3. Joe is considering pursuing an MBA degree. He has applied to t.pdf
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3. Joe is considering pursuing an MBA degree. He has applied to two different universities. The acceptance rate for applicants with similar qualifications is 25% for University A and 40% for University B. What is the probability that Joe will not be accepted at either university? (Points : 4) 0.75 0.45 0.90 0.65 0.60 Solution Given P(A)=0.25, P(B)=0.4 So the probability is P(A\' and B\') = P(A\')*P(B\') =[1-P(A)]*[1-P(B)] =(1-0.25)*(1-0.4) =0.45 Answer:0.45.
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![3. Joe is considering pursuing an MBA degree. He has applied to two different universities. The
acceptance rate for applicants with similar qualifications is 25% for University A and 40% for
University B.
What is the probability that Joe will not be accepted at either university? (Points : 4)
0.75
0.45
0.90
0.65
0.60
Solution
Given P(A)=0.25, P(B)=0.4 So the probability is P(A' and B') = P(A')*P(B')
=[1-P(A)]*[1-P(B)] =(1-0.25)*(1-0.4) =0.45 Answer:0.45](https://image.slidesharecdn.com/3-230323124432-83cb9dd4/85/3-Joe-is-considering-pursuing-an-MBA-degree-He-has-applied-to-t-pdf-1-320.jpg)
Recommended
4) Converting data into indices is... a) used only by governments.pdf
4) Converting data into indices is...
a) used only by governments
b) a way to show the change in the data over time
c) a way to show the data in percentage form
d) misleading if theres a sudden change in the datas trend
5) The table below reports the earnings per share of common stock for Home Dumpo Inc. for
recent years.
Solution
I have fitted a parabolic curve for the given data and found good correlation.
4) A wastewater treatment operator has been fined because the.pdf
4) A wastewater treatment operator has been fined because the DO at a point 1 km downstream
is below the 4.5 mg/L regulation. Assuming the plant can produce a wastewater effluent with a
BODu of 10mg/L, what is the lowest DO in the wastewater? DO in wastewater must remain
above 4.5 mg/L year round. QR = 50 m3/s Qw = 15 m3/s Tw Summer 25°C R Summer 18°C
kaonaar= 0.15 d-1 (assume KBOD is the same for the river and effluent) H =1 m (assume no
change due to wastewater flow) V = 0.04 m/s (assume no change due to wastewater flow) DOR
= 6 mg/L BODww=10mg/L BODa = 5 mg/L n = 0.3 DOs @ 20. = 9.2 mg/L Downstream
Distance = 1km
Solution
wastewater
(given)
Stream Given
Mixture of WW and River
flow (m3/s)
15
50
Qmixture=50 + 15 = 65
65
Dissolved oxygen, mg/L
9.2
6
DOmixture=(9.2* 15 + 6*50)/(50 + 15)
= 6.738 mg/L
6.738
Temperature, °C
25
18
Temperature Mixture = (25 x15 + 18 x 50)/(65
19.615
BOD5 at 20°C, mg/L
10
5
BOD mixture = (10 x 15 + 5 x 50)/65
6.153
Oxygen consumption rate (K1 at
°C) (1/day)
0.15
0.15
(0.15 x 15 + 0.15 x 50)/65
0.15
Oxygen reaeration rate (K2 at °C)
(1/day)
0.83
0.7
Mixture = (.83 x15 + 0.7 x 50)/65
0.73
L = L0e-k1t
So, K1s=
K2 = 3.9 x v1/2 x [ (1.025) T-20]1/2/H3/2
K2R = 3.9 x (0.04) 1/2 x [ (1.025) 18-20]1/2/H3/2 = 0.7
K2s = 3.9 x (0.04) 1/2 x [ (1.025) 25 -20]1/2/H3/2= 0.83
Ultimate BOD= Yultimate=L0 = (5-day BOD in mixture water)/[1-exp(-K1mixture*5)]
=(6.153 mg/L)/ [1-exp(-0.15*5)] = 11.66 mg/L
For 20°C stream water temperature, equilibrium concentration of oxygen =9.17 mg/L
D0 = 9.17 mg/L-6.738 mg/L = 2.432 mg/L
Initial deficit = 2.432 mg/L
D (t=2 days) = [K1*L0]*[exp (-K1*t)-exp (-K2*t)]/ (K2-K1) + D0 exp (-K2*t)
= (0.15 x 11.66) *[exp (-0.15*13.649)-exp (-0.73*13.649)]/ (0.73- 0.15) + 2.432 exp (-
0.73*13.649)
= 0.388 mg/L
DO (2day) = 9.17- 0.388 = 8.782 mg/L
Time for critical DO deficit
(tc)= 1/ (K2-K1)*ln [(K2/K1)*(1-D0 *(K2-K1)/ (K1L0))]
= 1/ (0.73-0.15)*ln [(0.73/0.15)*(1-2.432 *(0.73-0.15)/ (0.73* 11.66 ))]
= 2.41 days
Critical DO deficit (Dc)= (K1/K2)*L0 exp (-K1*tc)
= ( 0.15/0.73) x 11.66 exp( - 0.15 x 2.41)= 1.669 mg/L
DOcritical= 9.17- 1.669 = 7.501 mg/L
wastewater
(given)
Stream Given
Mixture of WW and River
flow (m3/s)
15
50
Qmixture=50 + 15 = 65
65
Dissolved oxygen, mg/L
9.2
6
DOmixture=(9.2* 15 + 6*50)/(50 + 15)
= 6.738 mg/L
6.738
Temperature, °C
25
18
Temperature Mixture = (25 x15 + 18 x 50)/(65
19.615
BOD5 at 20°C, mg/L
10
5
BOD mixture = (10 x 15 + 5 x 50)/65
6.153
Oxygen consumption rate (K1 at
°C) (1/day)
0.15
0.15
(0.15 x 15 + 0.15 x 50)/65
0.15
Oxygen reaeration rate (K2 at °C)
(1/day)
0.83
0.7
Mixture = (.83 x15 + 0.7 x 50)/65
0.73.
4.634.154.764.704.654.524.705.064.424.514.2.pdf
4.63
4.15
4.76
4.70
4.65
4.52
4.70
5.06
4.42
4.51
4.24
4.52
Listed below is the rate of return for one year (reported in percent) for a sample of 12 mutual
funds that are classified as taxable money market funds.
Solution
Critical t value is (DF=11) 1.796
Sample mean is 4.5717
Sample standard deviation is 0.2405
t= (4.5717- 4.5)/.2405/12= 1.033
Do not reject the null hypothesis (that the mean rate of return is 4.5)..
4. Let TL, t 2 be a colloclion of arbitrary vcclors in R. Is span(T.pdf
4. Let TL, t 2 be a colloclion of arbitrary vcclors in R\". Is span(TL, i2,...,im) aveclor subspace
of RA? Explain why or why not.
Solution
i dont know why the font size is too small cant even read it properly repost the
question with proper font size thanks.
4. Image and Kernel of homomorphism. Let f R S be a ring homomo.pdf
4. Image and Kernel of homomorphism. Let f: R ? > S be a ring homomorphism. Prove that ker
(f) is an ideal of R.
Solution
Consider ker(f) as a subring of R
Let y belong to ker(f) and x belong to R.
Then f(xy)=f(x)f(y)=f(x)0=0
Thus, f(xy)=0 that is xy belongs to ker(f)
Hence, ker(f) is an ideal of R.
4. Find all values of z that satisfy the equation sin z = -5. Express.pdf
4. Find all values of z that satisfy the equation sin z = -5. Express each in the form a + bi. Or,
explain why no such z exists.
Solution
If we try solving this we will get
sine inverse of (-5) =z
But as we know that sine inverse functionby definition takes the values [-1,1] Hence no such z is
defined for the above equation..
4. A steel channel made of 1040 HR is welded to a thick steel wall wi.pdf
4. A steel channel made of 1040 HR is welded to a thick steel wall with fillet welds around the
outside of the channel only. Select the least expensive fillet that will provide a FOS of 3 for the
static load shown.
Solution
The cheapest way of welding is fillet joint which is most commonly used in welding processs.it
is mainly used for the welding of steels..
4. A f-unction f D right arrow R is a Lipschitz function if there is.pdf
4. A f-unction f: D right arrow R is a Lipschitz function if there is a C > = 0 such that |f(x) f(y)|
Solution
a) Suppose f is Lipschitz on [a,b] and let e > 0. Set d = e/M. If P = {[xi, yi]} is any partition of
[a,b] with ||P|| < d, then ? |f(xi) - f(yi) ? ? M|xi - yi| = M||P|| < M(e/M) = e. Thus f is absolutely
continuous on [a,b].
b) If f is not Lipschitz, then are sequences {x_n} and {y_n} in [a,b] such that |f(x_n) - f(y_n)| >
n|x_n - y_n| for all n ? 1. Thus |f(x_n) - f(y_n)|/|x_n - y_n| > n for all n ? 1. Hence lim sup |f(x_n)
- f(y_n)|/|x_n - y_n| = ? and lim inf |f(x_n) - f(y_n)|/|x_n - y_n| = ?. This the one sided derivatives
of f are not bounded..
Recommended
4) Converting data into indices is... a) used only by governments.pdf
4) Converting data into indices is...
a) used only by governments
b) a way to show the change in the data over time
c) a way to show the data in percentage form
d) misleading if theres a sudden change in the datas trend
5) The table below reports the earnings per share of common stock for Home Dumpo Inc. for
recent years.
Solution
I have fitted a parabolic curve for the given data and found good correlation.
4) A wastewater treatment operator has been fined because the.pdf
4) A wastewater treatment operator has been fined because the DO at a point 1 km downstream
is below the 4.5 mg/L regulation. Assuming the plant can produce a wastewater effluent with a
BODu of 10mg/L, what is the lowest DO in the wastewater? DO in wastewater must remain
above 4.5 mg/L year round. QR = 50 m3/s Qw = 15 m3/s Tw Summer 25°C R Summer 18°C
kaonaar= 0.15 d-1 (assume KBOD is the same for the river and effluent) H =1 m (assume no
change due to wastewater flow) V = 0.04 m/s (assume no change due to wastewater flow) DOR
= 6 mg/L BODww=10mg/L BODa = 5 mg/L n = 0.3 DOs @ 20. = 9.2 mg/L Downstream
Distance = 1km
Solution
wastewater
(given)
Stream Given
Mixture of WW and River
flow (m3/s)
15
50
Qmixture=50 + 15 = 65
65
Dissolved oxygen, mg/L
9.2
6
DOmixture=(9.2* 15 + 6*50)/(50 + 15)
= 6.738 mg/L
6.738
Temperature, °C
25
18
Temperature Mixture = (25 x15 + 18 x 50)/(65
19.615
BOD5 at 20°C, mg/L
10
5
BOD mixture = (10 x 15 + 5 x 50)/65
6.153
Oxygen consumption rate (K1 at
°C) (1/day)
0.15
0.15
(0.15 x 15 + 0.15 x 50)/65
0.15
Oxygen reaeration rate (K2 at °C)
(1/day)
0.83
0.7
Mixture = (.83 x15 + 0.7 x 50)/65
0.73
L = L0e-k1t
So, K1s=
K2 = 3.9 x v1/2 x [ (1.025) T-20]1/2/H3/2
K2R = 3.9 x (0.04) 1/2 x [ (1.025) 18-20]1/2/H3/2 = 0.7
K2s = 3.9 x (0.04) 1/2 x [ (1.025) 25 -20]1/2/H3/2= 0.83
Ultimate BOD= Yultimate=L0 = (5-day BOD in mixture water)/[1-exp(-K1mixture*5)]
=(6.153 mg/L)/ [1-exp(-0.15*5)] = 11.66 mg/L
For 20°C stream water temperature, equilibrium concentration of oxygen =9.17 mg/L
D0 = 9.17 mg/L-6.738 mg/L = 2.432 mg/L
Initial deficit = 2.432 mg/L
D (t=2 days) = [K1*L0]*[exp (-K1*t)-exp (-K2*t)]/ (K2-K1) + D0 exp (-K2*t)
= (0.15 x 11.66) *[exp (-0.15*13.649)-exp (-0.73*13.649)]/ (0.73- 0.15) + 2.432 exp (-
0.73*13.649)
= 0.388 mg/L
DO (2day) = 9.17- 0.388 = 8.782 mg/L
Time for critical DO deficit
(tc)= 1/ (K2-K1)*ln [(K2/K1)*(1-D0 *(K2-K1)/ (K1L0))]
= 1/ (0.73-0.15)*ln [(0.73/0.15)*(1-2.432 *(0.73-0.15)/ (0.73* 11.66 ))]
= 2.41 days
Critical DO deficit (Dc)= (K1/K2)*L0 exp (-K1*tc)
= ( 0.15/0.73) x 11.66 exp( - 0.15 x 2.41)= 1.669 mg/L
DOcritical= 9.17- 1.669 = 7.501 mg/L
wastewater
(given)
Stream Given
Mixture of WW and River
flow (m3/s)
15
50
Qmixture=50 + 15 = 65
65
Dissolved oxygen, mg/L
9.2
6
DOmixture=(9.2* 15 + 6*50)/(50 + 15)
= 6.738 mg/L
6.738
Temperature, °C
25
18
Temperature Mixture = (25 x15 + 18 x 50)/(65
19.615
BOD5 at 20°C, mg/L
10
5
BOD mixture = (10 x 15 + 5 x 50)/65
6.153
Oxygen consumption rate (K1 at
°C) (1/day)
0.15
0.15
(0.15 x 15 + 0.15 x 50)/65
0.15
Oxygen reaeration rate (K2 at °C)
(1/day)
0.83
0.7
Mixture = (.83 x15 + 0.7 x 50)/65
0.73.
4.634.154.764.704.654.524.705.064.424.514.2.pdf
4.63
4.15
4.76
4.70
4.65
4.52
4.70
5.06
4.42
4.51
4.24
4.52
Listed below is the rate of return for one year (reported in percent) for a sample of 12 mutual
funds that are classified as taxable money market funds.
Solution
Critical t value is (DF=11) 1.796
Sample mean is 4.5717
Sample standard deviation is 0.2405
t= (4.5717- 4.5)/.2405/12= 1.033
Do not reject the null hypothesis (that the mean rate of return is 4.5)..
4. Let TL, t 2 be a colloclion of arbitrary vcclors in R. Is span(T.pdf
4. Let TL, t 2 be a colloclion of arbitrary vcclors in R\". Is span(TL, i2,...,im) aveclor subspace
of RA? Explain why or why not.
Solution
i dont know why the font size is too small cant even read it properly repost the
question with proper font size thanks.
4. Image and Kernel of homomorphism. Let f R S be a ring homomo.pdf
4. Image and Kernel of homomorphism. Let f: R ? > S be a ring homomorphism. Prove that ker
(f) is an ideal of R.
Solution
Consider ker(f) as a subring of R
Let y belong to ker(f) and x belong to R.
Then f(xy)=f(x)f(y)=f(x)0=0
Thus, f(xy)=0 that is xy belongs to ker(f)
Hence, ker(f) is an ideal of R.
4. Find all values of z that satisfy the equation sin z = -5. Express.pdf
4. Find all values of z that satisfy the equation sin z = -5. Express each in the form a + bi. Or,
explain why no such z exists.
Solution
If we try solving this we will get
sine inverse of (-5) =z
But as we know that sine inverse functionby definition takes the values [-1,1] Hence no such z is
defined for the above equation..
4. A steel channel made of 1040 HR is welded to a thick steel wall wi.pdf
4. A steel channel made of 1040 HR is welded to a thick steel wall with fillet welds around the
outside of the channel only. Select the least expensive fillet that will provide a FOS of 3 for the
static load shown.
Solution
The cheapest way of welding is fillet joint which is most commonly used in welding processs.it
is mainly used for the welding of steels..
4. A f-unction f D right arrow R is a Lipschitz function if there is.pdf
4. A f-unction f: D right arrow R is a Lipschitz function if there is a C > = 0 such that |f(x) f(y)|
Solution
a) Suppose f is Lipschitz on [a,b] and let e > 0. Set d = e/M. If P = {[xi, yi]} is any partition of
[a,b] with ||P|| < d, then ? |f(xi) - f(yi) ? ? M|xi - yi| = M||P|| < M(e/M) = e. Thus f is absolutely
continuous on [a,b].
b) If f is not Lipschitz, then are sequences {x_n} and {y_n} in [a,b] such that |f(x_n) - f(y_n)| >
n|x_n - y_n| for all n ? 1. Thus |f(x_n) - f(y_n)|/|x_n - y_n| > n for all n ? 1. Hence lim sup |f(x_n)
- f(y_n)|/|x_n - y_n| = ? and lim inf |f(x_n) - f(y_n)|/|x_n - y_n| = ?. This the one sided derivatives
of f are not bounded..
3. Use the compound interest formulas A F 1+ and A Pe to find the acc.pdf
3. Use the compound interest formulas A F 1+ and A Pe to find the accumulated value of an
investment of S12,850 for 8 years at an interest rate of 6 iftie money is... a. Compounded
Quarterly b. compounded Continuously
Solution
A=P[1+(R/N)]^(NT) P=12850 $ T=8 YRS R=6.5%=0.065 a).
3. Transportation technology (15 points) a. Explain both generally an.pdf
3. Transportation technology (15 points) a. Explain both generally and with specific examples
how changes in transportation technology have affected the development of urban areas. b.
Indicate two ways in which governmental policy has influenced the shape of urbanization in the
U.S. c. Do you believe that new telecommunication and other forms of technology will make
distance an unimportant factor in economic development and, thus, make cities unimportant
economic factors? Support your answer.
Solution
Transportation enbles activity of connecting people, businesses and resources.Transportation
improvement is necessary for the economicdevelopment and transportation policies help to
support the economic objectives.it helps to improve economic efficieny. Economic efficiency
refers to the ratio total benifit to cost. Increase economic efficieny, increase productivity which
leads to economic development.
The ultimate goal of transportation is accessabilty, access the people and the company abilities
including raw material, labor, worksites, proffessional services, distrbutors etc. Increased
accessibilty means reduction in time,money, risk require recouces reaches in less time so the
productivity of a company increases.
cost saving in an organisation---Economic efficiency more output per unit of input---productivity
more goods and services produced in less cost---Ecoonomis development.
Example- if in a company a manager has to submit an assignment cost company say 10 crores
and he has to submit it with in 2 days then the manager priority is to travel by Air not by train
etc. because the loss of the company is veyr high then the cost of transportation. In transportation
priortization is imortant point to improve productivity hence economic development.
Urbanisation means the population shifted rural to urban areas and how the economy adopt the
change.Recent development of commiucation technology in US, such as broadband services ,
package express delivery services, cheap telecommunication, courier services , easy ways to
transprotation, highways development promotes people to work in the different ares of the
country.
But now days the USA government initiating suburbanisation..
3. Use the compound interest formulas A=P(1+rn)^m and A= Pe^rt to fi.pdf
3. Use the compound interest formulas A=P(1+r/n)^m and A= Pe^rt to find the accumulated
value of an investment of $12,850 for 8 years at an interest rate of 6.5% if the money is a.
Compounded Quarterly b. Compounded Continuously
Solution
A=P[1+(R/N)]^(NT) P=12850 $ T=8 YRS R=6.5%=0.065 a).
3. The number N of bacteria present in a culture at time t, in ho.pdf
3. The number N of bacteria present in a culture at time t, in hours, obeys the law exponential
growth N(t) = 500e^0.02t a) How many bacteria are there at t = 0 hours? b) When will the
number of bacteria triple?
Solution
put t=0
N(0)= 500e0= 500
bacteria triple
N(t) = 500*3 = 1500
1500=500e0.02t
3=e0.02t
ln3=0.02t
t = ln3*50
=50ln3.
3. Suppose for f(z) holomorphic C there are constants a, b, e 0 Pr.pdf
3. Suppose for f(z) holomorphic C there are constants a, b, e > 0: Prove that f is a polynomial of
degree n less than equal to c .
Solution
A holomorphic function is a complex-valued function of one or more complex variables that is
complex differentiable in a neighborhood of every point in its domain.
Hence f(z) is holomorphic means
f\'(z) exists for all z
2x= z+z bar and 2iy = z-z bar
f(z,zbar) = =?m?0,n?0amnzmz.
3. Let On be the set of odd permutations in Sn. Is On a subgroup .pdf
3. Let On be the set of odd permutations in Sn. Is On a subgroup?
Solution
No, On is not a subgroup..
to see that the odd permutations don\'t form a subgroup of Sn is a bit easier to resolve.
Recall that any subgroup of Sn has the same identity element as the identity element in Sn. But
since the identity element in Sn is not an odd permutation (a result we saw in class), it follows
that
the collection of odd permutations cannot be a subgroup of Sn. (One could also argue that the
odd
elements aren\'t closed under the group operation.).
3. (4 points) Consider the function (in the domain {z i Solut.pdf
3. (4 points) Consider the function (in the domain {|z ? i|
Solution
Consider f(z)
It got a denominator of 1+(z-i)2
Equate to 0 the denominator to get singular points
(z-i)2 =-1
Take square root
z-i =i, -i
z=2i or 0
As |z-i|<1 we can expand f(z) by binomial series
f(z) = 1-(z-i)2+(z-i)4+...+(-1)n(z-i)2n+...
As odd powers are 0 we have
derivative of f^m(z) =0 for m = odd
For n even f\"m(z) = (-1)n(2n)(2n-1)....(1)(x2n-2n)
= (-1)n(2n)!
Proved..
3. Find currents (values and directions) in all resistors of the circ.pdf
3. Find currents (values and directions) in all resistors of the circuit given below. Find total
electric power developing in this circuit Neglect internal resistances of the batteries.
Solution
As all the Voltage sources are DC (Batteries) there by the capacitor acts as short circuit .
I1 = Current through R1 =E1-E2)/R1 = (6-3)/5 = 3/5 Amps (direction from E1 to E2)
I2= Current Throuh R3 = E2/R3 = 3/4 Amps (direction from Top to bottom)
I3 = Current Through R2 = E3-E2)/R2 = (5-3)/5 = 2/5 Amps (direction from E3 to E2)
Power delivered = (I1*E1) + (E3*I3) =(3/5)*6 + (2/5)*5 = 28/5 Watts.
3. Assume that the DS register contains 0100 and register SI contains.pdf
3. Assume that the DS register contains 0100 and register SI contains 0020. What is the result of
following instruction LES DI, [SI] Where the content of memory Location is as follows:
Solution
the command LES DI , [SI] does the following:
it takes the pointer of SI , i.e the address of the memory location SI and breaks it into lower bits
and upper bits.
then the lower bits are stored in DI and the upper bits are stored in ES.
so address of SI is 1027 .
lower bit is 27 and upper bit is 10.
27 will be stored in DI . and 10 will be saved in ES.
now DI is the lower bit of the register DS.
as it is said initially DS 0100 so, initially DI is 00.
which is changed to 27.
now DS is 0127..
2a Briefly describe the 1st law of thermodynamics. You can use either.pdf
2a Briefly describe the 1st law of thermodynamics. You can use either plain language or
equations with short explanation. Figures can also be used if necessary. (2b) Please demonstrate
that the unit of the specific kinetic energy 1/2 v^2 (where V is velocity with unit s /m) is Joule /
kg.
Solution.
3 (a) calculate thy potential at point P located a distance z above .pdf
3 (a) calculate thy potential at point P located a distance z above the center of a charged disk of
radius a and surface charge density p, Coulombs per square meter. Please show all work An
integral that you may need is The elemental area in cylindrical coordinates is dA = (b) Calculate
the electric field at P. Use back of page if necessary
Solution
function [Ef] = Loop_Ef() %(Main M-File function)
clc, clear all, close all
u.N1 = 1e21; %This is the number of states
u.NA = 1e15; %This is the number of acceptors
u.ND = 0; %This is the number of donors
u.Z1 = 281.73e19; %This term in front of the square root of the energy (this contains mass of
electron)
u.Z2 = 757.989e19; %This is the mass of hole
u.Eg = 1.12/2; %This is the energy gap
u.a = 0.0259;
%u.rat_a = 2 %scaling factor that modifies the steepness of the oxide exponential density of
states
u.es = 11.9 * 8.854187817e-14; %permittivity measured C / cm * V
u.qe = 1.602e-19; %This is the value of charge
u.psi_0_range = linspace(-0.4, 1, 160)\'; %This defines the range of Psi_0
u.display = true; %psi is in Volts Not ev?
%%
%ratN_range = 1e-2; ratS_range = [ 0.5: 0.2: 2];
ratN_range = 1; ratS_range = 1; %This doesn’t matter for Parabolic DOS
for iN = 1 : length(ratN_range)
for iS = 1 : length(ratS_range)
u.ratN = ratN_range(iN);
u.ratS = ratS_range(iS);
u.N2 = u.ratN * u.N1;
try
% Ef_(iN,iS) = Calculate_Ef(u); %Here its calculating the f_function
Ef_(iN,iS) = -0.356324 %REMOVE-TESTING
u.Ef_ = Ef_(iN,iS); %Here Ef equals
u.Q = Calculate_El(u);%Here is calculating the charge
DiffQ(u);
catch
Ef_(iN,iS) = NaN; %If there is nothing then NaN
end
end
end
clf,
plot(u.psi_0_range, u.Q) %Its plotting the graph
u.Ef_all = Ef_;
save \'u.mat\' %It saves all the data in u.file
save \'Ef_.dat\' Ef_ -ascii %Exports the data of Ef as ascii
keyboard %This command gives the control to the user
end
%================================================
function [Ef_] = Calculate_Ef(u)
clc
time = cputime;
%rescale volumes for numerical stability
u.Nscale = u.N1 ^ 2;
Nscale = u.Nscale;
u.N1 = u.N1 / Nscale;
u.N2 = u.N2 / Nscale;
u.Z1 = u.Z1 / Nscale;
u.Z2 = u.Z2 / Nscale;
%get rid of the u.\'\'
N1 = u.N1; N2 = u.N2;
Z1 = u.Z1; Z2 = u.Z2;
Eg = u.Eg;
a = u.a;
NA = u.NA/Nscale; ND = u.ND/Nscale;
P_1 = @(E) Z1.*sqrt(-Eg-E); %This is the parabolic function(Valance band)
P_2 = @(E) Z2.*sqrt(E-Eg); %This is the parabolic function (Conduction band)
Fn = @(E,Ef) 1 ./ ( exp((E-Ef)/a) ); %This is the Fermi-dirac for electrons
Fp = @(E,Ef) 1 ./ ( exp((Ef-E)/a) ); %This is the Fermi-dirac for holes
dp = @(E,Ef) P_1(E) .* Fp(E,Ef); %the integrands
dn = @(E,Ef) P_2(E) .* Fn(E,Ef);
p = @(Ef) Nscale * (IntegrateInf_p( @(E) dp(E,Ef), u ) + ND); %define our functions p & n
(Ef)
n = @(Ef) Nscale * (IntegrateInf_n( @(E) dn(E,Ef), u ) + NA);
%%we need to make sure that multiplication/division with Nscale is consistent all the way
through to DiffQ
Ef_ = Solve_Ef(-0.5, p, n); %Ef* is obtained here
fprintf(\'The value Ef*= %f, gives equal val.
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3. Use the compound interest formulas A F 1+ and A Pe to find the acc.pdf
3. Use the compound interest formulas A F 1+ and A Pe to find the accumulated value of an
investment of S12,850 for 8 years at an interest rate of 6 iftie money is... a. Compounded
Quarterly b. compounded Continuously
Solution
A=P[1+(R/N)]^(NT) P=12850 $ T=8 YRS R=6.5%=0.065 a).
3. Transportation technology (15 points) a. Explain both generally an.pdf
3. Transportation technology (15 points) a. Explain both generally and with specific examples
how changes in transportation technology have affected the development of urban areas. b.
Indicate two ways in which governmental policy has influenced the shape of urbanization in the
U.S. c. Do you believe that new telecommunication and other forms of technology will make
distance an unimportant factor in economic development and, thus, make cities unimportant
economic factors? Support your answer.
Solution
Transportation enbles activity of connecting people, businesses and resources.Transportation
improvement is necessary for the economicdevelopment and transportation policies help to
support the economic objectives.it helps to improve economic efficieny. Economic efficiency
refers to the ratio total benifit to cost. Increase economic efficieny, increase productivity which
leads to economic development.
The ultimate goal of transportation is accessabilty, access the people and the company abilities
including raw material, labor, worksites, proffessional services, distrbutors etc. Increased
accessibilty means reduction in time,money, risk require recouces reaches in less time so the
productivity of a company increases.
cost saving in an organisation---Economic efficiency more output per unit of input---productivity
more goods and services produced in less cost---Ecoonomis development.
Example- if in a company a manager has to submit an assignment cost company say 10 crores
and he has to submit it with in 2 days then the manager priority is to travel by Air not by train
etc. because the loss of the company is veyr high then the cost of transportation. In transportation
priortization is imortant point to improve productivity hence economic development.
Urbanisation means the population shifted rural to urban areas and how the economy adopt the
change.Recent development of commiucation technology in US, such as broadband services ,
package express delivery services, cheap telecommunication, courier services , easy ways to
transprotation, highways development promotes people to work in the different ares of the
country.
But now days the USA government initiating suburbanisation..
3. Use the compound interest formulas A=P(1+rn)^m and A= Pe^rt to fi.pdf
3. Use the compound interest formulas A=P(1+r/n)^m and A= Pe^rt to find the accumulated
value of an investment of $12,850 for 8 years at an interest rate of 6.5% if the money is a.
Compounded Quarterly b. Compounded Continuously
Solution
A=P[1+(R/N)]^(NT) P=12850 $ T=8 YRS R=6.5%=0.065 a).
3. The number N of bacteria present in a culture at time t, in ho.pdf
3. The number N of bacteria present in a culture at time t, in hours, obeys the law exponential
growth N(t) = 500e^0.02t a) How many bacteria are there at t = 0 hours? b) When will the
number of bacteria triple?
Solution
put t=0
N(0)= 500e0= 500
bacteria triple
N(t) = 500*3 = 1500
1500=500e0.02t
3=e0.02t
ln3=0.02t
t = ln3*50
=50ln3.
3. Suppose for f(z) holomorphic C there are constants a, b, e 0 Pr.pdf
3. Suppose for f(z) holomorphic C there are constants a, b, e > 0: Prove that f is a polynomial of
degree n less than equal to c .
Solution
A holomorphic function is a complex-valued function of one or more complex variables that is
complex differentiable in a neighborhood of every point in its domain.
Hence f(z) is holomorphic means
f\'(z) exists for all z
2x= z+z bar and 2iy = z-z bar
f(z,zbar) = =?m?0,n?0amnzmz.
3. Let On be the set of odd permutations in Sn. Is On a subgroup .pdf
3. Let On be the set of odd permutations in Sn. Is On a subgroup?
Solution
No, On is not a subgroup..
to see that the odd permutations don\'t form a subgroup of Sn is a bit easier to resolve.
Recall that any subgroup of Sn has the same identity element as the identity element in Sn. But
since the identity element in Sn is not an odd permutation (a result we saw in class), it follows
that
the collection of odd permutations cannot be a subgroup of Sn. (One could also argue that the
odd
elements aren\'t closed under the group operation.).
3. (4 points) Consider the function (in the domain {z i Solut.pdf
3. (4 points) Consider the function (in the domain {|z ? i|
Solution
Consider f(z)
It got a denominator of 1+(z-i)2
Equate to 0 the denominator to get singular points
(z-i)2 =-1
Take square root
z-i =i, -i
z=2i or 0
As |z-i|<1 we can expand f(z) by binomial series
f(z) = 1-(z-i)2+(z-i)4+...+(-1)n(z-i)2n+...
As odd powers are 0 we have
derivative of f^m(z) =0 for m = odd
For n even f\"m(z) = (-1)n(2n)(2n-1)....(1)(x2n-2n)
= (-1)n(2n)!
Proved..
3. Find currents (values and directions) in all resistors of the circ.pdf
3. Find currents (values and directions) in all resistors of the circuit given below. Find total
electric power developing in this circuit Neglect internal resistances of the batteries.
Solution
As all the Voltage sources are DC (Batteries) there by the capacitor acts as short circuit .
I1 = Current through R1 =E1-E2)/R1 = (6-3)/5 = 3/5 Amps (direction from E1 to E2)
I2= Current Throuh R3 = E2/R3 = 3/4 Amps (direction from Top to bottom)
I3 = Current Through R2 = E3-E2)/R2 = (5-3)/5 = 2/5 Amps (direction from E3 to E2)
Power delivered = (I1*E1) + (E3*I3) =(3/5)*6 + (2/5)*5 = 28/5 Watts.
3. Assume that the DS register contains 0100 and register SI contains.pdf
3. Assume that the DS register contains 0100 and register SI contains 0020. What is the result of
following instruction LES DI, [SI] Where the content of memory Location is as follows:
Solution
the command LES DI , [SI] does the following:
it takes the pointer of SI , i.e the address of the memory location SI and breaks it into lower bits
and upper bits.
then the lower bits are stored in DI and the upper bits are stored in ES.
so address of SI is 1027 .
lower bit is 27 and upper bit is 10.
27 will be stored in DI . and 10 will be saved in ES.
now DI is the lower bit of the register DS.
as it is said initially DS 0100 so, initially DI is 00.
which is changed to 27.
now DS is 0127..
2a Briefly describe the 1st law of thermodynamics. You can use either.pdf
2a Briefly describe the 1st law of thermodynamics. You can use either plain language or
equations with short explanation. Figures can also be used if necessary. (2b) Please demonstrate
that the unit of the specific kinetic energy 1/2 v^2 (where V is velocity with unit s /m) is Joule /
kg.
Solution.
3 (a) calculate thy potential at point P located a distance z above .pdf
3 (a) calculate thy potential at point P located a distance z above the center of a charged disk of
radius a and surface charge density p, Coulombs per square meter. Please show all work An
integral that you may need is The elemental area in cylindrical coordinates is dA = (b) Calculate
the electric field at P. Use back of page if necessary
Solution
function [Ef] = Loop_Ef() %(Main M-File function)
clc, clear all, close all
u.N1 = 1e21; %This is the number of states
u.NA = 1e15; %This is the number of acceptors
u.ND = 0; %This is the number of donors
u.Z1 = 281.73e19; %This term in front of the square root of the energy (this contains mass of
electron)
u.Z2 = 757.989e19; %This is the mass of hole
u.Eg = 1.12/2; %This is the energy gap
u.a = 0.0259;
%u.rat_a = 2 %scaling factor that modifies the steepness of the oxide exponential density of
states
u.es = 11.9 * 8.854187817e-14; %permittivity measured C / cm * V
u.qe = 1.602e-19; %This is the value of charge
u.psi_0_range = linspace(-0.4, 1, 160)\'; %This defines the range of Psi_0
u.display = true; %psi is in Volts Not ev?
%%
%ratN_range = 1e-2; ratS_range = [ 0.5: 0.2: 2];
ratN_range = 1; ratS_range = 1; %This doesn’t matter for Parabolic DOS
for iN = 1 : length(ratN_range)
for iS = 1 : length(ratS_range)
u.ratN = ratN_range(iN);
u.ratS = ratS_range(iS);
u.N2 = u.ratN * u.N1;
try
% Ef_(iN,iS) = Calculate_Ef(u); %Here its calculating the f_function
Ef_(iN,iS) = -0.356324 %REMOVE-TESTING
u.Ef_ = Ef_(iN,iS); %Here Ef equals
u.Q = Calculate_El(u);%Here is calculating the charge
DiffQ(u);
catch
Ef_(iN,iS) = NaN; %If there is nothing then NaN
end
end
end
clf,
plot(u.psi_0_range, u.Q) %Its plotting the graph
u.Ef_all = Ef_;
save \'u.mat\' %It saves all the data in u.file
save \'Ef_.dat\' Ef_ -ascii %Exports the data of Ef as ascii
keyboard %This command gives the control to the user
end
%================================================
function [Ef_] = Calculate_Ef(u)
clc
time = cputime;
%rescale volumes for numerical stability
u.Nscale = u.N1 ^ 2;
Nscale = u.Nscale;
u.N1 = u.N1 / Nscale;
u.N2 = u.N2 / Nscale;
u.Z1 = u.Z1 / Nscale;
u.Z2 = u.Z2 / Nscale;
%get rid of the u.\'\'
N1 = u.N1; N2 = u.N2;
Z1 = u.Z1; Z2 = u.Z2;
Eg = u.Eg;
a = u.a;
NA = u.NA/Nscale; ND = u.ND/Nscale;
P_1 = @(E) Z1.*sqrt(-Eg-E); %This is the parabolic function(Valance band)
P_2 = @(E) Z2.*sqrt(E-Eg); %This is the parabolic function (Conduction band)
Fn = @(E,Ef) 1 ./ ( exp((E-Ef)/a) ); %This is the Fermi-dirac for electrons
Fp = @(E,Ef) 1 ./ ( exp((Ef-E)/a) ); %This is the Fermi-dirac for holes
dp = @(E,Ef) P_1(E) .* Fp(E,Ef); %the integrands
dn = @(E,Ef) P_2(E) .* Fn(E,Ef);
p = @(Ef) Nscale * (IntegrateInf_p( @(E) dp(E,Ef), u ) + ND); %define our functions p & n
(Ef)
n = @(Ef) Nscale * (IntegrateInf_n( @(E) dn(E,Ef), u ) + NA);
%%we need to make sure that multiplication/division with Nscale is consistent all the way
through to DiffQ
Ef_ = Solve_Ef(-0.5, p, n); %Ef* is obtained here
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