5.2 % per dS/m Rel. Yield for Cotton ECe = 5.4 dS/m ECe <=> Rel. Yield Rel. Yield = 100 % 0 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 ECe (dS/m) [B] 39. Wheat is grown in a desert area where rainfall is negligible and ETc = 0.39 in/day. ECw = 1.4 dS/m. With good management, IE = 80 %. What is the required gross daily application? A) 0.37 in/day B) 0.51 in/day C) 0.58 in/day D) 0.28 in/day E) 0.44 in/day Solution: The threshold ECe* for this crop = 6.0 dS/m LR = ECw / [5 x ECe - ECw] = 1.4 / [5 x 6 - 1.4] = 0.049 Gross = net / [(IE/100 x (1-LR)] = 0.39 in/day / [(80/100) x (1-0.049)] = 0.51 in/day [B] 40. Potatoes are grown in a desert area where rainfall is negligible and ETc = 0.31 in/day. ECw = 1.3 dS/m. With good management, IE = 70 %. What is the required gross daily application? A) 0.29 in/day B) 0.54 in/day C) 0.59 in/day D) 0.48 in/day E) 0.38 in/day Solution: The threshold ECe* for this crop = 1.7 dS/m LR = ECw / [5 x ECe - ECw] = 1.3 / [5 x 1.7 - 1.3] = 0.181 Gross = net / [(IE/100 x (1-LR)] = 0.31 in/day / [(70/100) x (1-0.181)] = 0.54 in/day Answers to even numbered problems: 2-[A] 4-[A] 6-[D] 8-[I] 22-[D] 24-[C] 26-[C] 28-[T] 10-[I] 30-[F] 12-[I] 32-[F] 14-[D] 34-[A] 16-[C] 36-[D] 18-[C] 38-[B] 20-[A] 40-[B]

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Page |1 Robert L. Kerin SL1105D-26M Week 5 Individual Work Assignment Thesis Statement 9/12/2009Women and Self-Defense, we know that women have a slight disadvantage when it comes to the physical aspects of men, but do they really? A Women can be faster,
Everest University - SL - SL1105D
Page |1Robert L. Kerin SL1105D-26M Week 5 Individual Work Assignment Title Page 9/12/2009Women in Martial Arts.
Everest University - SL - SL1105D
Page |1Robert L. Kerin SL1105D-26M Week 6 Individual Work Assignment Opening Paragraph 9/12/2009Guys have John Wayne, and who do the women have? The first thing that comes to mind is Wonder Women, though she is just a made up super hero from the 70s. Wo
Everest University - SL - SL1105D
Page |1Robert L. Kerin SL1105D-26M Week 5 Individual Work Assignment Title Page 9/12/2009Women in Martial Arts.Page |2Robert L. Kerin SL1105D-26M Week 6 Individual Work Assignment Opening Paragraph 9/12/2009Guys have John Wayne, and who do the women
Everest University - SL - SL1105D
Page |1Robert L. Kerin SL1105D-26M Week 7 Critical Thinking Outline 9/30/2009IA Self Defense Program that has been developed to offer protection opportunities for all ages. Self Defense Program for Women, Youth Workshops, Family Courses, Senior Program
Everest University - SL - SL1105D
Page |1 Robert L. Kerin SL1105D-26M Week 12 Individual Work Assignment Final Thesis Paper 10/54/2009Self Defense for Women.Page |2Guys have John Wayne, and who do the women have? The first thing that comes to mind is Wonder Women, though she is just a
Georgia Perimeter - ACCT - 2102
Achievement Test 1: Chapters 1 and 2 Financial Accounting Kimmel, Weygandt, &amp; KiesoName _ Instructor _ Section # _ Date _Part Points ScoreI 42II 8III 10IV 10V 18VI 12Total 100PART I MULTIPLE CHOICE (42 points) Instructions: Designate the best an
Georgia Perimeter - ACCT - 2102
Achievement Test 2: Chapters 3 and 4 Financial Accounting Kimmel, Weygandt, &amp; KiesoName _ Instructor _ Section # _ Date _Part Points ScoreI 33II 24III 24IV 10V 9Total 100PART I MULTIPLE CHOICE (33 points) Instructions Designate the best answer fo
Georgia Perimeter - ACCT - 2102
Achievement Test 3: Chapters 5 and 6 Financial Accounting Kimmel, Weygandt, &amp; KiesoName _ Instructor _ Section # _ Date _Part Points ScoreI 45II 25III 12IV 10V 8Total 100PART I MULTIPLE CHOICE (45 points) Instructions Designate the best answer fo
Georgia Perimeter - ACCT - 2102
Achievement Test 4: Chapters 7 and 8 Financial Accounting Kimmel, Weygandt, &amp; KiesoName _ Instructor _ Section # _ Date _Part Points ScoreI 30II 12III 20IV 10V 16VI 12Total 100PART I MULTIPLE CHOICE (30 points) Instructions Designate the best an
Georgia Perimeter - ACCT - 2102
Achievement Test 5: Chapters 9 and 10 Financial Accounting Kimmel, Weygandt, &amp; KiesoName _ Instructor _ Section # _ Date _Part Points ScoreI 34II 6III 16IV 16V 10V 18Total 100PART I MULTIPLE CHOICE (32 points) Instructions Designate the best ans
Georgia Perimeter - ACCT - 2102
Achievement Test 6 &amp; 7 Chapters 11-13 Financial Accounting Kimmel, Weygandt, &amp; KiesoName _ Instructor _ Section # _ Date _Part Points ScoreI 36II 6III 6IV 10V 12VI 10VII 20Total 100PART I MULTIPLE CHOICE (32 points) Instructions Designate the b
Georgia Perimeter - ACCT - 2102
APPENDIX CTIME VALUE OF MONEYSUMMARY OF QUESTIONS BY OBJECTIVES AND BLOOMS TAXONOMYItem 1. 2. 3. 4. 21. 22. 23. 24. 25. 26. 27. 28. 29. 65. 66. 67. 78. 79. 86. SO 1 1 1 1 1 2 2 2 2 2 3 3 3 2 2 2,3 1 2 1-6 BT K K K K K K AP K K AP AP K AP AP AP AP K K K
Georgia Perimeter - ACCT - 2102
APPENDIX DREPORTING AND ANALYZING INVESTMENTSSUMMARY OF QUESTIONS BY STUDY OBJECTIVE AND BLOOMS TAXONOMYItem 1. 2. 3. 4. 5. 6. 7. 8. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 155. 156. 161. 162. 163. 17
Georgia Perimeter - ACCT - 2102
C HAPTER 1INTRODUCTION TO FINANCIAL STATEMENTSSUMMARY OF QUESTIONS BY STUDY OBJECTIVE AND BLOOMS TAXONOMYItem 1. 2. 3. 4. 5. 6. 7. 8. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 151. 152. 153. 154. 154. 168.
Georgia Perimeter - ACCT - 2102
C HAPTER 2A FURTHER LOOK AT FINANCIAL STATEMENTSSUMMARY OF QUESTIONS BY STUDY OBJECTIVE AND BLOOMS TAXONOMYItem 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81. 8
Georgia Perimeter - ACCT - 2102
CHAPTER 3THE ACCOUNTING INFORMATION SYSTEMSUMMARY OF QUESTIONS BY STUDY OBJECTIVE AND BLOOMS TAXONOMYItem 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80.
Georgia Perimeter - ACCT - 2102
CHAPTER 4ACCRUAL ACCOUNTING CONCEPTSSUMMARY OF QUESTIONS BY STUDY OBJECTIVE AND BLOOMS TAXONOMYItem 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81. 82. 8
Georgia Perimeter - ACCT - 2102
CHAPTER 5MERCHANDISING OPERATIONSSUMMARY OF QUESTIONS BY STUDY OBJECTIVESTrue-False StatementsItem 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. SO 1 1 1 1 1 1 1 1 1 1 1 Item 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. SO 1 1 2 2 2 2 2 2 2 2 2 Item 23. 24. 25. 2
Georgia Perimeter - ACCT - 2102
CHAPTER 5 Test Bank: Study Objective 1 TRUE-FALSE STATEMENTS AND SOLUTIONS1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. Retailers and wholesalers are both considered merchandising enterprises. The operating cycle of a merchandising company ordinarily is sho