show all work. A disc jockey has 10 songs to play. Seven are slow so.pdf
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show all work. A disc jockey has 10 songs to play. Seven are slow songs, and three are fast songs. Each songs is to be played only once. In how many ways can the disc jockey play the 10 songs if : The songs can be played in any order, The first song must be a slow song and the last song must be a slow song, The first two songs must be fast songs Solution 11! 9! 9!.
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SHOW ALL WORK.Apple reported net income of 1,000,000. In 2014, App.pdf
SHOW ALL WORK.
Apple reported net income of 1,000,000. In 2014, Apple had 10,000 shares of $100 par 6% non-
convertible, non cumulative preferred stock outstanding throughout 2014. In 2014, Apple
declared and paid $50,000 preferred dividends. Common stock activity for 2014 was follows
1/1= 100,000 shares of outstanding.
4/1 = 20,000 additional shared issue.
8/1 = 2 for 1 stock split
12/1 = 12,000 additional share issued.
Tax rate 30%
Compute basic: Earn Per Share (EPS)
Also, In which financial statement would EPS presented?
Solution.
Show full steps. sec 3x dxSolution Integrate.pdf
Show full steps. sec 3x dx
Solution
Integrate by parts: u = sec(x) dv = sec²(x) dx du = sec(x)*tan(x) dx v = tan(x) K =
sec(x)*tan(x) - ?sec(x)*tan²(x) dx Recall the identity 1 + tan²(x) = sec²(x). K = sec(x)*tan(x) -
?sec³(x) dx + ?sec(x) dx K = sec(x)*tan(x) - K + ln|sec(x) + tan(x)| + C K = ?sec³(x) dx =
[sec(x)*tan(x) + ln|sec(x) + tan(x)|]/2 + C.
Show that the product of 114 by 1 + 12 + 14 is equal to 18 (prob.pdf
Show that the product of 1/14 by 1 + 1/2 + 1/4 is equal to 1/8 (problem 12 of the Rhind Papyrus)
(hint have to use fractions table and doubling )
Solution
first calculate 1+1/2+1/4 = 4/4 + 2/4 +1/4 = 7/4 Now multiply 1/14 by 7/4 = 1/8
(1/14) * (7/4) = 7/56 = 1/8 please, rate. thank you..
Show that the following intervals of real numbers are equivalent;tha.pdf
Show that the following intervals of real numbers are equivalent;that is, there exists a 1-1
correspondence between them.
(0, 1) and (4, )
(0, 1) and (4, )
Solution
Consider the 1-1 correspondence function from (0,1)->(4,) f(x)=4/x.
Show that the composition of two single-rooted relations is single-r.pdf
Show that the composition of two single-rooted relations is single-rooted. Conclude that the
composition of two one-to-one functions is one-to-one
Solution
Sorry, but I do not understand the problem. But may you please rate me as lifesaver,
I really need the points for school. Thank you very much..
Show that the closure of S is closed.Solution .pdf
Show that the closure of S is closed.
Solution
Let T be the complement of the closure. For any point t in T, thereis a ball around t
that does not intersect S. (Else t would be inthe closure, not its complement.) Hence the ball is in
T. We just showed T is open. Hence its complement is closed..
Show that if z is an n-th root of one, then so is 1zSol.pdf
Show that if \"z\" is an \"n-th\" root of one, then so is 1/z
Solution
we know that when z is a complex number and is a root of an equation which has natural
coefficients , then the complex roots always exist in conjugate pairs , so if Z is a root , then 1/Z
should also be a root of the given equation i.e n-th root of 1 ..
Show that if A is a fixed event of positive probability, then the fu.pdf
Show that if A is a fixed event of positive probability, then the function Q[B]=P[B|A] taking
events B into R satisfies the three defining axioms of probability.
Here is the three defining axioms of probability.
A probability measure P is a function taking the family of events H to the real numbers such that
(i) P[Pi]=1
(ii) For all A includes in H, P[A] 0.
(iii) If A1, A2,.....is a sequence of pairwise disjoint events then
P[A1 U A2 U.....]=P[Ai]
Solution
say A1 , A2 , . . ., are called mutually disjoint or pairwise disjoint if Ai n A j = 0 for
any two of the events Ai and A j ; that is, no two of the events overlap. According to
Kolmogorov’s axioms, each event A has a probability P(A), which is a number. These numbers
satisfy three axioms: Axiom 1: For any event A, we have P(A) = 0. Axiom 2: P(S ) = 1. 4
Axiom 3: If the events A1 , A2 , . . . are pairwise disjoint, then CHAPTER 1. BASIC IDEAS
P(A1 ? A2 ? · · ·) = P(A1 ) + P(A2 ) + · · · Note that in Axiom 3, we have the union of events
and the sum of numbers. Don’t mix these up; never write P(A1 ) ? P(A2 ), for example.
Sometimes we sep- arate Axiom 3 into two parts: Axiom 3a if there are only ?nitely many events
A1 , A2 , . . . , An , so that we have P(A1 ? · · · ? An ) = ? P(Ai ), n i=1 and Axiom 3b for
in?nitely many. We will only use Axiom 3a, but 3b is important later on. Notice that we write n
? P(Ai) i=1 for P(A1 ) + P(A2 ) + · · · + P(An ). 1.4 Proving things from the axioms You can
prove simple properties of probability from the axioms. That means, every step must be justi?ed
by appealing to an axiom. These properties seem obvious, just as obvious as the axioms; but the
point of this game is that we assume only the axioms, and build everything else from that. Here
are some examples of things proved from the axioms. There is really no difference between a
theorem, a proposition, and a corollary; they all have to be proved. Usually, a theorem is a big,
important statement; a proposition a rather smaller statement; and a corollary is something that
follows quite easily from a theorem or proposition that came before. Proposition 1.1 If the event
A contains only a ?nite number of outcomes, say A = {a1 , a2 , . . . , an }, then P(A) = P(a1 ) +
P(a2 ) + · · · + P(an ). To prove the proposition, we de?ne a new event Ai containing only the
out- come ai , that is, Ai = {ai }, for i = 1, . . . , n. Then A1 , . . . , An are mutually disjoint 1.4.
PROVING THINGS FROM THE AXIOMS (each contains only one element which is in none
of the others), and A1 ? A2 ? · · · ? An = A; so by Axiom 3a, we have P(A) = P(a1 ) + P(a2 ) + · ·
· + P(an ). Corollary 1.2 If the sample space S is ?nite, say S = {a1 , . . . , an }, then P(a1 ) +
P(a2 ) + · · · + P(an ) = 1. For P(a1 ) + P(a2 ) + · · · + P(an ) = P(S ) by Proposition 1.1, and P(S )
= 1 by Axiom 2. Notice that once we have proved something, we can use it on the same basis as
an axiom to prove further facts. Now we see that, if all the n outcomes are equa.
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SHOW ALL WORK.Apple reported net income of 1,000,000. In 2014, App.pdf
SHOW ALL WORK.
Apple reported net income of 1,000,000. In 2014, Apple had 10,000 shares of $100 par 6% non-
convertible, non cumulative preferred stock outstanding throughout 2014. In 2014, Apple
declared and paid $50,000 preferred dividends. Common stock activity for 2014 was follows
1/1= 100,000 shares of outstanding.
4/1 = 20,000 additional shared issue.
8/1 = 2 for 1 stock split
12/1 = 12,000 additional share issued.
Tax rate 30%
Compute basic: Earn Per Share (EPS)
Also, In which financial statement would EPS presented?
Solution.
Show full steps. sec 3x dxSolution Integrate.pdf
Show full steps. sec 3x dx
Solution
Integrate by parts: u = sec(x) dv = sec²(x) dx du = sec(x)*tan(x) dx v = tan(x) K =
sec(x)*tan(x) - ?sec(x)*tan²(x) dx Recall the identity 1 + tan²(x) = sec²(x). K = sec(x)*tan(x) -
?sec³(x) dx + ?sec(x) dx K = sec(x)*tan(x) - K + ln|sec(x) + tan(x)| + C K = ?sec³(x) dx =
[sec(x)*tan(x) + ln|sec(x) + tan(x)|]/2 + C.
Show that the product of 114 by 1 + 12 + 14 is equal to 18 (prob.pdf
Show that the product of 1/14 by 1 + 1/2 + 1/4 is equal to 1/8 (problem 12 of the Rhind Papyrus)
(hint have to use fractions table and doubling )
Solution
first calculate 1+1/2+1/4 = 4/4 + 2/4 +1/4 = 7/4 Now multiply 1/14 by 7/4 = 1/8
(1/14) * (7/4) = 7/56 = 1/8 please, rate. thank you..
Show that the following intervals of real numbers are equivalent;tha.pdf
Show that the following intervals of real numbers are equivalent;that is, there exists a 1-1
correspondence between them.
(0, 1) and (4, )
(0, 1) and (4, )
Solution
Consider the 1-1 correspondence function from (0,1)->(4,) f(x)=4/x.
Show that the composition of two single-rooted relations is single-r.pdf
Show that the composition of two single-rooted relations is single-rooted. Conclude that the
composition of two one-to-one functions is one-to-one
Solution
Sorry, but I do not understand the problem. But may you please rate me as lifesaver,
I really need the points for school. Thank you very much..
Show that the closure of S is closed.Solution .pdf
Show that the closure of S is closed.
Solution
Let T be the complement of the closure. For any point t in T, thereis a ball around t
that does not intersect S. (Else t would be inthe closure, not its complement.) Hence the ball is in
T. We just showed T is open. Hence its complement is closed..
Show that if z is an n-th root of one, then so is 1zSol.pdf
Show that if \"z\" is an \"n-th\" root of one, then so is 1/z
Solution
we know that when z is a complex number and is a root of an equation which has natural
coefficients , then the complex roots always exist in conjugate pairs , so if Z is a root , then 1/Z
should also be a root of the given equation i.e n-th root of 1 ..
Show that if A is a fixed event of positive probability, then the fu.pdf
Show that if A is a fixed event of positive probability, then the function Q[B]=P[B|A] taking
events B into R satisfies the three defining axioms of probability.
Here is the three defining axioms of probability.
A probability measure P is a function taking the family of events H to the real numbers such that
(i) P[Pi]=1
(ii) For all A includes in H, P[A] 0.
(iii) If A1, A2,.....is a sequence of pairwise disjoint events then
P[A1 U A2 U.....]=P[Ai]
Solution
say A1 , A2 , . . ., are called mutually disjoint or pairwise disjoint if Ai n A j = 0 for
any two of the events Ai and A j ; that is, no two of the events overlap. According to
Kolmogorov’s axioms, each event A has a probability P(A), which is a number. These numbers
satisfy three axioms: Axiom 1: For any event A, we have P(A) = 0. Axiom 2: P(S ) = 1. 4
Axiom 3: If the events A1 , A2 , . . . are pairwise disjoint, then CHAPTER 1. BASIC IDEAS
P(A1 ? A2 ? · · ·) = P(A1 ) + P(A2 ) + · · · Note that in Axiom 3, we have the union of events
and the sum of numbers. Don’t mix these up; never write P(A1 ) ? P(A2 ), for example.
Sometimes we sep- arate Axiom 3 into two parts: Axiom 3a if there are only ?nitely many events
A1 , A2 , . . . , An , so that we have P(A1 ? · · · ? An ) = ? P(Ai ), n i=1 and Axiom 3b for
in?nitely many. We will only use Axiom 3a, but 3b is important later on. Notice that we write n
? P(Ai) i=1 for P(A1 ) + P(A2 ) + · · · + P(An ). 1.4 Proving things from the axioms You can
prove simple properties of probability from the axioms. That means, every step must be justi?ed
by appealing to an axiom. These properties seem obvious, just as obvious as the axioms; but the
point of this game is that we assume only the axioms, and build everything else from that. Here
are some examples of things proved from the axioms. There is really no difference between a
theorem, a proposition, and a corollary; they all have to be proved. Usually, a theorem is a big,
important statement; a proposition a rather smaller statement; and a corollary is something that
follows quite easily from a theorem or proposition that came before. Proposition 1.1 If the event
A contains only a ?nite number of outcomes, say A = {a1 , a2 , . . . , an }, then P(A) = P(a1 ) +
P(a2 ) + · · · + P(an ). To prove the proposition, we de?ne a new event Ai containing only the
out- come ai , that is, Ai = {ai }, for i = 1, . . . , n. Then A1 , . . . , An are mutually disjoint 1.4.
PROVING THINGS FROM THE AXIOMS (each contains only one element which is in none
of the others), and A1 ? A2 ? · · · ? An = A; so by Axiom 3a, we have P(A) = P(a1 ) + P(a2 ) + · ·
· + P(an ). Corollary 1.2 If the sample space S is ?nite, say S = {a1 , . . . , an }, then P(a1 ) +
P(a2 ) + · · · + P(an ) = 1. For P(a1 ) + P(a2 ) + · · · + P(an ) = P(S ) by Proposition 1.1, and P(S )
= 1 by Axiom 2. Notice that once we have proved something, we can use it on the same basis as
an axiom to prove further facts. Now we see that, if all the n outcomes are equa.
Show that Greenwoods formula reduces to the binomial variance formul.pdf
Show that Greenwoods formula reduces to the binomial variance formula.
Solution
Let S(t) be the probability that an item from a given population will have a lifetime
exceeding t. For a sample from this population of size N let the observed times until death of N
sample members be Corresponding to each ti is ni, the number \"at risk\" just prior to time ti,
and di, the number of deaths at time ti. Note that the intervals between each time typically are not
uniform. For example, a small data set might begin with 10 cases, have a death at Day 3, a loss
(censored case) at Day 9, and another death at Day 11. Then we have (t1 = 3, t2 = 11), (n1 = 10,
n2 = 8), and (d1 = 1, d2 = 1). The Kaplan–Meier estimator is the nonparametric maximum
likelihood estimate of S(t). It is a product of the form When there is no censoring, ni is just the
number of survivors just prior to time ti. With censoring, ni is the number of survivors less the
number of losses (censored cases). It is only those surviving cases that are still being observed
(have not yet been censored) that are \"at risk\" of an (observed) death.[3] There is an alternative
definition that is sometimes used, namely The two definitions differ only at the observed event
times. The latter definition is right-continuous whereas the former definition is left-continuous.
Let T be the random variable that measures the time of failure and let F(t) be its cumulative
distribution function. Note that Consequently, the right-continuous definition of may be
preferred in order to make the estimate compatible with a right-continuous estimate of F(t)..
Show that every subset of a totally bounded set is totally bounded..pdf
Show that every subset of a totally bounded set is totally bounded.
Solution
Let S be a totally bounded set and F a subset of it.
AS S is totally bounded, by definition, for every epsilon >0, there exists a final collection of
open balls in S of radius epsilon such that their union contain S.
Or there exists a finite epsilon net.
As S is totally bounded we have S is the union of small subsets
Consider F, a subset of S
S is contained in A1UA2UA3...An where each Ai is a subset with epsilon radius.
Select Ais such that they contain all elements of F
Let Al, Am, As,...contain elements of F
Then F is obviously contained in AlUAmU...
From this it follows that F has an epilon net for itself
Hence F is totally bounded..
Should the benefit to the criminal be included in the social welfare.pdf
Should the benefit to the criminal be included in the social welfare function?
Solution
Answer: Criminal is inlvoved in a lot of activities that are not acceptable to society as whole. In
other words these activites are anti-law. But the criminal is also part of society. If any kinds of
benefits that accrueing to him from sources other then the criminal activites, must be given
ample place in the social welfare funciton. But if he benefits are attributable to criminal activites,
then it should not form the part of social welfare function as these benefits are being reaped by
causing harm to someone else..
Show that every integer m is of one of the forms 2n, 2n+1 foranother.pdf
Show that every integer m is of one of the forms 2n, 2n+1 foranother integer n by quoting
the Division Theorem.
Solution
let m be an integer
2 is the first even integer
if 2 divides m then m=2n for some integern
if 2 does not divide m then the remainder is 1. Hencem=2n+1 for some integer n.
SHOW LIST OF ACCOUNTSLINK TO TEXTFIFOLIFOAVERAGE-COSTOpen .pdf
SHOW LIST OF ACCOUNTS
LINK TO TEXT
FIFO
LIFO
AVERAGE-COST
Open Show Work
SHOW LIST OF ACCOUNTS
LINK TO TEXTGroves Company Inc. had a beginning inventory of 141 units of Product MLN
at a cost of $11 per unit. During the year, purchases were:
Feb. 20845units at$13Aug. 12564units at$16May 5705units at$14Dec. 8141units at$17
Groves Company uses a periodic inventory system. Sales totaled 2,114 units.
Solution
(a)
Calculation of Cost of goods available for sale:
Units
Rate
Cost
A
B
A*B
Beginning Inventory
141
$ 11
$ 1,551
Feb. 20
845
$ 13
$ 10,985
May. 5
705
$ 14
$ 9,870
Aug. 12
564
$ 16
$ 9,024
Dec. 8
141
$ 17
$ 2,397
Cost of goods available for sale
2396
$ 33,827
(b)
Calculation of Average cost per unit
Cost of goods available for sale
$ 33,827
Total Units Available
2396
Average cost per unit = 33827 / 2396
$ 14.118
Calculations of Ending Cost of goods sold and inventory under FIFO periodic inventory system
Cost of Goods Sold:
Units
Rate
Cost
A
B
A*B
Beginning Inventory
141
$ 11
$ 1,551
Feb. 20
845
$ 13
$ 10,985
May. 5
705
$ 14
$ 9,870
Aug. 12
423
$ 16
$ 6,768
Cost of Goods Sold
2114
$ 29,174
Ending inventory :
Cost of goods available for sale
$ 33,827
Less: Cost of goods sold
$ (29,174)
Ending inventory
$ 4,653
Calculations of Ending Cost of goods sold and inventory under LIFO periodic inventory system
Cost of Goods Sold:
Units
Rate
Cost
A
B
A*B
Dec. 8
141
$ 17
$ 2,397
Aug. 12
564
$ 16
$ 9,024
May. 5
705
$ 14
$ 9,870
Feb. 20
704
$ 13
$ 9,152
Cost of Goods Sold
2114
$ 30,443
Ending inventory :
Cost of goods available for sale
$ 33,827
Less: Cost of goods sold
$ (30,443)
Ending inventory
$ 3,384
Calculations of Ending Cost of goods sold and inventory under Average Cost periodic inventory
system
Cost of Goods Sold:
Units Sold (A)
2114
Average cost per unit (B)
$ 14.118
Cost of Goods Sold = A*B
$ 29,845
Ending inventory :
Cost of goods available for sale
$ 33,827
Less: Cost of goods sold
$ (29,845)
Ending inventory
$ 3,982
(a)
Calculation of Cost of goods available for sale:
Units
Rate
Cost
A
B
A*B
Beginning Inventory
141
$ 11
$ 1,551
Feb. 20
845
$ 13
$ 10,985
May. 5
705
$ 14
$ 9,870
Aug. 12
564
$ 16
$ 9,024
Dec. 8
141
$ 17
$ 2,397
Cost of goods available for sale
2396
$ 33,827
(b)
Calculation of Average cost per unit
Cost of goods available for sale
$ 33,827
Total Units Available
2396
Average cost per unit = 33827 / 2396
$ 14.118
Calculations of Ending Cost of goods sold and inventory under FIFO periodic inventory system
Cost of Goods Sold:
Units
Rate
Cost
A
B
A*B
Beginning Inventory
141
$ 11
$ 1,551
Feb. 20
845
$ 13
$ 10,985
May. 5
705
$ 14
$ 9,870
Aug. 12
423
$ 16
$ 6,768
Cost of Goods Sold
2114
$ 29,174
Ending inventory :
Cost of goods available for sale
$ 33,827
Less: Cost of goods sold
$ (29,174)
Ending inventory
$ 4,653
Calculations of Ending Cost of goods sold and inventory under LIFO periodic inventory system
Cost of Goods Sold:
Units
Rate
Cost
A
B
A*B
Dec. 8
141
$ 17
$ 2,397
Aug. 12
564
$ 16
$ 9,024
May. 5
705
$ 14
$ 9,870
Feb. 20
704
$ 13.
Should preferred stock be classified as debt or equityDoes it matte.pdf
Should preferred stock be classified as debt or equity/Does it matter
if the classification is being made by the firm\'s a. management b.
creditors, or c.equity investors? Please explain.
Solution
The preference shares should be classified into Equity since because those shareholders will not
get the fixed income when there are no sufficient profits on hand to distribute.
In any case like Managemnet , creditors & Equity investors the preference stock shall be
classified as Equity and not as Debt..
Show all work. A disc jockey has 11 songs to play. Seven are slow so.pdf
Show all work. A disc jockey has 11 songs to play. Seven are slow songs, and four are fast
songs. Each song is to be played only once. In how many ways can the disc jockey play the 11
songs if
The songs can be played in any order.
The first song must be a slow song and the last song must be a slow song.
The first two songs must be fast songs.
Solution
11! 9! 9!.
Show all work. A disc jockey has 9 songs to play. Five are slow song.pdf
Show all work. A disc jockey has 9 songs to play. Five are slow songs, and 4 are fast songs. Each
song is to be played only once. In how many ways can the disc jockey play the 9 songs if
The songs can be played in any order.
The first song must be a slow song and the last song must be a slow song.
The first two songs must be fast songs.
Solution
11! 9! 9!.
Should healthcare use CHINS -community health information networks o.pdf
Should healthcare use CHINS -community health information networks or RHIOs? What are the
privacy and security issues.provide articles
Solution
A) CHINs- provides technology-based information services to help maintain optimal health for
all the residents of a community.its services are available to all stakeholders and that competitors
collaborate to share information. It is usually to facilitate paying for health care or to share
clinical records.
B) RHIOs- is a group of organizations within a specific geographical area that share healthcare-
related information electronically according to national standards. RHIO is one model toward
achieving the proposed National Health Information Network (NHIN) in the United States.
*The last 18 months will likely be remembered as the turning point in which Regional Health
Information Organizations (RHIOs) gained widespread consensus. Today, few would argue that
Health Information Networks (HINs), and especially RHIOs, lie at the top of the healthcare
agenda in the United States. With the possibility to dramatically increase the quality of
healthcare, all stakeholders.
Home assignment II on Spectroscopy 2024 Answers.pdf
Answers to Home assignment on UV-Visible spectroscopy: Calculation of wavelength of UV-Visible absorption
2024.06.01 Introducing a competency framework for languag learning materials ...
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Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
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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
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https://app.box.com/s/hqnndn05v4q5a4k4jd597rkdbda0fniiSynthetic Fiber Construction in lab .pptx
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
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Show that Greenwoods formula reduces to the binomial variance formul.pdf
Show that Greenwoods formula reduces to the binomial variance formula.
Solution
Let S(t) be the probability that an item from a given population will have a lifetime
exceeding t. For a sample from this population of size N let the observed times until death of N
sample members be Corresponding to each ti is ni, the number \"at risk\" just prior to time ti,
and di, the number of deaths at time ti. Note that the intervals between each time typically are not
uniform. For example, a small data set might begin with 10 cases, have a death at Day 3, a loss
(censored case) at Day 9, and another death at Day 11. Then we have (t1 = 3, t2 = 11), (n1 = 10,
n2 = 8), and (d1 = 1, d2 = 1). The Kaplan–Meier estimator is the nonparametric maximum
likelihood estimate of S(t). It is a product of the form When there is no censoring, ni is just the
number of survivors just prior to time ti. With censoring, ni is the number of survivors less the
number of losses (censored cases). It is only those surviving cases that are still being observed
(have not yet been censored) that are \"at risk\" of an (observed) death.[3] There is an alternative
definition that is sometimes used, namely The two definitions differ only at the observed event
times. The latter definition is right-continuous whereas the former definition is left-continuous.
Let T be the random variable that measures the time of failure and let F(t) be its cumulative
distribution function. Note that Consequently, the right-continuous definition of may be
preferred in order to make the estimate compatible with a right-continuous estimate of F(t)..
Show that every subset of a totally bounded set is totally bounded..pdf
Show that every subset of a totally bounded set is totally bounded.
Solution
Let S be a totally bounded set and F a subset of it.
AS S is totally bounded, by definition, for every epsilon >0, there exists a final collection of
open balls in S of radius epsilon such that their union contain S.
Or there exists a finite epsilon net.
As S is totally bounded we have S is the union of small subsets
Consider F, a subset of S
S is contained in A1UA2UA3...An where each Ai is a subset with epsilon radius.
Select Ais such that they contain all elements of F
Let Al, Am, As,...contain elements of F
Then F is obviously contained in AlUAmU...
From this it follows that F has an epilon net for itself
Hence F is totally bounded..
Should the benefit to the criminal be included in the social welfare.pdf
Should the benefit to the criminal be included in the social welfare function?
Solution
Answer: Criminal is inlvoved in a lot of activities that are not acceptable to society as whole. In
other words these activites are anti-law. But the criminal is also part of society. If any kinds of
benefits that accrueing to him from sources other then the criminal activites, must be given
ample place in the social welfare funciton. But if he benefits are attributable to criminal activites,
then it should not form the part of social welfare function as these benefits are being reaped by
causing harm to someone else..
Show that every integer m is of one of the forms 2n, 2n+1 foranother.pdf
Show that every integer m is of one of the forms 2n, 2n+1 foranother integer n by quoting
the Division Theorem.
Solution
let m be an integer
2 is the first even integer
if 2 divides m then m=2n for some integern
if 2 does not divide m then the remainder is 1. Hencem=2n+1 for some integer n.
SHOW LIST OF ACCOUNTSLINK TO TEXTFIFOLIFOAVERAGE-COSTOpen .pdf
SHOW LIST OF ACCOUNTS
LINK TO TEXT
FIFO
LIFO
AVERAGE-COST
Open Show Work
SHOW LIST OF ACCOUNTS
LINK TO TEXTGroves Company Inc. had a beginning inventory of 141 units of Product MLN
at a cost of $11 per unit. During the year, purchases were:
Feb. 20845units at$13Aug. 12564units at$16May 5705units at$14Dec. 8141units at$17
Groves Company uses a periodic inventory system. Sales totaled 2,114 units.
Solution
(a)
Calculation of Cost of goods available for sale:
Units
Rate
Cost
A
B
A*B
Beginning Inventory
141
$ 11
$ 1,551
Feb. 20
845
$ 13
$ 10,985
May. 5
705
$ 14
$ 9,870
Aug. 12
564
$ 16
$ 9,024
Dec. 8
141
$ 17
$ 2,397
Cost of goods available for sale
2396
$ 33,827
(b)
Calculation of Average cost per unit
Cost of goods available for sale
$ 33,827
Total Units Available
2396
Average cost per unit = 33827 / 2396
$ 14.118
Calculations of Ending Cost of goods sold and inventory under FIFO periodic inventory system
Cost of Goods Sold:
Units
Rate
Cost
A
B
A*B
Beginning Inventory
141
$ 11
$ 1,551
Feb. 20
845
$ 13
$ 10,985
May. 5
705
$ 14
$ 9,870
Aug. 12
423
$ 16
$ 6,768
Cost of Goods Sold
2114
$ 29,174
Ending inventory :
Cost of goods available for sale
$ 33,827
Less: Cost of goods sold
$ (29,174)
Ending inventory
$ 4,653
Calculations of Ending Cost of goods sold and inventory under LIFO periodic inventory system
Cost of Goods Sold:
Units
Rate
Cost
A
B
A*B
Dec. 8
141
$ 17
$ 2,397
Aug. 12
564
$ 16
$ 9,024
May. 5
705
$ 14
$ 9,870
Feb. 20
704
$ 13
$ 9,152
Cost of Goods Sold
2114
$ 30,443
Ending inventory :
Cost of goods available for sale
$ 33,827
Less: Cost of goods sold
$ (30,443)
Ending inventory
$ 3,384
Calculations of Ending Cost of goods sold and inventory under Average Cost periodic inventory
system
Cost of Goods Sold:
Units Sold (A)
2114
Average cost per unit (B)
$ 14.118
Cost of Goods Sold = A*B
$ 29,845
Ending inventory :
Cost of goods available for sale
$ 33,827
Less: Cost of goods sold
$ (29,845)
Ending inventory
$ 3,982
(a)
Calculation of Cost of goods available for sale:
Units
Rate
Cost
A
B
A*B
Beginning Inventory
141
$ 11
$ 1,551
Feb. 20
845
$ 13
$ 10,985
May. 5
705
$ 14
$ 9,870
Aug. 12
564
$ 16
$ 9,024
Dec. 8
141
$ 17
$ 2,397
Cost of goods available for sale
2396
$ 33,827
(b)
Calculation of Average cost per unit
Cost of goods available for sale
$ 33,827
Total Units Available
2396
Average cost per unit = 33827 / 2396
$ 14.118
Calculations of Ending Cost of goods sold and inventory under FIFO periodic inventory system
Cost of Goods Sold:
Units
Rate
Cost
A
B
A*B
Beginning Inventory
141
$ 11
$ 1,551
Feb. 20
845
$ 13
$ 10,985
May. 5
705
$ 14
$ 9,870
Aug. 12
423
$ 16
$ 6,768
Cost of Goods Sold
2114
$ 29,174
Ending inventory :
Cost of goods available for sale
$ 33,827
Less: Cost of goods sold
$ (29,174)
Ending inventory
$ 4,653
Calculations of Ending Cost of goods sold and inventory under LIFO periodic inventory system
Cost of Goods Sold:
Units
Rate
Cost
A
B
A*B
Dec. 8
141
$ 17
$ 2,397
Aug. 12
564
$ 16
$ 9,024
May. 5
705
$ 14
$ 9,870
Feb. 20
704
$ 13.
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Show all work. A disc jockey has 11 songs to play. Seven are slow so.pdf
Show all work. A disc jockey has 11 songs to play. Seven are slow songs, and four are fast
songs. Each song is to be played only once. In how many ways can the disc jockey play the 11
songs if
The songs can be played in any order.
The first song must be a slow song and the last song must be a slow song.
The first two songs must be fast songs.
Solution
11! 9! 9!.
Show all work. A disc jockey has 9 songs to play. Five are slow song.pdf
Show all work. A disc jockey has 9 songs to play. Five are slow songs, and 4 are fast songs. Each
song is to be played only once. In how many ways can the disc jockey play the 9 songs if
The songs can be played in any order.
The first song must be a slow song and the last song must be a slow song.
The first two songs must be fast songs.
Solution
11! 9! 9!.
Should healthcare use CHINS -community health information networks o.pdf
Should healthcare use CHINS -community health information networks or RHIOs? What are the
privacy and security issues.provide articles
Solution
A) CHINs- provides technology-based information services to help maintain optimal health for
all the residents of a community.its services are available to all stakeholders and that competitors
collaborate to share information. It is usually to facilitate paying for health care or to share
clinical records.
B) RHIOs- is a group of organizations within a specific geographical area that share healthcare-
related information electronically according to national standards. RHIO is one model toward
achieving the proposed National Health Information Network (NHIN) in the United States.
*The last 18 months will likely be remembered as the turning point in which Regional Health
Information Organizations (RHIOs) gained widespread consensus. Today, few would argue that
Health Information Networks (HINs), and especially RHIOs, lie at the top of the healthcare
agenda in the United States. With the possibility to dramatically increase the quality of
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