Discuss the two group classifications that exist in Windows Server 2012 R2 describing both the
group type and group scope. Is there a reason you would use one scope over another?
Solution
A group is a collection of users and accouts that can be managed as a single unit. The 2 group
classification in Windows 2012 are
1. Distribution group
2. Security group
Distribution groups are only created to use email applications (like Exchange Server) to send
emails to a group of users. It is a feature introduced by Office 365 and allows sending emails to
large number of people both inside and outside the organization. It eliminates sending individual
emails to users. Bulk emails can be sent to a large group of users. The only purpose of
distribution group is to share information
Security groups are sets of users who have the same access rights, permission to resources and
performs same kind of tasks. There can be many security groups and a user can belong to more
than 1 security group. It facilitates administrators to assign same set of permissions to users
rather than individually assigning permissions to each one of them.
There are 3 group scopes as follows:
1. Universal scope: It can have users, accounts, groups or other universal groups from any
domain. Users can add members from any domain to an universal group and can assign
permission to access any resource located in any domain.
2. Global scope: These are containers for users and accounts which can be assigned permission
to objects in any domain. However, it can only have members from the domain in which it is
created.
3. Domain local group: These can have users and accounts from any domain as member.
However, only domain local groups can assign permissions to local resources or to resources that
reside in the domain in which the domain local group was created.
It is preferable to choose domain local group scope as it provides more security than the other 2
scopes. Also, the applied restrictions mean limited resources are utilized by the group..
Exploitation and distribution of setuid and setgid binaries on Linux systemsZero Science Lab
Abstract—In an era of internet freedom, lack of control and supervision, every system is exposed to various attackers and malicious users which, given the right circumstances, are able to cause colossal damage. A single security vulnerability can be the reason for a business’ downfall, therefore significant attention needs to be paid to said systems’ security to avoid such issues. Unix-like filesystems define certain access rights flags, named setuid and setgid, which allow users to execute files with the permissions of the file’s owner or group. This can be exploited to gain unprivileged access using buffer overflow attacks. I performed tests by running a script to collect the files in Ubuntu, Debian, Slackware, Fedora and CentOS to find the files with the setuid and setgid bits set. My aim is to determine which distribution is the most secure one and whether Slackware, considering it’s known for its’ secure design and characteristics, will prove its’ reputation. The results show that Debian and CentOS have e least amount of exploitable binaries, while Slackware and Fedora have the most.
Working with user accounts,modification,deletion and creating a group its policies and share and printer sharing over a network and windows server backup 2008
A self destruction system for dynamic group data sharing in cloudeSAT Publishing House
IJRET : International Journal of Research in Engineering and Technology is an international peer reviewed, online journal published by eSAT Publishing House for the enhancement of research in various disciplines of Engineering and Technology. The aim and scope of the journal is to provide an academic medium and an important reference for the advancement and dissemination of research results that support high-level learning, teaching and research in the fields of Engineering and Technology. We bring together Scientists, Academician, Field Engineers, Scholars and Students of related fields of Engineering and Technology.
IJRET : International Journal of Research in Engineering and Technology is an international peer reviewed, online journal published by eSAT Publishing House for the enhancement of research in various disciplines of Engineering and Technology. The aim and scope of the journal is to provide an academic medium and an important reference for the advancement and dissemination of research results that support high-level learning, teaching and research in the fields of Engineering and Technology. We bring together Scientists, Academician, Field Engineers, Scholars and Students of related fields of Engineering and Technology
Security and LDAP integration in InduSoft Web StudioAVEVA
With cybersecurity threat vectors increasing and attacks on industrial control systems on the rise, it’s more important than ever to take proper safety precautions when developing HMI or SCADA applications. In this webinar, we’ll go over how your application can be integrated with LDAP, and some best practices for developing more secure SCADA/HMI systems.
If 0.150 mL of 10^-7.5 dilution yielded an average of 256 coloniesp.pdfkesav24
If 0.150 mL of 10^-7.5 dilution yielded an average of 256 colonies/plate, calculate the number of
bacteria per mL of the original culture. Show all calculations.
Solution
The number of colony-forming units (CFU) = [no of cells]/ [volume of dilution x volume of
diluted sample added to the plate]
Now CFU per ml of dilution is 256 colonies divided by 0.150 ml plated = 1706.7 colony-
forming units/ml of dilution.
Dividing this number by the 10-7.5 dilution = 1706.7 / 10-7.5 = 1706.7 x 107.5 colony-forming
units/ml of original culture sample.
In the equation below f is a frequency, and L is a length. What are .pdfkesav24
In the equation below f is a frequency, and L is a length. What are the dimensions of a? In the
equation below f is a frequency, and L is a length. What are the dimensions of a 1/f = 2.pi.root
L/a
Solution
On the left hand side:
Dimension of frequency is T-1. So, dimension of 1/f will be T.
Now on the right hand side,
2pi is dimension less constant and L is length
To obtain the dimension of T on the right side, we have to divide the length dimension by
another length dimension and multiply by T2.
Hence, dimension of 1/a = T2/L
So, dimensions of a = L/T2 = LT-2.
More Related Content
Similar to Discuss the two group classifications that exist in Windows Server 2.pdf
Exploitation and distribution of setuid and setgid binaries on Linux systemsZero Science Lab
Abstract—In an era of internet freedom, lack of control and supervision, every system is exposed to various attackers and malicious users which, given the right circumstances, are able to cause colossal damage. A single security vulnerability can be the reason for a business’ downfall, therefore significant attention needs to be paid to said systems’ security to avoid such issues. Unix-like filesystems define certain access rights flags, named setuid and setgid, which allow users to execute files with the permissions of the file’s owner or group. This can be exploited to gain unprivileged access using buffer overflow attacks. I performed tests by running a script to collect the files in Ubuntu, Debian, Slackware, Fedora and CentOS to find the files with the setuid and setgid bits set. My aim is to determine which distribution is the most secure one and whether Slackware, considering it’s known for its’ secure design and characteristics, will prove its’ reputation. The results show that Debian and CentOS have e least amount of exploitable binaries, while Slackware and Fedora have the most.
Working with user accounts,modification,deletion and creating a group its policies and share and printer sharing over a network and windows server backup 2008
A self destruction system for dynamic group data sharing in cloudeSAT Publishing House
IJRET : International Journal of Research in Engineering and Technology is an international peer reviewed, online journal published by eSAT Publishing House for the enhancement of research in various disciplines of Engineering and Technology. The aim and scope of the journal is to provide an academic medium and an important reference for the advancement and dissemination of research results that support high-level learning, teaching and research in the fields of Engineering and Technology. We bring together Scientists, Academician, Field Engineers, Scholars and Students of related fields of Engineering and Technology.
IJRET : International Journal of Research in Engineering and Technology is an international peer reviewed, online journal published by eSAT Publishing House for the enhancement of research in various disciplines of Engineering and Technology. The aim and scope of the journal is to provide an academic medium and an important reference for the advancement and dissemination of research results that support high-level learning, teaching and research in the fields of Engineering and Technology. We bring together Scientists, Academician, Field Engineers, Scholars and Students of related fields of Engineering and Technology
Security and LDAP integration in InduSoft Web StudioAVEVA
With cybersecurity threat vectors increasing and attacks on industrial control systems on the rise, it’s more important than ever to take proper safety precautions when developing HMI or SCADA applications. In this webinar, we’ll go over how your application can be integrated with LDAP, and some best practices for developing more secure SCADA/HMI systems.
Similar to Discuss the two group classifications that exist in Windows Server 2.pdf (20)
If 0.150 mL of 10^-7.5 dilution yielded an average of 256 coloniesp.pdfkesav24
If 0.150 mL of 10^-7.5 dilution yielded an average of 256 colonies/plate, calculate the number of
bacteria per mL of the original culture. Show all calculations.
Solution
The number of colony-forming units (CFU) = [no of cells]/ [volume of dilution x volume of
diluted sample added to the plate]
Now CFU per ml of dilution is 256 colonies divided by 0.150 ml plated = 1706.7 colony-
forming units/ml of dilution.
Dividing this number by the 10-7.5 dilution = 1706.7 / 10-7.5 = 1706.7 x 107.5 colony-forming
units/ml of original culture sample.
In the equation below f is a frequency, and L is a length. What are .pdfkesav24
In the equation below f is a frequency, and L is a length. What are the dimensions of a? In the
equation below f is a frequency, and L is a length. What are the dimensions of a 1/f = 2.pi.root
L/a
Solution
On the left hand side:
Dimension of frequency is T-1. So, dimension of 1/f will be T.
Now on the right hand side,
2pi is dimension less constant and L is length
To obtain the dimension of T on the right side, we have to divide the length dimension by
another length dimension and multiply by T2.
Hence, dimension of 1/a = T2/L
So, dimensions of a = L/T2 = LT-2.
how are the axial and appendicular skeletons attachedSolutionA.pdfkesav24
how are the axial and appendicular skeletons attached
Solution
Answer : Axial and appendicular skeleton are the two principal kinds of bones found in the adult
humans in which both divisions join to form a complete skeleton. Axial skeleton consists of the
bones that lie around the longitudinal axis of the human body. Examples of axial bons are skull
bones, ear bones, ribs, sterum and bones of vertebral column. Appendicular skeleton consists of
the bones of the upper and the lower limbs. Appendicular bones also consists of bones forming
the girdles that connects the limbs to the axial skeleton.
The pectoral girdle which is also called as shoulder girdle connects the upper limbs to the body
where as the pelvic girdle/Hip girdle of the bony pelvis connects the bones of the lower limbs to
the axial skeleton.
How many base pairs of DNA represent the haploid human genome How m.pdfkesav24
How many base pairs of DNA represent the haploid human genome? How many base pairs of
DNA represent the haploid chimpanzee genome? Assume the similarity between human and
chimpanzee genomes is 97.5%. Based on 97.5% similarity how many base pairs do the
chimpanzee and human genomes differ by?
Solution
3 billion DNA base pairs represent haploid genome
there are 24 base pairs of DNA in the haploid chimpanzee genome.
humans and chimpanzee differ by 1 base pair .The humans have 23 base pairs and chimpanzee
have 24.
From a group of cards that has 13 hearts, 6 spades, 2 diamonds and 4.pdfkesav24
From a group of cards that has 13 hearts, 6 spades, 2 diamonds and 4 clubs, the number of ways
that all four dealt cards could be from different suits is
Solution
there will be total 25 cards and it will be a problem without replacement type
therefore getting 4 card of different suit will be
the total ways in wich this can ve done is
13*6*2*4 ( as there can be 1 card from13 cards at 1st and similarly 6,2,4 cards at respective
times)
= 624 ways.
Find and classify all singularities of the gamma and riemann zeta fu.pdfkesav24
Find and classify all singularities of the gamma and riemann zeta function
Determine the residues at any isolated singularities. Hence find the value of
please answer this with step by step information or in clear image
Solution
sin(z)(z)(z) = i 2 e iz Z C w z1 ew 1 dw. On the other hand, we can calculate using the residue
theorem the integral along the path drawn in the following figure. There the outside square
passes thru points (2m + 1)i and (2m + 1)i. -1 2i -2i 2mi 2(m+1)i -2mi -2(m+1)i NOTES ON
RIEMANN’S ZETA FUNCTION 9 We have Z w z1 ew 1 dw = 2iXm n=1 (2n) z1 e i (z1) 2 +
(2n) z1 e 3i (z1) 2 = 2z z iei(z1) e i (z1) 2 + e i (z1) 2 Xm n=1 n z1 = 2 z+1 z ieiz cos (z 1) 2 Xm
n=1 1 n1z = 2 z+1 z ieiz sin z 2 Xm n=1 1 n1z . Now we want to estimate the integral along the
sides of the square for Re z < 0. For a fixed z, we see as before that there exists M > 0, such that
on the right vertical side of the square the integrand satisfies the estimate w z1 ew 1 M |w| Re
z1 |ew 1| . Moreover, we have |e w 1| ||e w| 1| = |e Re w 1| for any w, hence we have w z1 ew
1 M |w| Re z1 |eRe w 1| M | Re w| Re z1 |eRe w 1| and the right side of the square. This
expression clearly tends to zero faster than 1 Re w as Re w +. Hence, the integral over the right
side tends to zero as the square grows. On the left side of the square we have the same estimate
w z1 ew 1 M | Re w| Re z1 |eRe w 1| . Since Re w is negative in this case, for large | Re w| we
have w z1 ew 1 2M| Re w| Re z1 . Since Re z < 0, this bound goes to zero faster that 1 | Re w| .
On the other hand, the length of the side of the square is 2| Re w|. Hence, the integral along the
left side also tends to zero as the square grows. It remains to treat the top and bottom side. As
before, we see that on these sides we have w z1 ew 1 M |w| Re z1 |ew 1| M |Im w| Re z1 |ew
1| . On the other hand, the function w 7 |e w 1| on the horizontal lines tends to as Re w + and to
1 as Re w . Hence it is bounded from below. Moreover, since it is periodic with period 2i,
shifting the line up and down by 2i doesn’t change that bound. Hence, in our estimates we can
assume that 1 |ew 1| C 10 D. MILICI C´ on top and bottom sides of all squares. This implies
that w z1 ew 1 MC|Im w| Re z1 on top and bottom sides of all squares. Since Re z < 0 and the
length of these sides is 2|Im w|, we see that the integrals over these sides tend to zero as the
squares grow. Hence, for Re z < 0, as the squares grow, i.e., as m tends to infinity, the integral
over the sides of the square tends to 0 and the integral along converges to the integral along the
path C from 2.2.2. In addition, the sum on the right side of the above expression converges to the
series for (1 z), i.e., we have 2ieiz sin(z)(z)(z) = 2 z+1 z ieiz sin z 2 (z 1). Hence, we have
sin(z)(z)(z) = 2z z sin z 2 (1 z). This yields 2 sin z 2 cos z 2 (z)(z) = 2z z sin z 2 (1 z). and cos z
2 (z)(z) = 2z1 z (1 z). for all z C. By substituting 1 z for z we get 2 z 1z (z) = cos (1 z) .
Describe the differences between allelic heterogeneity and locus het.pdfkesav24
Describe the differences between allelic heterogeneity and locus heterogeneity
Solution
Ans.) The differences between allelic heterogeneity and locus heterogeneity
In general terms, allelic heterogeneity defines the different mutation in the same gene locus that
displays different phenotype. The variation occur due to natural selection as a consequence of
genetic drift, mutation etc.
The genetic or locus heterogeneity is the single trait that caused due to mutation in different gene
loci. Only a single mutant locus is needed to display the phenotype.
Allelic heterogeneity is the mutation in same loci that showed different phenotypic character
whereas locus heterogeneity is the only one trait occurs due to mutation in different loci..
Decision Tree Table for choosing present and future value equations. .pdfkesav24
Decision Tree Table for choosing present and future value equations. ltem to Number of Present
or Annual or Perpetual or evaluate occurrences Future Equation to use Periodi Terminal Series
na na 0 value payment:s Future Present na na Periodic Annual Periodic Vn Vo(1+i)\"
Vo=Vn/(1+i)\" Vn=pl(1+i)n-1)/[(1+1)t-1)] Single Revenue Future Terminal Perpetual Vo=
p/[(1+i)\'-1] Series of Equal payments Present Terminal Vo (1-1(1+)-1](1+i)) Cost PerpetualVo
Annual Terminal Vo=p[(1+i)n-1)/[i(1+i)\") interest rate per year (in decimals) Vo = initial value
(in time 0) V- value in time n n total number of years of compounding or discounting considered
p = amount of payment occurring every t years t number of years in one multiyear group or
interval of years (length of the interval) Payments occur at the end of the period or multiyear
interval Please note: these equations can be used for months or weeks or any other length of time
as long as the interest rate is defined at that period, n is defined as total number of periods, and t
is the number of periods in one multiperiod group or interval. In this class, we will stick
primarily to using the annual rates. Source: Figure 7.3 from Davis, et al. 2001 Forest
Management, p. 341
Solution
a. As seen from the table,
Present value of perpetual periodic CF is = P/[(1+i)^t - 1]= 50000/[1.07^45 - 1] = 50000/(21.002
- 1)
= $2,499.69
b. Present value of a perpetual $100 tax = P/i = 100/0.07 = $1,428.57
Present value of the cash flow = 2,499.69 - 1,428.57 = $1,071.12
c. AS the asking price of the land is $1000 and is less than the NPV of $1,071.12 you should buy
it..
Consider this experiment I have 20-sided die where each face has a .pdfkesav24
Consider this experiment: I have 20-sided die where each face has a number 1 through 20. I roll
the die one time and record the number that comes up on the top face.
What is the sample space of this experiment?
Solution
The sample space is all the possible outcomes of the roll, that is:
S:{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}.
congenital malformations. why do you think they occurSolution.pdfkesav24
congenital malformations. why do you think they occur?
Solution
Congenital malformations
It is a morphological defect of an organ, part of an organ or a larger region of the body resulting
from an intrinsically abnormal developmental process. Structural, functional, metabolic or
behavioral disorders those are present at birth
The causes of congenital malformations divided into five broad groups.
Considering the cultures used to inoculate each medium in this labora.pdfkesav24
Considering the cultures used to inoculate each medium in this laboratory, how many different
microbial types should you expect to see in/on each medium? Which medium was most difficult
for you to transfer from? which medium was most difficult for you to inoculate? Explain your
difficulties. What is the purpose of flaming in the aseptic techniques? Provide 3 (three) reasons
why the use of aseptic techniques is essential when handling microbial cultures. Methoos of
isolation What is a microbial colony? Can a pure culture by prepared from a mixd broth or
mixed agar slant? Explain.
Solution
Answers are :
When the cultures are inoculated onto the surface of the medium, different colonies are
obtained, based on the mother culture we have chosen colonies are obtained. Until unless it is not
contaminated so.
Preparation of agar slants is difficult to transfer, and inoculate, because it needs to place
correct angle while solidifying, less surface area. Though among the medium we can say some
difficult but not so rather difficult. Nutrient broth is also difficult to transfer because more prone
of contamination and determination of the bacterial growth is also difficult.
Purpose of flaming during aseptic transfer :- Flaming causes direct burning of the cultures
samples which stick onto the inoculating loop to avoid from contamination, which technique is
called Incineration i.e., Flame sterilization.
Aseptic techniques are safe transfer of culture onto the fresh medium this is very essential
because :-
To avoid contamination to grow unwanted cultures and spoiling the medium components.
To know the perfect study of the unknown characteristics, contamination shouldn’t be seen .
To avoid infecting the person who handling the sample, some may cause severe infections and
death of the investigator.
To obtain the pure culture techniques to study the characteristics of the cultures.
Microbial colony means, group of the similar cells which shows same characteristic features ,
which obtained on the surface of the solid agar medium. These are varied in shapes and surface
but shows same morphological and physiological characteristics.
Yes, we can prepare pure culture from mixed sample either broth or solid agar, in case of
broth prepare serial dilution of the sample to obtain the pure cultures. In case of the solid slant
prepare quadrant streaking on the fresh agar plate..
Are the transfermations linear Are they an isomorphism WhySol.pdfkesav24
Are the transfermations linear? Are they an isomorphism? Why?
Solution
1. T(f)=f\'
T(f+g)=(f+g)\'=f\'+g\'
T(cf)=(cf)\'=cf\'
So, T is linear
P2 is space of polynomials of degree atmost 2 so T(P2) can have polynomials of degree atmost 1
So it cannot be an isomorphism because P2 has dimension 2 and T(P2) has dimension atmost 1
2.
T(f+g)=(f+g)\'+(f+g)\'\'=f\'+f\'\'+g\'+g\'\'=T(f)+T(g)
T(cf)=(cf)\'+(cf)\'\'=cf\'+cf\'\'=cT(f)
HEnce, T is linear
But T is not an isomorphism
As T maps a polynomial to a polynomial of degree stricly smaller than itself.
HEnce P3 and T(P3) have different dimension so T cannot be an isomorphism..
bessavilla
bessavilla
bessavilla
bessavilla
Solution
Besavilla Review Center, formerly called Besavilla Engineering Review Center was the oldest
review center (32 years) for civil engineering up to now. It was founded by Engr. Venancio I.
Besavilla, Jr. at Cebu City on 1980 and expanded its operations in Manila and other key cities in
the country (Baguio, Cagayan de Oro, General Santos City and Davao)
Besavilla Review Center offer comprehensive review courses to Civil and Geodetic Engineering
Students using innovative and updated teaching technique. It is an intensive lecture/discussion
type covering basic theories, principles, problems and applications..
ciny Univer Laboratory es and crossing over assortment of independent.pdfkesav24
ciny Univer Laboratory es and crossing over assortment of independent the processes of Explain
Meiosis. 3 Compare and contrast meiosis with mitosis. Number of times Number of cell
divisions Number of daughter cells Genetic relationship of daughter cells to What structures
separate during anaphase (Anaphase I in Meiosis)? Name the function(s) each process performs
4. Describe the events ofspermatogenesis and oogenesis. s. Define the followin Study suggestion
1. Explain
Solution
Answer for Question 3.
A population of rabbits quadruples every 2 years. If the initial num.pdfkesav24
A population of rabbits quadruples every 2 years. If the initial number of rabbits was 25, how
long will it take for the population to reach 700??
I believe I understand how to solve the logarithm, I just don\'t understand how to extract the log
it from the question.
Solution
The population of rabbits quadruples every 2 years. Let the number of years taken to reach 700
be T. The initial population is 25. This gives an equation:
700 = 25*(4)^(T/2)
=> 700/25 = 4^(T/2)
=> 28 = 4^(T/2)
take the log of both the sides
=> log 28 = (T/2)*log 4
=> T/2 = log 28/log 4
=> T = 2*(log 28/log 4)
=> T = 4.807 years.
The population of rabbits grows to 700 from 25 in 4.807 years..
7. describe and correct the error. Describe and correct the error. .pdfkesav24
7. describe and correct the error. Describe and correct the error. (-2)^2/43 middot 2^5 = (-
2)^2/(2^2)^3 middot 2^5 (-2)^2/2^6 middot 2^5 =(-2)^2/2^11 = 1/2^2 middot 2^11 1/2^13
Solution
the error is at step 1/2^2 .2^11
since (-2)^2 is not equal to 1/2^2
so correcting (-2)^2/2^11 = 2^2/2^11 = 1/2^9
ANSWER: 1/2^9.
3. Where would you expect to find fenestrated capillaries (More tha.pdfkesav24
3. Where would you expect to find fenestrated capillaries? (More than one answer may be
correct)
1. Thyroid
2. Skeletal muscle
3. Cardiac muscle
4. Spleen
Solution
Answer : The correct answer is 1 ( Thyroid glands)
Explanation : The other location where fenestrated capillaries found are kidneys, villi of small
instetine, choroid plexus of the ventricles of the brain and in some endocrine glands..
Assembly LanguageReverse an array.Use a loop with indirect or i.pdfkesav24
Assembly Language:
Reverse an array.
Use a loop with indirect or indexed addressing to reverse the elements of an integer array in
place. Do not copy the elements to any other array. Do not use stack operations. Use the
SIZEOF, TYPE, LENGTHOF operators.
Fibonacci Numbers.
Write a code that fills the array of 15 elements with consecutive Fibonacci numbers: Fib(1) =
Fib(2) = 1; Fib (n) = Fib(n-1) + Fib(n-2) for each N > 2.
corrent code:
Solution
data barray BYTE 6 DUP (?) dwarray DWORD 3 DUP (?) TOTAL_LENGTH = $-barray
bdmessage BYTE \"Today as stored in a double word\", 0dh, 0ah, 0 dwmessage BYTE 0dh,
0ah, \"Seeing bytes in a single word\", 0dh, 0ah, 0 lengthmessage BYTE 0dh, 0ah, \"total bytes
used: \", 0 finalmessage BYTE 0dh, 0ah, \"Press any key to continue ...\", 0 .code main PROC
CALL Crlf MOV edx, OFFSET bdmessage CALL WriteString MOV eax, DWORD
PTR [barray + 2] CALL WriteDec MOV edx, OFFSET dwmessage CALL WriteString
MOVZX eax, WORD PTR [dwarray + 6] CALL WriteDec MOV edx, OFFSET
lengthmessage CALL WriteString MOV eax, TOTAL_LENGTH CALL WriteDec
MOV edx, OFFSET finalmessage CALL WriteString CALL ReadChar INVOKE
ExitProcess, 0 main ENDP END main.
wo alternative investment proposals are under consideration for a va.pdfkesav24
wo alternative investment proposals are under consideration for a vacant owner by Urban
Development Corporation. Plan A would require an immediate investment of $120,000 and first-
year expenditure for property taxes, maintenance, and insurance of $4,000, with this amount
expected to increase at a rate of $1,000 per year. Plan B would have a first cost of $170,000 and
total first-year expenses of $9,000, with an increase of $1,000 per year. The economic life of
each project is forecast to be 10 years; and at the end of this time, only the facilities from Plan B
with a value of $50,000 are expected to salvage. During the life of the project, the facility in plan
A is expected to produce $34,000 annually, whereas Plan B is expected to produce $42,000 . a)
Determine the rate of return of each plan. b) Determine the rate of return of the Additional
investment required in Plan B compared with Plan A. c) Which plan should Urban Development
select if the company uses a MARR of 12 percent?
Solution
Working notes:
(1) For each plan,
Net Annual Benefit = Annual Benefit - Annual Expense
(2) For plan B, Net Annual Benefit in terminal year (year 10) is higher by $50,000 (Salvage
Value).
(3) Rate of Return is the Internal Rate of Return (IRR) computed using Excel function
=IRR(Value 1, Value 2,.....Value n) where \"Value\" is the Net Annual Benefit, undiscounted.
(4) Based on MARR of 12%, the plan choice will be made based on NPV of the plans. NPV is
the sum of all cash outflows and inflows discounted at 12%. The plan with higher positive NPV
should be chosen.
Calculations as follow:
Plan A
Year
Initial Cost ($)
Annual Benefit ($)
Annual Cost ($)
Net Annual Benefit ($)
Net Cash Flow ($)
Discount Factor @12%
Discounted Net Annual Benefit ($)
(A)
(B)
(C)
(D) = (B) - (C)
(E) = (D) + (A)
(F)
(G) = (E) x (F)
0
-1,20,000
-1,20,000
1.0000
-1,20,000
1
34,000
4,000
30,000
30,000
0.8929
26,786
2
34,000
5,000
29,000
29,000
0.7972
23,119
3
34,000
6,000
28,000
28,000
0.7118
19,930
4
34,000
7,000
27,000
27,000
0.6355
17,159
5
34,000
8,000
26,000
26,000
0.5674
14,753
6
34,000
9,000
25,000
25,000
0.5066
12,666
7
34,000
10,000
24,000
24,000
0.4523
10,856
8
34,000
11,000
23,000
23,000
0.4039
9,289
9
34,000
12,000
22,000
22,000
0.3606
7,933
10
34,000
13,000
21,000
21,000
0.3220
6,761
IRR (%) =
18.12%
NPV ($) =
29,253
Plan B
Year
Initial Cost ($)
Annual Benefit ($)
Annual Cost ($)
Net Annual Benefit ($)
Net Cash Flow ($)
Discount Factor @12%
Discounted Net Annual Benefit ($)
(A)
(B)
(C)
(D) = (B) - (C)
(E) = (D) + (A)
(F)
(G) = (E) x (F)
0
-1,70,000
-1,70,000
1.0000
-1,70,000
1
42,000
9,000
33,000
33,000
0.8929
29,464
2
42,000
10,000
32,000
32,000
0.7972
25,510
3
42,000
11,000
31,000
31,000
0.7118
22,065
4
42,000
12,000
30,000
30,000
0.6355
19,066
5
42,000
13,000
29,000
29,000
0.5674
16,455
6
42,000
14,000
28,000
28,000
0.5066
14,186
7
42,000
15,000
27,000
27,000
0.4523
12,213
8
42,000
16,000
26,000
26,000
0.4039
10,501
9
42,000
17,000
25,000
25,000
0.3606
9,015
10
92,000
18,.
You are searching for new species that are highly tolerant to acidic.pdfkesav24
You are searching for new species that are highly tolerant to acidic conditions and find a new
single celled organism that you call WERID. The organism is ~5 um, does not have a cell wall,
but your portable field microscope is broken, and you are unable to look at the internal
structures. You do have portable equipment that will allow you to analyze the chemical content
of the WEIRD cells along with control prokaryotic and eukaryotic cells. Out of protein, DNA,
and lipid, which analysis would be most useful if you were interested in figuring out if your cells
are prokaryotic or eukaryotic? Explain your answer.
Solution
The organism that we have discovered is single celled, and devoid of a cell wall. It is ~5
micrometers in length. Since the organism has no cell wall, we cannot reach a conclusion based
or the presence/absence of peptidoglycan. Also, morphological analysis cannot be carried out
since the microscope is broken.
However, we do have portable equipment to analyze the chemical content of the cells: protein,
DNA and lipid, and compare it with control prokaryotic and eukaryotic cells. The protein and
lipid type and content may vary between prokaryotes and eukaryotes.
However, the DNA has certain differences that are absolutely unique to prokaryotes and
eukaryotes. For example, the ribosomes-part of the protein synthesis machinery differ between
the prokaryotes and eukaryotes in that the prokaryotes have a 16S rRNA while the eukaryotes
have an 18S rRNA. This difference can be exploited using our portable equipment to check for
presence of genes in the WEIRD DNA that code for either of these rRNAs, and comparing them
with the controls. Also, eukaryotes usually have DNA in the form of multiple chromosomes,
while the prokaryotes have a single circular DNA molecule.
The above differences are irrefutable, and unique to prokaryotes and eukaryotes. Therefore, our
choice for identification of the WEIRD cells would be DNA analysis.
P.S. It is also important to take into account that eukaryotes are usually multicellular, while
prokaryotes are unicellular. Also, eukaryotic cells are generally larger in size than that of the
prokaryotes..
How to Make a Field invisible in Odoo 17Celine George
It is possible to hide or invisible some fields in odoo. Commonly using “invisible” attribute in the field definition to invisible the fields. This slide will show how to make a field invisible in odoo 17.
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...Levi Shapiro
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
A Strategic Approach: GenAI in EducationPeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
Model Attribute Check Company Auto PropertyCeline George
In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
Francesca Gottschalk - How can education support child empowerment.pptxEduSkills OECD
Francesca Gottschalk from the OECD’s Centre for Educational Research and Innovation presents at the Ask an Expert Webinar: How can education support child empowerment?
Embracing GenAI - A Strategic ImperativePeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
Introduction to AI for Nonprofits with Tapp NetworkTechSoup
Dive into the world of AI! Experts Jon Hill and Tareq Monaur will guide you through AI's role in enhancing nonprofit websites and basic marketing strategies, making it easy to understand and apply.
The Roman Empire A Historical Colossus.pdfkaushalkr1407
The Roman Empire, a vast and enduring power, stands as one of history's most remarkable civilizations, leaving an indelible imprint on the world. It emerged from the Roman Republic, transitioning into an imperial powerhouse under the leadership of Augustus Caesar in 27 BCE. This transformation marked the beginning of an era defined by unprecedented territorial expansion, architectural marvels, and profound cultural influence.
The empire's roots lie in the city of Rome, founded, according to legend, by Romulus in 753 BCE. Over centuries, Rome evolved from a small settlement to a formidable republic, characterized by a complex political system with elected officials and checks on power. However, internal strife, class conflicts, and military ambitions paved the way for the end of the Republic. Julius Caesar’s dictatorship and subsequent assassination in 44 BCE created a power vacuum, leading to a civil war. Octavian, later Augustus, emerged victorious, heralding the Roman Empire’s birth.
Under Augustus, the empire experienced the Pax Romana, a 200-year period of relative peace and stability. Augustus reformed the military, established efficient administrative systems, and initiated grand construction projects. The empire's borders expanded, encompassing territories from Britain to Egypt and from Spain to the Euphrates. Roman legions, renowned for their discipline and engineering prowess, secured and maintained these vast territories, building roads, fortifications, and cities that facilitated control and integration.
The Roman Empire’s society was hierarchical, with a rigid class system. At the top were the patricians, wealthy elites who held significant political power. Below them were the plebeians, free citizens with limited political influence, and the vast numbers of slaves who formed the backbone of the economy. The family unit was central, governed by the paterfamilias, the male head who held absolute authority.
Culturally, the Romans were eclectic, absorbing and adapting elements from the civilizations they encountered, particularly the Greeks. Roman art, literature, and philosophy reflected this synthesis, creating a rich cultural tapestry. Latin, the Roman language, became the lingua franca of the Western world, influencing numerous modern languages.
Roman architecture and engineering achievements were monumental. They perfected the arch, vault, and dome, constructing enduring structures like the Colosseum, Pantheon, and aqueducts. These engineering marvels not only showcased Roman ingenuity but also served practical purposes, from public entertainment to water supply.
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
Discuss the two group classifications that exist in Windows Server 2.pdf
1. Discuss the two group classifications that exist in Windows Server 2012 R2 describing both the
group type and group scope. Is there a reason you would use one scope over another?
Solution
A group is a collection of users and accouts that can be managed as a single unit. The 2 group
classification in Windows 2012 are
1. Distribution group
2. Security group
Distribution groups are only created to use email applications (like Exchange Server) to send
emails to a group of users. It is a feature introduced by Office 365 and allows sending emails to
large number of people both inside and outside the organization. It eliminates sending individual
emails to users. Bulk emails can be sent to a large group of users. The only purpose of
distribution group is to share information
Security groups are sets of users who have the same access rights, permission to resources and
performs same kind of tasks. There can be many security groups and a user can belong to more
than 1 security group. It facilitates administrators to assign same set of permissions to users
rather than individually assigning permissions to each one of them.
There are 3 group scopes as follows:
1. Universal scope: It can have users, accounts, groups or other universal groups from any
domain. Users can add members from any domain to an universal group and can assign
permission to access any resource located in any domain.
2. Global scope: These are containers for users and accounts which can be assigned permission
to objects in any domain. However, it can only have members from the domain in which it is
created.
3. Domain local group: These can have users and accounts from any domain as member.
However, only domain local groups can assign permissions to local resources or to resources that
reside in the domain in which the domain local group was created.
It is preferable to choose domain local group scope as it provides more security than the other 2
scopes. Also, the applied restrictions mean limited resources are utilized by the group.
