LS30A - Final
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Economics
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Feb 20, 2024
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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
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)
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AUG
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90
80
70
60
ATC
50
40
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MC
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20
30
40
50
60
70
80
90
100
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TOTAL
MARGINAL
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PRODUCT
PRODUCT
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COST
AVERAGE
VARIABLE
COST
TOTAL FIXED
TOTAL
AVERAGE
TOTAL
MARGINAL
COST
COST
COST
COST
012345678
0
$1
$2
6
$1
2
15
$1
27
$1
37
$1
45
$1
50
$1
52
$1
50
$1
ଖ ଖ ଖ ଖ ଖ ଖ ଖ ଖ ଖ
$2
$2
$2
$2
$2
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Offices
Entrances
Age
AssessedValue ($'000)
4790
4
2
8
1796
4720
3
2
12
1544
5940
4
2
2
2094
5720
4
2
34
1968
3660
3
2
38
1567
5000
4
2
31
1878
2990
2
1
19
949
2610
2
1
48
910
5650
4
2
42
1774
3570
2
1
4
1187
2930
3
2
15
1113
1280
2
1
31
671
4880
3
2
42
1678
1620
1
2
35
710
1820
2
1
17
678
4530
2
2
5
1585
2570
2
1
13
842
4690
2
2
45
1539
1280
1
1
45
433
4100
3
1
27
1268
3530
2
2
41
1251
3660
2
2
33
1094
1110
1
2
50
638
2670
2
2
39
999
1100
1
1
20
653
5810
4
3
17
1914
2560
2
2
24
772
2340
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1
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890
3690
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2
15
1282
3580
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2
27
1264
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ISBN:9781259290619
Author:Michael Baye, Jeff Prince
Publisher:McGraw-Hill Education
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- Note:- Do not provide handwritten solution. Maintain accuracy and quality in your answer. Take care of plagiarism.Answer completely.You will get up vote for sure.arrow_forwardWhat is the Consistency Ratio of the GEAR Matrix? This question is related to BIKE and not fruit. Use BIKE MATRIX. Please have it be original work and highlight the answer!! this is the 3rd time I've posted this and both have been the same incorrect answer.arrow_forwardNote:- Do not provide handwritten solution. Maintain accuracy and quality in your answer. Take care of plagiarism. Answer completely. You will get up vote for sure.arrow_forward
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