Philosophy
Ethics Test
Jacob Silber
Test Nov 26
1.
Subjectivism: there are no facts to the matter about whether an act is right or wrong that exist
independently of anyone's beliefs or feelings about whether the act is right or wrong
Objectivism: solid fa
Philosophy Final Exam
Jacob Silber
5. A person who is claiming that the salt is not soluble based on the fact that the salt will never be
placed in water is not recognizing the fact that salt has a power that diamonds do not. A power
is what would happen
7
Name:_ ID:_
Problem 5 (50 points)
Synthesize compounds A and B using any combination of the starting materials 17 as the
only sources of carbons. Note that you do not have to use all the starting materials. Feel
free to use any reagents you want, in or
Economics
Semester 1
Answers to Sample Examinations
2009
UNSW Foundation Studies
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Sydney NSW 2052
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All rights reserved. Except under the
conditions described in the Copyright
Act 1968 of Australia and subsequent
a
TUTORIAL
SAMPLE
(SOLUTIONS)
Mathematics C
Semester 1 Examination
Time Allowed: 1.5 hours
Reading Time: 5 minutes
UNSW Foundation Studies
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UNSW
Sydney NSW 2052
Copyright 2009
All rights reserved. Except under the
conditions described
Name:
5
ANSWER KEY_
Problem 4 (30 points)
Complete the following transformations. Pay careful attention to the stereochemistry of the
starting materials and/or products. Mechanisms are not required.
O
O
OH H+
,
HO
PCC
O
or CrO3
HO
HO
H+, H2O
O
O
1. CH3CH2
5
Name:_ ID:_
Problem 3 (50 points)
Diene 1 can produce progesterone derivative 3 via a sequence of two steps that includes:
(a) ozonolysis of 1 to form 2 and (b) treatment of 2 with water and KOH to form 3.
What is the structure of 2?
10 points
Me
Me
Me
UNIVERSITY OF WATERLOO
DEPARTMENT OF MANAGEMENT SCIENCES
MSCI603  Principles of Operations Research Fall 2014
Problem set 1
Problem 1
Use the graphical method to solve the following problem:
Solution
Problem 2
Use the graphical method to solve the follow
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3
Name:_ ID:_
Problem 1 cont. (50 points)
In compounds 1 and 2 indicate the expected splitting patterns for each peak in the 1H NMR.
(Consider only 3 bonds apart couplings)
C1 protons= triplet (due to C2)
4
Br
H3C
example
H3C
2
3
5
1
CH3
C2 protons= quart
4. (ii)
(g)
There are two solutions as the graph of y = 11e 2 x cuts the
above graph in two places.
y
(3,92)
(2,59)
(0,11)
x
5.
(i)
Altitude of climber
1000+(1800+1200+800+)
1800
= 1000 +
2
1
3
= 1000 + 5400
= 6400
If he climbs forever he will only reach
4.
(a)
Domain: x
Range: y 1
(b)
(i)
( )
( )
(c)
The horizontal line is y = 1
The oblique line is y = x + 1
1
f ( x) =
x + 1
(ii)
16+18+20+
25
(2 16 + 24 2)
2
= 1000
S 25 =
x0
x<0
4. (iii)
(a)
Vertical asymptote
2x 3 = 0
2x = 3
3
x=
2
xintercept when y
University of Wollongong
School of Computer Science and Software Engineering
CSCI124
Applied Programming
Spring 2012
Quiz 1 Revision
100 points
C+ Programming
ANSWERS
Write all answers in the spaces provided.
Time allowed 60 minutes.
Question 1 (20 points
3.
(ii)
(d)
y
max profit
(16,160)
(20,140)
x
4686
breakeven point
(1,12125)
4.
(i)
13
(2n 3)
n =1
= 1 + 1 + 3 + KK + 23
13
(1 + 23)
2
= 143
=
(ii)
(a)
(b)
y = 12 x 3 3 x 4 + 11
dy
= 36 x 2 12 x 3
dx
d2y
= 72 x 36 x 2
2
dx
Stationary points when
dy
=0
d
ECONOMICS SAMPLE SEMESTER EXAMS
ANSWERS TO MULTIPLE CHOICE
SAMPLE SEMESTER EXAM A
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
D
A
B
D
B
D
C
C
C
D
B
C
C
A
A
B
A
D
B
D
SAMPLE SEMESTER EXAM B
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
N1 = quantity of Naphtha used to produce gasoline
N2 = quantity of Naphtha used to produce jet fuel
R11 = quantity of cracked 1 used to produce gasoline
R12 = quantity of cracked 1 used to produce jet fuel
R21 = quantity of cracked 2 used to produce gasol
Solution
Problem 5
Each week, Chemco can purchase unlimited quantities of raw material at $6/lb. Each pound of
purchased raw material can be used to produce either input 1 or input 2. Each pound of raw
material can yield 2 oz of input 1, requiring 2 hours
PART II. (15 points for each problem)
Instructions for Part II: For each question, show all of your work.
1.
The quantile function Q(u) is to be estimated from a random sample of 20
data values. Their order statistics are :
7.50 5.45 5.00 4.60 4.21 
Philosophy
Jacob Silber
Exam 2
Due Oct. 31
1.) When one says that people should accept God's existence on faith it means that the
person believing in God only on faith holds that belief with no evidence or reasoning for
it but still believes it. When some
Philosophy 101
Jacob Silber
Take Home Test: Part 1
9/19/13
Due:
1.) Socrates argues with Theaetetus that knowledge cannot be a list of examples of things that are
not knowledge. The flaw in this argument is that when one gives a definition and uses the wo
Philosophy
Take Home Test Part 2
Jacob Silber
Due: 10/3/13
1.) The second argument that Decartes gives that we should not trust beliefs based on the senses is that
that we ought to trust beliefs that are based on the senses in optimal perceptual condition
4. (iii)
(c)
() From the sketch y < 0 when x > 2 x < 1
()
Draw y = ln x on same diagram as the hyperbola.
It will intersect twice with the hyperbola.
2 solutions to
5.
(i)
1
2
4 2x
= ln x
2x 3
2 + x, 6 + x, 13 + x
6 + x 13 + x
=
2+ x
6+ x
2
(6 + x ) = (2
Name:_ ID:_
CHEMISTRY 140B
FINAL Exam
1
03/17/05
Instructions:
1: This exam consists of 10 pages, including this page. (Pages #9,10 are blank). At the
beginning of the examination period, put your name at the top of each page.
2. This exam accounts for 50
2. (iv)
9 x 10 3 x + 9 = 0
Put m = 3 x
m 2 10m + 9 = 0
(m 9)(m 1) = 0
m = 9 m =1
3x = 9
3x = 1
x=2
x=0
(a)
f ( 0) = 4
(b)
f ( x ) 0 for x 3
(c)
(v)
g (2) = f (2) + 10
= 1 + 10
= 11
(d)
(e)
3. (i)
y=5
(a)
x 1> 0
Df : x >1
3.
(b)
(c)
y = ln( x 1)
interchang
LECTURE
SAMPLE
(SOLUTIONS)
Mathematics C
Semester 1 Examination
Time Allowed: 1.5 hours
Reading Time: 5 minutes
UNSW Foundation Studies
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Sydney NSW 2052
Copyright 2009
All rights reserved. Except under the
conditions described
2.
(3 x + 1) 2 ( x + 1)( x + 7)
(ii)
9x 2 + 6x + 1 x 2 + 8x + 7
8x 2 2 x 6 0
2(4 x 2 x 3) 0
2(4 x + 3)( x 1) 0
Solution is cfw_x :
(iii)
3
x 1
4
log b 12 = log b 2 2 3 = 2 log b 2 + log b 3 = 2 x + y
(iv)
y= x=
1
x2
1
dy 1 2
1
= x =
dx 2
2 x
1
1 1 1
= =
5.
(iii)
(a)
y = g (x)
(b)
Stationary points at x = 2 and x = 1
x
y'
x=2
1
2
0
3
+
x
y'
min at x = 2
(c)
x = 1
2 1
+
0
0

max at x = 1
Min at x = 2
Max at x = 1
Crosses x axis once.
or
y
1
2
x
UNSW Foundation Year
UNSW Foundation Studies UNSW Sydney N
STUDENT
SAMPLE
(SOLUTIONS)
Mathematics C
Semester 1 Examination
Time Allowed: 1.5 hours
Reading Time: 5 minutes
UNSW Foundation Studies
UNSW Global Pty Limited
UNSW
Sydney NSW 2052
Copyright 2009
All rights reserved. Except under the
conditions described