LS30A - Final

pdf

School

California State University, East Bay *

*We aren’t endorsed by this school

Course

30A

Subject

Economics

Date

Feb 20, 2024

Type

pdf

Pages

24

Uploaded by elizabetheg

Report
22F-LIFESCI-30A-LEC-3 / 22F-LIFESCI-30A-LEC-4 Final Exam CASSIDY ALLEN TOTAL POINTS 108.5 / 112 QUESTION 1 1 Protein modeling 8 / 8 ✓ - 0 pts Correct - 2 pts Assumption 1 - Incorrect - 1 pts Assumption 1 - Partially Correct - 0.5 pts Assumption 1 - Incorrect Sign - 2 pts Assumption 2 - Incorrect - 1 pts Assumption 2 - Partially Correct - 0.5 pts Assumption 2 - Incorrect Sign - 2 pts Assumption 3 - Incorrect - 1 pts Assumption 3 - Partially Correct - 0.5 pts Assumption 3 - Incorrect Sign - 0.5 pts Assumption 3 - Missing k - 0.5 pts Assumption 3 - Partially Correct, Reasonable Explanation - 2 pts Assumption 4 - Incorrect - 1 pts Assumption 4 - Partially Correct QUESTION 2 2 Ecological modeling 9 / 10 - 0 pts Correct - 1.5 pts Partially Correct- L assumptions - 3.5 pts Incorrect - L assumptions - 1.5 pts Partially Correct- C assumptions - 3.5 pts Incorrect- C assumptions - 1.5 pts Partially Correct- S assumptions - 3 pts Incorrect- S assumptions - 1 pts Minor errors- L assumptions - 1 pts Minor errors- C assumptions ✓ - 1 pts Minor errors- S assumptions QUESTION 3 3 Sketch trajectory 8 / 8 ✓ - 0 pts Correct - 8 pts Didn't attempt/Left blank/Graph is not trajectory - 6 pts We were looking for a spiral pattern. The pattern isn't spiral in your answer. - 4 pts Wrong axis representation. The spiral should not meet at the origin or have negative values. - 1 pts Axes not labeled - 4 pts Spiraling in rather than out QUESTION 4 Cricket bifurcations 8 pts 4.1 How many bifurcations? 2 / 2 ✓ - 0 pts Correct - 2 pts Correct answer is 3 bifurcations. 4.2 Ammonium 11 2 / 2 ✓ - 0 pts Correct - 2 pts Since the ammonium level is 11 and
there is a stable equilibrium at 3000 and unstable equilibrium at 6500. Any value of the crickets (between 3000 to 6500) over long run will move away from the unstable equilibrium(6500) and towards the stable equilibrium(3000). Therefore, the final answer is 3000. - 1.5 pts Answer relies on distance instead of stability - 2 pts Answer seems to assume ammonium concentration will change - 0.5 pts Misreads scale - 2 pts Does not go to stable equilibrium - 2 pts Incorrect; reasoning unclear - 2 pts Says population will stay at unstable equilibrium - 1 pts Two answers; wrong one plausible - 2 pts No answer given - 0.5 pts Forgot a 0 4.3 Reduce to 3000 2 / 2 ✓ - 0 pts Correct - 2 pts Since the ammonium level is 12 and there is a stable equilibrium at 3000 and unstable equilibrium at 7000. Any value of the crickets (between 3000 to 7000) over long run will move away from the unstable equilibrium(7000) and towards the stable equilibrium(3000).Since the given cricket level is 9000, we have to kill atleast 2000 crickets to bring the cricket level to 7000, thus ensuring that it eventually become 3000(since we have the stable equilibrium at 3000). - 1.5 pts Does not use stability of equilibria, giving a larger number than necessary - 0 pts Manipulates ammonium in addition to or instead of population. Not the intended answer but strictly speaking correct. - 1.5 pts Relies on distance instead of stability - 1 pts Gives ammonium value instead of population change - 2 pts Incorrect; reasoning unclear - 0.5 pts Apparently misreads scale - 2 pts No answer given - 2 pts Multiple answers given 4.4 Eradication 2 / 2 ✓ - 0 pts Correct - 2 pts Correct Option is not selected. Correct option is B Option A: Kill 6000 crickets. = This brings the cricket population to (8500-6000 = 2500). At ammonium nitrate level 11, we have unstable equilibrium at 1500 and stable equilibrium at 3000. Therefore, it moves from 2500 to 3000 in the long run. Final level of crickets = 3000. Option B: Reduce the amount of ammonium nitrate to 5 and then kill 1000 crickets = Reducing the nitrate level to 5 brings the cricket level to 3500 (since there is stable equilibrium at 3500). Killing 1000 crickets will take the cricket level to 2500, which will then move to 0 as the stable equilibrium is at 0. Final level of crickets = 0. Option C = Reduce the amount of ammonium
nitrate to 9 and then kill 4000 crickets = Reducing the nitrate level to 9 brings the cricket level to 7000 (since there is stable equilibrium at 7000). Killing 4000 crickets will take the cricket level to 3000, which will stay at 3000 as the stable equilibrium is at 3000. Final level of crickets = 3000. Option D = Reduce the amount of ammonium nitrate to 1 and then kill 1000 crickets = Reducing the nitrate level to 1 brings the cricket level to 3000 (since there is stable equilibrium at 3000). Killing 1000 crickets will take the cricket level to 2000, which will move at 3000 as the stable equilibrium is at 3000. Final level of crickets = 3000. Option E = It is impossible to eradicate the crickets = Incorrect as option B results in complete eradication. - 1 pts One additional incorrect option chosen. Correct option is B Option A: Kill 6000 crickets. = This brings the cricket population to (8500-6000 = 2500). At ammonium nitrate level 11, we have unstable equilibrium at 1500 and stable equilibrium at 3000. Therefore, it moves from 2500 to 3000 in the long run. Final level of crickets = 3000. Option B: Reduce the amount of ammonium nitrate to 5 and then kill 1000 crickets = Reducing the nitrate level to 5 brings the cricket level to 3500 (since there is stable equilibrium at 3500). Killing 1000 crickets will take the cricket level to 2500, which will then move to 0 as the stable equilibrium is at 0. Final level of crickets = 0. Option C = Reduce the amount of ammonium nitrate to 9 and then kill 4000 crickets = Reducing the nitrate level to 9 brings the cricket level to 7000 (since there is stable equilibrium at 7000). Killing 4000 crickets will take the cricket level to 3000, which will stay at 3000 as the stable equilibrium is at 3000. Final level of crickets = 3000. Option D = Reduce the amount of ammonium nitrate to 1 and then kill 1000 crickets = Reducing the nitrate level to 1 brings the cricket level to 3000 (since there is stable equilibrium at 3000). Killing 1000 crickets will take the cricket level to 2000, which will move at 3000 as the stable equilibrium is at 3000. Final level of crickets = 3000. Option E = It is impossible to eradicate the crickets = Incorrect as option B results in complete eradication. - 2 pts More than 2 incorrect options chosen. Correct option is B Option A: Kill 6000 crickets. = This brings the cricket population to (8500-6000 = 2500). At ammonium nitrate level 11, we have
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
unstable equilibrium at 1500 and stable equilibrium at 3000. Therefore, it moves from 2500 to 3000 in the long run. Final level of crickets = 3000. Option B: Reduce the amount of ammonium nitrate to 5 and then kill 1000 crickets = Reducing the nitrate level to 5 brings the cricket level to 3500 (since there is stable equilibrium at 3500). Killing 1000 crickets will take the cricket level to 2500, which will then move to 0 as the stable equilibrium is at 0. Final level of crickets = 0. Option C = Reduce the amount of ammonium nitrate to 9 and then kill 4000 crickets = Reducing the nitrate level to 9 brings the cricket level to 7000 (since there is stable equilibrium at 7000). Killing 4000 crickets will take the cricket level to 3000, which will stay at 3000 as the stable equilibrium is at 3000. Final level of crickets = 3000. Option D = Reduce the amount of ammonium nitrate to 1 and then kill 1000 crickets = Reducing the nitrate level to 1 brings the cricket level to 3000 (since there is stable equilibrium at 3000). Killing 1000 crickets will take the cricket level to 2000, which will move at 3000 as the stable equilibrium is at 3000. Final level of crickets = 3000. Option E = It is impossible to eradicate the crickets = Incorrect as option B results in complete eradication. - 1.5 pts 2 Additional incorrect options chosen Correct option is B Option A: Kill 6000 crickets. = This brings the cricket population to (8500-6000 = 2500). At ammonium nitrate level 11, we have unstable equilibrium at 1500 and stable equilibrium at 3000. Therefore, it moves from 2500 to 3000 in the long run. Final level of crickets = 3000. Option B: Reduce the amount of ammonium nitrate to 5 and then kill 1000 crickets = Reducing the nitrate level to 5 brings the cricket level to 3500 (since there is stable equilibrium at 3500). Killing 1000 crickets will take the cricket level to 2500, which will then move to 0 as the stable equilibrium is at 0. Final level of crickets = 0. Option C = Reduce the amount of ammonium nitrate to 9 and then kill 4000 crickets = Reducing the nitrate level to 9 brings the cricket level to 7000 (since there is stable equilibrium at 7000). Killing 4000 crickets will take the cricket level to 3000, which will stay at 3000 as the stable equilibrium is at 3000. Final level of crickets = 3000. Option D = Reduce the amount of ammonium nitrate to 1 and then kill 1000 crickets = Reducing the nitrate level to 1 brings the cricket
level to 3000 (since there is stable equilibrium at 3000). Killing 1000 crickets will take the cricket level to 2000, which will move at 3000 as the stable equilibrium is at 3000. Final level of crickets = 3000. Option E = It is impossible to eradicate the crickets = Incorrect as option B results in complete eradication. QUESTION 5 Tortoise stability 16 pts 5.1 How many equilibria? 2 / 2 ✓ - 0 pts Correct answer is 4. - 1 pts Correctly identified the three but potentially missed the one at x = 0 - 2 pts Incorrect / Didn't attempt 5.2 Finding equilibria 4 / 4 ✓ - 0 pts Correct. The equilibrium exists at x = 0, 1, 3, & 4. - 1 pts Identified three of the four equilibria - 2 pts Identified two of the four equilibria - 3 pts Identified one of the four equilibria - 4 pts Incorrect - 1 pts Correctly identified all equilibria but multiplied answer with 1000 - 1.5 pts Correctly identified all equilibria but reported additional points also as equilibria - 2 pts Correctly identified three of the four equilibria but multiplied answer with 1000 5.3 Start at 1500 1.5 / 2 - 0 pts 3000 is a semi-stable equilibrium. Points on the left of 3000 are attracted towards it and the points on the right diverge from it. ✓ - 0.5 pts 4000 can be considered as partially correct, since detecting that the semi-stable equilibrium at 3000 attracts points on it's left is difficult. - 2 pts Incorrect - 0.5 pts Correct (3) but reported answer in terms of X and not in terms of number of tortoises - 1 pts Partially correct (4) but reported answer in terms of X and not in terms of number of tortoises - 0.5 pts Misreads equilibrium point at X=3 as a different value - 1 pts Includes incorrect answer 5.4 Increase population 3 / 4 - 0 pts Since 3000 is a semi-stable equilibrium, which attracts points to it's left but repels points to it's right. Hence, we need tortoises population to be atleast 3001 to get a different answer. Thus answer is 3001 - 1500 = 1501 ✓ - 1 pts Found 4000 as the answer to the previous question as a result reported N/A. Since detecting that the semi-stable equilibrium at 3000 attracts only points to it's left is difficult, hence partial points. - 0.5 pts Correct answer is 1501 but reported 1500 - 4 pts Incorrect - 1 pts Forgot to subtract 1500 from 3001. Also, considered 3000 as the minimum tortoise population whereas it should be 3001 - 0.5 pts Forgot to subtract 1500 from 3001.
- 1 pts Correct answer is 1501 but reported 1500. Also reported answer in terms of X and not in terms of tortoise population. - 1 pts Calculation error - 1 pts Starts at previous answer rather than 1500 - 2 pts Confuses slope at equilibrium with slope elsewhere - 4 pts No answer given - 0 pts Incorrect but consistent with previous answer - 1 pts Calculation error 5.5 Disease 4 / 4 - 0 pts E should be a correct answer, if one considers long term. As the population will reach 3000 because 3000 is a semi-stable equilibrium and 1000 is an unstable equilibrium. ✓ - 0 pts D can also be considered as correct, since the population will increase to 1500 but it wont stay there for long - 4 pts Incorrect QUESTION 6 Linear stability 12 pts 6.1 How many equilibria? 4 / 4 - 4 pts Number of equilibria marked is not 3 ✓ - 0 pts Correct 6.2 Stability 4 / 4 - 4 pts Marked Answer is not SMU ✓ - 0 pts Correct - 3 pts First S is correct - 2 pts first s and m are correct - 0 pts Incorrect but consistent with previous - 1 pts Ignored eq pt w/ X'=0 6.3 New change vectors 4 / 4 - 4 pts Marked Answer is not D ✓ - 0 pts Correct QUESTION 7 Zebras and gazelles 18 pts 7.1 Nullclines 8 / 8 ✓ - 0 pts Correct - 8 pts incorrect/missing - 2 pts missing z=0/g=0 or not obvious to see on graph - 2 pts missing/incorrect (on graph) z=-0.5*g+10 - 2 pts missing/incorrect (on graph) z=4-4/3*g - 1 pts some mistakes in drawing or minor mistake on coordinates - 4 pts missing nullclines on graph - 2 pts missing/incorrect details(coordinates of intersection points) on graph - 4 pts incorrect graph - 1 pts redundant nullcline shown 7.2 How many feasible eqs? 2 / 2 ✓ - 0 pts Correct - 2 pts incorrect 7.3 Find equilibria 6 / 6 ✓ - 0 pts Correct - 6 pts missing/incorrect - 1.5 pts missing (0,0) - 1.5 pts missing (0,4) - 1.5 pts missing (20,0)
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
- 0.5 pts incorrect(redundant) EP shown 7.4 Coexistence 2 / 2 ✓ - 0 pts Correct - 2 pts incorrect QUESTION 8 8 Total energy use 8 / 8 ✓ - 0 pts Correct - 6 pts Not using integration concept - 2 pts Improper explanation for Riemann sum QUESTION 9 Derivatives 8 pts 9.1 Rate of change 4 / 4 ✓ - 0 pts Correct $$3x^2cos(x^3)=12cos(8)$$ $$12cos(8 radian) = -1.746$$ $$12cos(8 degrees) = 11.883$$ - 0.5 pts Minor Calculation Error for $$12cos(8)$$ - 1 pts Calculation Error $$12cos(8)$$ - 2 pts Wrong approach (Euler's method) - 4 pts Wrong answer 9.2 Linear approximation 4 / 4 ✓ - 0 pts ``Correct: Need to use linear approximation with the answer of 9(a). $$f' \approx \frac{\triangle f}{\triangle x} $$ $$f'(2) \approx \frac{f(2.05)-f(2)}{2.05-2}$$ $$f(2.05) \approx f(2)+f'(2)(2.05-2)$$ $$f(2.05) \approx sin(2^3)+(3*2^2)cos(2^3)*0.05$$ - 1 pts Wrong answer: Need to find $$f(2.05)$$ instead of $$\triangle f$$ - 1 pts Minor Error - 2 pts Wrong equation/ Wrong answer - 3.5 pts Wrong answer: Need to use linear approximation with the answer of 9(a). - 4 pts No answer /Wrong answer QUESTION 10 Osciillations 8 pts 10.1 Concentric trajectories 4 / 4 ✓ - 0 pts Correct - 4 pts Incorrect - 2 pts Partially correct 10.2 Shaking hands 4 / 4 ✓ - 0 pts Correct (AD) - 1 pts 1 correct - 1 pts 2 correct, 1 wrong - 2 pts 1 correct, 1 wrong - 3 pts 1 correct, 2 wrong - 3 pts 2 correct, 2 wrong - 4 pts Incorrect QUESTION 11 11 Encrypting number 7 / 8 - 0 pts Correct - 8 pts No answer - 1 pts Hardcoded length of key (should have used len(digitList) instead) ✓ - 1 pts Did not define encrypted as an empty list - 1 pts Did not use indexing while adding numbers inside the for loop / incorrect indexing - 1 pts Chose the wrong list to loop over - 1 pts Added the wrong variable inside loop /
did not use the loop variable "i" - 1 pts Returned the wrong variable, should have returned "encrypted" - 1 pts Appended the wrong variable to "encrypted" - 2 pts Incorrect definition for "prod" - 1 pts Appended "prod" to the wrong list - 2 pts Incorrect variables for ".append()" - 8 pts Incorrect - 1 pts Added the same number to every digit - 0.5 pts Used digitList instead of len(digitList) / wrong function Page 8
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
0T For
to E
My Esq
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Ʃ See
7 5 2
Isar
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
i I I
F
inner
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
safe 7 4
EBBETS
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help

Related Documents

IMG_6996.png
SCI200_Section_016_Week_03_Quizz_4.docx
business marketing.docx
SA3_RSolution_Econ140.pdf
Practice Exam#1.pdf
Assignment 1.pdf
Midterm1_solution.pdf
Excel Pie Charts Directions-1.pdf
Discussion Case #2 - Craft Beer.docx
Supply+Chain+Case+Study_Master+Tag.docx
Chapter 23 Review Questions.docx
Chapter 16 Problems.docx
Recommended textbooks for you
Text book image
Managerial Economics: A Problem Solving Approach
Economics
ISBN:9781337106665
Author:Luke M. Froeb, Brian T. McCann, Michael R. Ward, Mike Shor
Publisher:Cengage Learning
Text book image
Managerial Economics: Applications, Strategies an...
Economics
ISBN:9781305506381
Author:James R. McGuigan, R. Charles Moyer, Frederick H.deB. Harris
Publisher:Cengage Learning
Text book image
Economics:
Economics
ISBN:9781285859460
Author:BOYES, William
Publisher:Cengage Learning
Text book image
Microeconomics: Principles & Policy
Economics
ISBN:9781337794992
Author:William J. Baumol, Alan S. Blinder, John L. Solow
Publisher:Cengage Learning
Text book image
Economics Today and Tomorrow, Student Edition
Economics
ISBN:9780078747663
Author:McGraw-Hill
Publisher:Glencoe/McGraw-Hill School Pub Co
Text book image
Micro Economics For Today
Economics
ISBN:9781337613064
Author:Tucker, Irvin B.
Publisher:Cengage,
SEE MORE TEXTBOOKS
Recommended textbooks for you
Text book image
Managerial Economics: A Problem Solving Approach
Economics
ISBN:9781337106665
Author:Luke M. Froeb, Brian T. McCann, Michael R. Ward, Mike Shor
Publisher:Cengage Learning
Text book image
Managerial Economics: Applications, Strategies an...
Economics
ISBN:9781305506381
Author:James R. McGuigan, R. Charles Moyer, Frederick H.deB. Harris
Publisher:Cengage Learning
Text book image
Economics:
Economics
ISBN:9781285859460
Author:BOYES, William
Publisher:Cengage Learning
Text book image
Microeconomics: Principles & Policy
Economics
ISBN:9781337794992
Author:William J. Baumol, Alan S. Blinder, John L. Solow
Publisher:Cengage Learning
Text book image
Economics Today and Tomorrow, Student Edition
Economics
ISBN:9780078747663
Author:McGraw-Hill
Publisher:Glencoe/McGraw-Hill School Pub Co
Text book image
Micro Economics For Today
Economics
ISBN:9781337613064
Author:Tucker, Irvin B.
Publisher:Cengage,
SEE MORE TEXTBOOKS