The five most popular art museums in the world are Mus The five most.pdf
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The five most popular art museums in the world are Mus The five most popular art museums in the world are Musée du Louvre, the Metropolitan Museum of Art, British Museum, National Gallery, and Tate Modern (The Art Newspaper, April 2012). Which of these five museums would visitors most frequently rate as spectacular? Samples of recent visitors of each of these museums were taken, and the results of these samples follow. Let p1 = population proportion of visitors who rate the Musée du Louvre as spectacular p2 = population proportion of visitors who rate the Metropolitan Museum of Art as spectacular p3 = population proportion of visitors who rate the British Museum as spectacular p4 = population proportion of visitors who rate the National Gallery as spectacular p5 = population proportion of visitors who rate the Tate Modern as spectacular Using a .05 level of significance, what is the p-value. a. less than .005 b. between .005 and .010 c. between .01 and .025 d. between.025 and .05 e. between .05 and .10 f. higher than .10 g. none Solution p1= 113/150 = 0.75 p2= 94/140 = 0.67 p3= 96/160 = 0.6 p4= 78/120 = 0.65 p5= 88/110 = 0.8 Museum that visitors most frequently rate as spectacular is : Tate Modern Using a .05 level of significance, p-value : less than 0.05.
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Poisson distribution] The number of data packets arriving in 1 sec at.pdf
Poisson distribution] The number of data packets arriving in 1 sec at a particular switch (in
millions) is modeled by random variable X which is Poisson distributed with parameter A = 2.
Find the probability that X = 3 million packets arrive in a given second. kind the probability that
X = 0 packets arrive.
Solution.
Problem 1 Create Node class (or use what you have done in Lab4)• .pdf
Problem 1: Create Node class (or use what you have done in Lab4)
• Add methods to get and set, Data and Link
• Here the data is an object. (java object) your Linked List can hold any java class.
a) Create MyLinkedList2 class and write methods to
• Methods to insert nodes at the beginning, at the end and at a specific position. insertFirst(Node
x), insertLast(Node x) , insertLast(Node x, int pos)
• Methods to delete node from the beginning and from the end. delFirst(), delLast()
• Method to search the Linked List for a specific object. searchList(x Object)
• Method to Sort the Linked List in ascendant or descendant order according to the object
value(s). for example sorted by name, by age, by gpa. sortAsc(),sortDesc().
• Method to display all nodes objects (printing an object is related to its toString() method).
• Method to return the largest and smallest object in the list – maxList(), minList() according to
some specific object members.
. • Method to return the number of nodes – listSize()
• Method to merge two linked Lists into one. MyLinkedList2 list3= mergeList(MyLinkedList2
list1, MyLinkedList2 list2)
b) Create Main class : MyLinkedList2Demo
• Create a linked List LL1 and insert some objects to it. Use any a student java class.
• Fill LL1 by adding students from an input file.
• Display all the objects (students) in the List LL1
• Search for a particular object (Student by name or by Id or both) in LL1 and print true or false
• Sort LL1 and display it ( using asc and desc on multiple members)
• Perform all the delete methods on LL1.
• Perform all the insert methods on LL1.
• Call and run all the implemented methods in a sequential order. NB: Organize a nice output for
all the questions above.
Problem 2: Same as Pb1 using double linked lists. Create Nod3e and MyLinkedList3 classes.
--------------------------------------------------------------------------------------------------------------------
---------------------------------------------------
The input file :
id fname lname age gpa year college cuCourses totalCourses
201412345 Amna chamisi 17 2.1 2014 CIT 5 30
201112345 Amna chamisi 17 2.1 2014 ENG 31 32
201412345 Amna chamisi 17 2.1 2014 COE 5 29
201412345 Amna chamisi 17 2.1 2014 CIT 5 30
201412345 Amna chamisi 17 2.1 2014 CIT 5 30
201412345 Amna chamisi 17 2.1 2014 CIT 5 30
201412345 Amna chamisi 17 2.1 2014 CIT 5 30
Solution
/**
* Node.java
* @author
*
*/
public class Node {
public Object getObj() {
return obj;
}
public Node setObj(Object obj) {
this.obj = obj;
return this;
}
public Node getNext() {
return next;
}
public Node setNext(Node next) {
this.next = next;
return this;
}
private Object obj;
private Node next;
}
import java.util.Collections;
import java.util.Comparator;
import java.util.LinkedList;
/**
* MyLinkedList2.java
* @author
*
*/
public class MyLinkedList2 {
public void insertFirst(Node x) {
if (head == null) {
head = x;
} else {
x.setNext(head);
}
}
public void insertLast(Node x) {
if (head == null) {
head = x;
r.
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Poisson distribution] The number of data packets arriving in 1 sec at.pdf
Poisson distribution] The number of data packets arriving in 1 sec at a particular switch (in
millions) is modeled by random variable X which is Poisson distributed with parameter A = 2.
Find the probability that X = 3 million packets arrive in a given second. kind the probability that
X = 0 packets arrive.
Solution.
Problem 1 Create Node class (or use what you have done in Lab4)• .pdf
Problem 1: Create Node class (or use what you have done in Lab4)
• Add methods to get and set, Data and Link
• Here the data is an object. (java object) your Linked List can hold any java class.
a) Create MyLinkedList2 class and write methods to
• Methods to insert nodes at the beginning, at the end and at a specific position. insertFirst(Node
x), insertLast(Node x) , insertLast(Node x, int pos)
• Methods to delete node from the beginning and from the end. delFirst(), delLast()
• Method to search the Linked List for a specific object. searchList(x Object)
• Method to Sort the Linked List in ascendant or descendant order according to the object
value(s). for example sorted by name, by age, by gpa. sortAsc(),sortDesc().
• Method to display all nodes objects (printing an object is related to its toString() method).
• Method to return the largest and smallest object in the list – maxList(), minList() according to
some specific object members.
. • Method to return the number of nodes – listSize()
• Method to merge two linked Lists into one. MyLinkedList2 list3= mergeList(MyLinkedList2
list1, MyLinkedList2 list2)
b) Create Main class : MyLinkedList2Demo
• Create a linked List LL1 and insert some objects to it. Use any a student java class.
• Fill LL1 by adding students from an input file.
• Display all the objects (students) in the List LL1
• Search for a particular object (Student by name or by Id or both) in LL1 and print true or false
• Sort LL1 and display it ( using asc and desc on multiple members)
• Perform all the delete methods on LL1.
• Perform all the insert methods on LL1.
• Call and run all the implemented methods in a sequential order. NB: Organize a nice output for
all the questions above.
Problem 2: Same as Pb1 using double linked lists. Create Nod3e and MyLinkedList3 classes.
--------------------------------------------------------------------------------------------------------------------
---------------------------------------------------
The input file :
id fname lname age gpa year college cuCourses totalCourses
201412345 Amna chamisi 17 2.1 2014 CIT 5 30
201112345 Amna chamisi 17 2.1 2014 ENG 31 32
201412345 Amna chamisi 17 2.1 2014 COE 5 29
201412345 Amna chamisi 17 2.1 2014 CIT 5 30
201412345 Amna chamisi 17 2.1 2014 CIT 5 30
201412345 Amna chamisi 17 2.1 2014 CIT 5 30
201412345 Amna chamisi 17 2.1 2014 CIT 5 30
Solution
/**
* Node.java
* @author
*
*/
public class Node {
public Object getObj() {
return obj;
}
public Node setObj(Object obj) {
this.obj = obj;
return this;
}
public Node getNext() {
return next;
}
public Node setNext(Node next) {
this.next = next;
return this;
}
private Object obj;
private Node next;
}
import java.util.Collections;
import java.util.Comparator;
import java.util.LinkedList;
/**
* MyLinkedList2.java
* @author
*
*/
public class MyLinkedList2 {
public void insertFirst(Node x) {
if (head == null) {
head = x;
} else {
x.setNext(head);
}
}
public void insertLast(Node x) {
if (head == null) {
head = x;
r.
One of your friends has several brothers and sisters, each quite dif.pdf
One of your friends has several brothers and sisters, each quite different in appearance, interests,
and abilities, even though all have the same parents. Explain in detail how these siblings reflect
crossing over and independent assortment.
Solution
Meiosis process increases the genetic diversity in a species. In prophase I, crossing over takes
place and in metaphase I and anaphase I, independent assortment occurs. When homologous
chromosomes are paired together, during meiosis, the chromosomes make contact with the other
pair of chromosomes. This point of contact is responsible for the exchange of genetic
information between the chromosomes and is called the chiasmata. Although the same genetic
information is contained by the two homologous chromosomes, the assortment of the
chromosomes determine which information is present in each of the four gametes produced. For
a particular phenotype, an allele can determine which characteristic will be expressed by the
organism. For example, Chromosome 1 contains an allele for brown eyes while chromosome 2
contains an allele for black eyes. Thus, crossing over and independent assortment are responsible
for different appearence, abilities and interest..
Malpighiaceae luonanthaceae Ochnaceae Flacourtiaceae O Monilophytes G.pdf
Malpighiaceae luonanthaceae Ochnaceae Flacourtiaceae O Monilophytes Gymnosperms O
Basal Angiosperms Monocots Basal Eudicots Asterids Rosids All Plots Escaloniacea
Pittosporaceae Araliaceae Caprifoliaceae Adoxaceae Garry aceae Rubiaceae Apocynaceae
Solution
There are 7 groups present in the legend around the phylogeny.Monophyletic taxon is one that
include a group of organism desecended from a single ancestor.A monophyletic taxon is called a
clade.Monophyletic groups are important to understand how animals are classified. It is
commonly known that all animals evolved over time.
The monophyletics in the legends are Cyatheaceae Commonly known as the scaly tree
ferns,Poaceae,Arecaceae are a botanical family of perennial climbers, shrubs, acaules and trees
commonly known as palm tree,Liliaceae are mostly perennial herbs from starchy rhizomes,
corms, or bulbs comprising about 280 genera and 4,000 species. earthworms, birds, and snakes
all have a common ancestor. Since all these organisms share a common ancestor, they also share
a synapomorphy, which is that earthworms, birds, and snakes all reproduce sexually. These are
all come under phylogeny of monophyletic..
In JavaWrite a program that reads a text file that contains a gra.pdf
In Java:
Write a program that reads a text file that contains a grade (for instance, 87) on each line.
Calculate and print the average of the grades.
Solution
import java.io.BufferedReader;
import java.io.FileReader; // importing filereader
import java.io.IOException; // importing file io exceptions
public class grade {
public static void main(String[] args) {
BufferedReader br = null;
double count=0;
int noOfgrades=0;
try {
String sLine;
br = new BufferedReader(new FileReader(\"sample.txt\")); // reading file using buffer
reader
while ((sLine = br.readLine()) != null) { // rading line by line
noOfgrades=noOfgrades+1; // getting number of grades
int num=Integer.parseInt(sLine); // convert string into integer
count=count+num; // adding each grade to count
}
System.out.println(\"count:\\t\"+count);
System.out.println(\"noOfgrades:\\t\"+noOfgrades);
System.out.println(\"Average of the grades:\\t\"+(count/noOfgrades));
} catch (IOException e) {
e.printStackTrace(); // catch exception
} finally {
try {
if (br != null)br.close(); // closing buffer
} catch (IOException ex) {
ex.printStackTrace();// printing exceptions
}
}
}
}
/********** out file******
fileName:sample.txt
file contains \"
23
34
56
7
88
89
52
34
65
\"
*/
/************output*************
count: 448.0
noOfgrades: 9
Average of the grades: 49.77
*/.
How does Pb and TCE partition between components of soil (mineral, o.pdf
How does Pb and TCE partition between components of soil (mineral, organic matter, air, and
water fractions)? Where can each pollutant can be found if it is spilled on to a loam soil with
about a neutral ph, and about 2% organic matter content?
Solution
lead is heavy metal as a mineral in soil while trichloroethylene is an ogranic pollutant present as
organic matter
lead will be found in as a mineral into the soil while TCE is present in air as organic pollutant..
Hi, please I need help with the correct answer to the above multiple.pdf
Hi, please I need help with the correct answer to the above multiple questions. Thanks. Some
animals, such as snakes, can see infrared waves. The un human eye can not detect these waves,
however, we can aided detect some wavelengths of light. Considering the following statements,
which, if any, would be true? (SELECT ALL TRUE STATEMENTS) O if an object is seen as
red, red light is absorbed by the object and not reflected O Visible light waves exist in the 100-
300nm wavelength range C ight is a form of electromagnetic energy that slows down as it enters
a medium that is less dense O if an object is green, the object is absorbing non-green waves and
reflecting green waves. since humans can not see all wavelengths of the electromagnetic
spectrum, only waves within the visible light spectrum will damage the eye e. UV light can not
damage the retina because we can not see UV light
Solution
1. If an object is green, the object is absorbing non-green waves and reflecting green waves. The
reflected green waves are captured by our eyes and the object appears \'green\'.
The first statement is the exact opposite of this and, hence, incorrect. 100-300 nm wavelengths
correspond to the UV spectrum. Light speeds up in a rarer medium. And, UV light can
(seriously) damage our eyes.
2. Superior colliculus: This part of the brain is responsible for saccades - the process by which
our eyes scan different images of an object with different focus and \'knits\' them together to
perceive useful information.
3. Lens cannot be transplanted! All other options are invalid.
4. Tympanic membrane: Commonly known as the \'ear drum\', this thin membranous structure is
present at the end of the external ear canal. All other structures listed are present beyond the
tympanic membrane. When something such as a Q-tip or earbud is inserted too deep in the ear
canal, it will first perforate the tympanic membrane before injuring any other part..
For which phyla are megaspores a shared trait Bryophyta Monilophyt.pdf
For which phyla are megaspores a shared trait? Bryophyta Monilophyta Coniferophyta
Anthophyta For which phyla are microspores a shared trait? Bryophyta Monilophyta
Coniferophyta Anthophyta For which phyla are sporophyte a shared trait? Bryophyta
Monilophyta Coniferophyta Q Anthophyta
Solution
13.Bryophta (Bryophytes do not have flowers or seeds and reproduce through spores.)
14.Coniferophyta( A pine tree produces two types of spores- the microspores seen on male cones
and megaspores seen on female cones. Hence, they are heterosporic.).
15.Bryphyta(The sporophyte dries and releases spores that grow into a new generation of
gametophytes. The sporophyte is the organ for asexual reproduction and consists of a foot, seta
and capsule. The capsule is a beak like structure found on top of a thin stalk and produces the
spores. Mosses show alternation of generation between gametophytic and sporophytic
generations though the plant has a gametophyte dominant cycle.).
For Programming Embedded SystemsQ-07 Whcih variable and condition.pdf
For Programming Embedded Systems
Q-07 *Whcih variable and condition is sued to detect unfinished tasks (interrupted interrupt)?
A) Line 28
B) Line 36
C) Line 25
D) Line 22 I. #include \"rims.h\" 2 3 4. typedef struct task state BL LedOff; break; default: int
state; unsigned long period; unsigned long elapsedTime: int (*TickFct) (int); taski state ) I
Transitions switch (state) case BL Ledoff: 11. task tasks [2] break; case BL Ledon: 13. const
unsigned char tasksNum 2; 14. const unsigned long periodBlinkLed = 1500; 15. const unsigned
long periodThreeLeds 500; break; default: break; ) // State actions 17, const unsigned long
tasksPeriodGCD = 500; return state 19. int TickFct BlinkLed (int state) 20. int TickFct
ThreeLeds (int state) 22. unsigned char process!ngRdyTasks = 0; 23. void TimerISR 98. enum
TL States TL TO, TL T1, TL T2 TL State; 100 99. int TickFct ThreeLeds (int state) ( 101
VARIABLES MUST BE DECLARED STATIC/ unsigned char i; if (processingRdyTasks)
/*e.g., static int x = 0; */ /*variables for state machine go here. printf (\"Period too short \ \"): 103
switch (state) case -1: process!ngRdyTasks = 1; for (1 = 0; 1 = 107 108 109 tasks [i].period) case
TL TO: tasks [1]. state = 1 tasks[i].TickFct (tasks[i].state); state = TL T1; tasks[i].elapsedTime0
break; case TL T1: 112 113 tasks [1].elapsedTime += tasksPeriodGCD; processingRdyTasks - 0;
state = TL T2 115 116 38. int main) break; case TL T2: unsigned char i=0; tasks [1]. state =-1;
tasks [1].period = periodBlinkLed; tasks [1] .elapsedTime tasks [1] ·period; tasks [1].TickFct
6T1ckFct BlinkLed; 118 119 120 121 122 123 1 state = TL T0; break; default: state --1; tasks [1]
, state = -1; tasks [1].period periodThreeLeas; tasks [1].elapsedTime = tasks [1].period; tasks
[1].TICkFct = &TICkfct; ThreeLeas; 125 switch (state) 127 128 129 130 case TL TO: get
Timerset (tasksPeriodGCD); break case TL T1: TimerOn () while (1) Sleep return 0 133 135 137
139 h break case TL T2: 61. enum BL States BL Ledoff, BL Ledon BL State; 62. int TickFct
BlinkLed (int state) *VARIABLES MUST BE DECLARED STATIC break; 64. /*e.g., static Int
X = 0; */ 65. /*variables for state machine go here/ default: break switch (state) case-1: return
state state = BL LedOff break; 146 case BL Ledoff: state = BL Ledon; break; case BL Ledon: 1
Solution
c) Line 25 is correct
Line 25 contains if condition and processingRdyTasks variable is used to detect the unfinished
tasks.
For analysis of Thermodynamic Cycle, we draw T-s or P-h diagram. Why.pdf
For analysis of Thermodynamic Cycle, we draw T-s or P-h diagram. Why don’t we draw T-h or
P-t diagram?
Solution
For analysis of thermodynamics cycle we draw various diagrams, In all these diagram T-s,P-h,T-
h and P-T diagram are some of them. Every diagram give some specific informations about
system working like T-s diagram give information about Heat transfer, P-h diagram use in air
conditioning because air is mixture so pressure affect more and enthalphy is measure of energy.
like these two diagram we use P-t diagram use in pure substance to see the transformation at
triple point. and T-h diagram use for visualise how energy change with temprature..
Fill in the blank. A contingency table relates only continuous rando.pdf
Fill in the blank. A contingency table relates only continuous random variables the difference
in the means of two random variables two categories of data a particular response with order in
which that response should be applied
Solution
A contingency table relates
Two categories of data
Contingency table (cross tabulation or crosstab) is a type of table format that displays the
frequency distribution of the two or more categorical variables..
Discuss the dangers associated with hydrogen sulfide in the petro.pdf
** Discuss the dangers associated with hydrogen sulfide in the petrochemical environment.
*** Your original post should be between 100 - 150 words in length.
Solution
1.on humans:Hydrogen sulfide enters your body primarily through the air you breathe. Much
smaller amounts can enter your body through the skin. Hydrogen sulfide is a gas, so you would
not likely be exposed to it by ingestion. When you breathe air containing hydrogen sulfide or
when hydrogen sulfide comes into contact with skin, it is absorbed into the blood stream and
distributed throughout the body.Exposure to low concentrations of hydrogen sulfide may cause
irritation to the eyes, nose, or throat. It may also cause difficulty in breathing for some
asthmatics.Respiratory distress or arrest has been found in people exposed to very high
concentrations of hydrogen sulfide.
The health effects of hydrogen sulfide depend on several factors such as how much hydrogen
sulfide you are exposed to and the length of that exposure. Studies in workers, communities
living near industrial sources of hydrogen sulfide, and volunteers suggest that the respiratory
tract and nervous system are the most sensitive targets of hydrogen sulfide toxicity.It is not
known whether exposure to hydrogen sulfide causes birth defects in humans. The results of
studies in animals suggest that exposure to low concentrations of hydrogen sulfide during
pregnancy does not cause birth defects.
Exposure to low concentrations of hydrogen sulfide may cause headaches, poor memory,
tiredness, and balance problems.Brief exposures to high concentrations of hydrogen sulfide
(greater than 500 ppm) can cause a loss of consciousness. In most cases, the person appears to
regain consciousness without any other effects. However, in some individuals, there may be
permanent or long-term effects such as headaches, poor attention span, poor memory, and poor
motor function..
Complete the following paragraph to describe the various reproductive.pdf
Complete the following paragraph to describe the various reproductive strategies seen in the
phylum Ascomycota. Some choices may be used more than once.
Solution
1. The Ascomycota contains two main groups; Asexual sac fungi which include bakers yeast and
Sexual sac fungi, which include candida yeast and molds.
2. Most Ascomyta are conidospores and reproduced by meiosis, as seen in candida yeast, or
through the production of fruiting body as seen in molds.
3. In contrast in haploid a(n) diploid containing two nuclei develops within a(n) Conidiospores
.
4. The nuclei fuse creating haploid cell that undergose mitosis to produce diploid cells.
5. Typically at this point, the ascospores swells and burst, releasing the ascus in to the air to
germinate once landed..
CommunicationsSystems have four basic properties (holism, equi-fi.pdf
Communications:
Systems have four basic properties (holism, equi-finality, negative entropy). List the principle
behind each property. What does it mean that “a system is more than the sum of its parts?” How
does this concept apply to organizations?
Solution
Four basic properties of systems are:
Principle of Holism:
It emphasizes that system is more than sum of its part due to interdependence upon each other.
Principle of Equi-finality:
It states that outcome can be achieved through many different paths due to interdependence of
components.
Principle of Negative Entropy:
It is based upon openness of the system that says that system owns the ability to prevent
deterioration and thrive.
Principle of Requisite Variety:
It is also based upon openness of the system and it says that internal complexity should be
maintained to deal with external complexity.
Statement of “a system is more than the sum of its parts” means that system achieves more when
all of its parts complement each other and work together. Each component / part of the system
cannot do much in isolation but they can achieve big goals when grouped together as a team.
It is very important in organizational perspective as it consists of many different functions as
components. When, these functions work together, organizational objective is achieved. If not,
organization falls apart..
Based on your reading of the GCU introduction and the textbooks, wha.pdf
Based on your reading of the GCU introduction and the textbooks, what is the main difference
between the modern western worldview and the postmodern worldview? Which parts, if any, do
you identify with? Explain. Cite references from your reading to support your answer.
Solution
Postmodernism is a response against modernism. According to everyone has a point of view
through which they see the world.
Premodern (the worldview before modernism)- says that natural events are caused by
supernatural causes. God, angels, demons, and the like are a reality to these folks..
Answer using Giardias, Shiga-toxin producing E.coli1.Causative Agn.pdf
Answer using Giardias, Shiga-toxin producing E.coli
1.Causative Agnet of Disease :
2.Type of Agent ( eg. Virus,bacterium) :
3.Further Classification ( eg.Gram type,viral serotype) :
4.How is disease diagnosed?
5.How is the disease treated?
6.How can the disease be prevented?
7.Distinguishing characteristics of organism (what makes the organism unique such as virulence
factors that are required for disease, unique attributes that aid or prohibit the identificaiton of the
organism ):
Solution
1). Giardiasis is an infectious disease caused by a parasite called Giardia lamblia. It is spread
through eating contaminated food, contact with the infected people, contaminated water and
through pets. This organism is generally found thought the worlds, particularly in the developing
countries..
Abraham Shine recently opaned his own accounting fiem on April 1, whi.pdf
Abraham Shine recently opaned his own accounting fiem on April 1, which he operates as a
corporation. The name of the new entity is Abraham Shine CPA Shine xparienced the following
events during the organizing phase of the new business and ies Srst month of operations in 2018
(Cick the icon to view the transactions) Read the teaments Requirement 1. Analyze the effects of
the events on the accounšing equation of Abraham Shine, CPA Anal ze the events chronolo caly
one transaction at a rme egning with the transaction on the 6th caklate the balance in each
account after analyzing the efect of the transaction on the account g equation Aher calculating
the ending balance of each account on the 30m caloulate total assets and totall Babilities and
equity (Complete only the nacessary answer boxes for your transaction ines Do not enter any
zeros for your transaction lines 1 Car down all balances to the al line including zero balance
accounts entering \"Q for any zero balances Enter decrease in ar account with 4 minus sign or
pare theses Abbreviations used Ap Accounts Payable A Accounts Receivable. Com Common
Contr “Conttuted. Div . Dividends. Exp . Expense. Fum . Fumitu, Liab. Liablies Rev-Revenue
Sup . Supplies, Uni-Uilties) ASSETS LIAB+ EQUITY Contr. Retained Earnings ServiceRent
Exp Cap Cash + AR + Ofice + Furn AP + Com Die UtI Sup Stock Rev Exp Enter any inumber
in the edit fhelds and then click Check Answer 13-taring Clear Al Check Answer
Solution
Answer
Assets
=
Liabilities
+
Equity
Contributed Capital
+
Retained earnings
Date
Cash
+
A/R
+
Office Sup
+
Furn
=
A/P
+
Com Stock
-
Div
+
Service Revenue
-
Rent Exp
-
Util Exp
Apr-05
80000
+
+
+
=
+
80000
-
+
-
-
Apr-06
-200
+
+
200
+
=
+
-
+
-
-
Apr-07
+
+
+
7500
=
7500
+
-
+
-
-
Apr-10
3500
+
+
+
=
+
-
+
3500
-
-
Apr-11
-390
+
+
+
=
+
-
+
-
-
390
Apr-12
+
12000
+
+
=
+
-
+
12000
-
-
Apr-18
-1600
+
+
+
=
+
-
+
-
1600
-
Apr-25
12000
+
-12000
+
+
=
+
-
+
-
-
Apr-27
-7500
+
+
+
=
-7500
+
-
+
-
-
Apr-30
-4000
+
+
+
=
+
-
4000
+
-
-
TOTAL
81810
0
200
7500
0
80000
4000
15500
1600
390
Income
Service Revenue
15500
Expenses
Rent Expense
1600
Utilities expense
390
Total expense
1990
Net Income
13510
Opening Bal
0
Add: Net Income
13510
Less: Dividend paid
4000
9510
Closing balance
9510
A. Owner\'s Equity
Common Stock
80000
Retained Earnings
9510
89510
B. Liabilities
Accounts payable
0
Total Owners\' Equity and Liabilities
89510
C. Assets
Cash
81810
Accounts Receivables
0
Office Supplies
200
Furniture
7500
Total Assets
89510
Assets
=
Liabilities
+
Equity
Contributed Capital
+
Retained earnings
Date
Cash
+
A/R
+
Office Sup
+
Furn
=
A/P
+
Com Stock
-
Div
+
Service Revenue
-
Rent Exp
-
Util Exp
Apr-05
80000
+
+
+
=
+
80000
-
+
-
-
Apr-06
-200
+
+
200
+
=
+
-
+
-
-
Apr-07
+
+
+
7500
=
7500
+
-
+
-
-
Apr-10
3500
+
+
+
=
+
-
+
3500
-
-
Apr-11
-390
+
+
+
=
+
-
+
-
-
390
Apr-12
+
12000
+
+
=
+
-
+
12000
-
-
Apr-18
-1600
+
+
+
=
+
-
+
-
1600
-
Apr-25
12000
+
-12000
+
+
=
+
-
+
-
-
Apr-27
-7500
+
+
+
=
-7500
+
-
+
-
-
Apr-3.
29. This figure shows the principal coronary blood vessels. Which one.pdf
29. This figure shows the principal coronary blood vessels. Which one is the left coronary artery
(LCA)? B. 8 C. 5 D. 3 E. 1 will cause a more severe myocardial infarction (MI) 30. Obstruction
of the than the obstruction of any of the others A. left marginal vein B. left coronary artery
(LCA) C. posterior interventricular vein D. anterior interventricular branch E. circumflex branch
Solution
29. D. 3
30. B. left coronary artery
31. A. cardiac muscle fibers have striations
32. C. Sinatrial (SA) node.
33. B. . The tendinous cords.
34. e. They have about the same endurance as skeletal muscle fibers.
35. SA node, AV node, AV bundle, Purkinje fibers, cardiocyte in LV
36. A. Na+ inflow
37. E. slow CA2+ channels.
14. Suppose I have collected the names of people in several separate.pdf
14. Suppose I have collected the names of people in several separate files based on their starting
letter. The names in the first file begin with A through G, the names in the second file begin with
H through N, the names in the third file beginning with O through T, and the names in the fourth
file contain names beginning with U through Z. None of the files are sorted. instead, the names
that appear in each file occur in any order. Describe in detail how I could take advantage of
multithreading to create one sorted list of all the names. Make sure you include details such as
what your Thread class would do and what your main class that uses the Threads would do. (3
points)
Solution
We can slice strings with negative indexes as well. This IDLE transcript should clarify the rules
for slicing where both indexes are specified:
>>> word = \"PYTHONISCOOL\"
>>> print(word[2:7])
THONI
>>> print(word[6:6])
>>> print(word[4:2])
>>> print(word[5:11])
NISCOO
>>> print(word[8:12])
COOL
>>>
>>> print(word[-7:-2])
NISCO
>>> print(word[-9:8])
HONIS
>>> print(word[-3:-7])
>>>
Notice if you attempt to slice a string where the starting index refers to a position that occurs at
or after the ending index, the slice is the empty string, containing no characters.
A string can also be sliced using only one index. If only one index is given, Python must know if
it’s the start or end of the slice. It assumes that the omitted index must refer to the beginning or
end of the string, accordingly. These examples should clarify slicing using only one index:
>>> print(word[8:])
COOL
>>> print(word[-6:])
ISCOOL
>>> print(word[:6])
PYTHON
>>> print(word[:-4])
PYTHONIS.
4. The size of instructions can be fixed or variable. What are advant.pdf
4. The size of instructions can be fixed or variable. What are advantage and disadvantage of
fixed compared to variable? point) eft 2 20 0 C-4131-28 Add suction 131-26 nvaction 125-2 PC
egister daiess mancion 120-16 date 1 regnter 2 nstructien 31-00 eg stcrs Read U ALLu dela tlata
dvts 02 rstend In the above figure (single-cycle implementation), current instruction is SW S1,
4(S2) (SW: Store Word; current addrss for this instruction is 100 in decimal; S1 and $2 are
initially 8). For redundancy, you should write \"X\" instead of 0 or . You should write the reason
for cach question as well. 5. What is the value of RegDst? (1 point) 6. What is the value of
RegWrite? (I point) 7. What is the value after sign-extension? 1 point) 8. What is the value of
MemRead? (1 point) 9. What is the value of MemtoReg? (1 point) In the above figure, current
instruction is ADDI $1. S2, #-1(which is minus l) (ADDI is ADD Immediate; current address for
this instruction s 100 in decimal and S2 are initially1). For redundancy, you should write “X\',
instead of 0 or 1 . You should write the reason for each question as well. 10. What is the value
(immediate value) of instruction [15-0)? Write the value in 6-bit binary (1 point) 11. What is the
value after sign extension? Write the value in 32-bit binary. (I point) 12. What is the value of
Reg Write? (1 point) 13. What is the value of MemtoReg? (1 point) 2/4
Solution
import java.util.AbstractSet; import java.util.Arrays; import java.util.Iterator; import
java.util.NoSuchElementException; /** * Implementation of an abstract set using an array-
based binary tree. This is * used to help teach binary tree\'s and will have more details explained
in * future lectures. * * @author William J. Collins * @author Matthew Hertz * @param
Data type (which must be Comparable) of the elements in this tree. */ public class
ArrayBinaryTree> extends AbstractSet { /** Entry in the data store where the root node can be
found. */ private static final int ROOT = 0; /** Array used to store the nodes which consist of
this binary tree. */ protected Node[] store; /** Number of elements within the tree. */
protected int size; /** * Initializes this ArrayBinaryTree object to be empty. This creates the
array * in which items will be stored. */ @SuppressWarnings(\"unchecked\") public
ArrayBinaryTree() { store = new Node[63]; size = 0; } /** * Initializes this
ArrayBinaryTree object to contain a shallow copy of a * specified ArrayBinaryTree object.
The worstTime(n) is O(n), where n is the * number of elements in the specified
ArrayBinaryTree object. * * @param otherTree The tree which will be copied to create our
new tree, */ @SuppressWarnings(\"unchecked\") public ArrayBinaryTree(ArrayBinaryTree
otherTree) { store = (Node[]) Arrays.copyOf(otherTree.store, otherTree.store.length); size =
otherTree.size; } public int countLeaves() { } /** * Returns the size of this
ArrayBinaryTree object. * * @return the size of this ArrayBinaryTree object. */
@Overrid.
13. What is the difference between post-consumer recycled materials a.pdf
13. What is the difference between post-consumer recycled materials and post- industrial
recycled content materials?
Solution
Postconsumer material means a material or finished product that has served its intended use and
has been diverted or recovered from waste destined for disposal, having completed its life as a
consumer item.
Post-consumer recycled content is defined as waste materials generated by households or by
commercial, industrial and institutional facilities in their role as end-users of a product, which
can no longer be used for its initial purpose..
1a.MinC and MinD promote even bacterial cell division by suppre.pdf
1a.MinC and MinD promote even bacterial cell division by
suppressing FtsZ polymerization at the cell poles
forming a ring at the midline
promoting FtsZ polymerization at the cell poles
localizing to the septum
1b.Biological membranes (phospholipid bilayers)
are impermeable to protons (H+)
are freely premeable to protons (H+)
transduce energy into different forms
both (a) and (c)
1c.The \"super-germ\" Methicillin-resistant Staphylococcus aureus (MRSA) cannot utilize light
as an energy source nor utilize inorganic carbon sources like CO2, making it a
photoheterotroph
chemoheterotroph
photoautotroph
chemoautotroph
Solution
1a. MinC and MinD promote even bacterial cell division by suppressing FtsZ polymerization at
the cell poles.
1b. Biological membranes areimpermeable to protons (H+)and transduceenergy into different
forms. so correct option is both a and c
1c. .The \"super-germ\" Methicillin-resistant Staphylococcus aureus (MRSA) cannot utilize light
as an energy source nor utilize inorganic carbon sources like CO2, making it a chemoheterotroph.
Biological Screening of Herbal Drugs in detailed.
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
Antifertility, Toxicity studies as per OECD guidelines
More Related Content
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One of your friends has several brothers and sisters, each quite dif.pdf
One of your friends has several brothers and sisters, each quite different in appearance, interests,
and abilities, even though all have the same parents. Explain in detail how these siblings reflect
crossing over and independent assortment.
Solution
Meiosis process increases the genetic diversity in a species. In prophase I, crossing over takes
place and in metaphase I and anaphase I, independent assortment occurs. When homologous
chromosomes are paired together, during meiosis, the chromosomes make contact with the other
pair of chromosomes. This point of contact is responsible for the exchange of genetic
information between the chromosomes and is called the chiasmata. Although the same genetic
information is contained by the two homologous chromosomes, the assortment of the
chromosomes determine which information is present in each of the four gametes produced. For
a particular phenotype, an allele can determine which characteristic will be expressed by the
organism. For example, Chromosome 1 contains an allele for brown eyes while chromosome 2
contains an allele for black eyes. Thus, crossing over and independent assortment are responsible
for different appearence, abilities and interest..
Malpighiaceae luonanthaceae Ochnaceae Flacourtiaceae O Monilophytes G.pdf
Malpighiaceae luonanthaceae Ochnaceae Flacourtiaceae O Monilophytes Gymnosperms O
Basal Angiosperms Monocots Basal Eudicots Asterids Rosids All Plots Escaloniacea
Pittosporaceae Araliaceae Caprifoliaceae Adoxaceae Garry aceae Rubiaceae Apocynaceae
Solution
There are 7 groups present in the legend around the phylogeny.Monophyletic taxon is one that
include a group of organism desecended from a single ancestor.A monophyletic taxon is called a
clade.Monophyletic groups are important to understand how animals are classified. It is
commonly known that all animals evolved over time.
The monophyletics in the legends are Cyatheaceae Commonly known as the scaly tree
ferns,Poaceae,Arecaceae are a botanical family of perennial climbers, shrubs, acaules and trees
commonly known as palm tree,Liliaceae are mostly perennial herbs from starchy rhizomes,
corms, or bulbs comprising about 280 genera and 4,000 species. earthworms, birds, and snakes
all have a common ancestor. Since all these organisms share a common ancestor, they also share
a synapomorphy, which is that earthworms, birds, and snakes all reproduce sexually. These are
all come under phylogeny of monophyletic..
In JavaWrite a program that reads a text file that contains a gra.pdf
In Java:
Write a program that reads a text file that contains a grade (for instance, 87) on each line.
Calculate and print the average of the grades.
Solution
import java.io.BufferedReader;
import java.io.FileReader; // importing filereader
import java.io.IOException; // importing file io exceptions
public class grade {
public static void main(String[] args) {
BufferedReader br = null;
double count=0;
int noOfgrades=0;
try {
String sLine;
br = new BufferedReader(new FileReader(\"sample.txt\")); // reading file using buffer
reader
while ((sLine = br.readLine()) != null) { // rading line by line
noOfgrades=noOfgrades+1; // getting number of grades
int num=Integer.parseInt(sLine); // convert string into integer
count=count+num; // adding each grade to count
}
System.out.println(\"count:\\t\"+count);
System.out.println(\"noOfgrades:\\t\"+noOfgrades);
System.out.println(\"Average of the grades:\\t\"+(count/noOfgrades));
} catch (IOException e) {
e.printStackTrace(); // catch exception
} finally {
try {
if (br != null)br.close(); // closing buffer
} catch (IOException ex) {
ex.printStackTrace();// printing exceptions
}
}
}
}
/********** out file******
fileName:sample.txt
file contains \"
23
34
56
7
88
89
52
34
65
\"
*/
/************output*************
count: 448.0
noOfgrades: 9
Average of the grades: 49.77
*/.
How does Pb and TCE partition between components of soil (mineral, o.pdf
How does Pb and TCE partition between components of soil (mineral, organic matter, air, and
water fractions)? Where can each pollutant can be found if it is spilled on to a loam soil with
about a neutral ph, and about 2% organic matter content?
Solution
lead is heavy metal as a mineral in soil while trichloroethylene is an ogranic pollutant present as
organic matter
lead will be found in as a mineral into the soil while TCE is present in air as organic pollutant..
Hi, please I need help with the correct answer to the above multiple.pdf
Hi, please I need help with the correct answer to the above multiple questions. Thanks. Some
animals, such as snakes, can see infrared waves. The un human eye can not detect these waves,
however, we can aided detect some wavelengths of light. Considering the following statements,
which, if any, would be true? (SELECT ALL TRUE STATEMENTS) O if an object is seen as
red, red light is absorbed by the object and not reflected O Visible light waves exist in the 100-
300nm wavelength range C ight is a form of electromagnetic energy that slows down as it enters
a medium that is less dense O if an object is green, the object is absorbing non-green waves and
reflecting green waves. since humans can not see all wavelengths of the electromagnetic
spectrum, only waves within the visible light spectrum will damage the eye e. UV light can not
damage the retina because we can not see UV light
Solution
1. If an object is green, the object is absorbing non-green waves and reflecting green waves. The
reflected green waves are captured by our eyes and the object appears \'green\'.
The first statement is the exact opposite of this and, hence, incorrect. 100-300 nm wavelengths
correspond to the UV spectrum. Light speeds up in a rarer medium. And, UV light can
(seriously) damage our eyes.
2. Superior colliculus: This part of the brain is responsible for saccades - the process by which
our eyes scan different images of an object with different focus and \'knits\' them together to
perceive useful information.
3. Lens cannot be transplanted! All other options are invalid.
4. Tympanic membrane: Commonly known as the \'ear drum\', this thin membranous structure is
present at the end of the external ear canal. All other structures listed are present beyond the
tympanic membrane. When something such as a Q-tip or earbud is inserted too deep in the ear
canal, it will first perforate the tympanic membrane before injuring any other part..
For which phyla are megaspores a shared trait Bryophyta Monilophyt.pdf
For which phyla are megaspores a shared trait? Bryophyta Monilophyta Coniferophyta
Anthophyta For which phyla are microspores a shared trait? Bryophyta Monilophyta
Coniferophyta Anthophyta For which phyla are sporophyte a shared trait? Bryophyta
Monilophyta Coniferophyta Q Anthophyta
Solution
13.Bryophta (Bryophytes do not have flowers or seeds and reproduce through spores.)
14.Coniferophyta( A pine tree produces two types of spores- the microspores seen on male cones
and megaspores seen on female cones. Hence, they are heterosporic.).
15.Bryphyta(The sporophyte dries and releases spores that grow into a new generation of
gametophytes. The sporophyte is the organ for asexual reproduction and consists of a foot, seta
and capsule. The capsule is a beak like structure found on top of a thin stalk and produces the
spores. Mosses show alternation of generation between gametophytic and sporophytic
generations though the plant has a gametophyte dominant cycle.).
For Programming Embedded SystemsQ-07 Whcih variable and condition.pdf
For Programming Embedded Systems
Q-07 *Whcih variable and condition is sued to detect unfinished tasks (interrupted interrupt)?
A) Line 28
B) Line 36
C) Line 25
D) Line 22 I. #include \"rims.h\" 2 3 4. typedef struct task state BL LedOff; break; default: int
state; unsigned long period; unsigned long elapsedTime: int (*TickFct) (int); taski state ) I
Transitions switch (state) case BL Ledoff: 11. task tasks [2] break; case BL Ledon: 13. const
unsigned char tasksNum 2; 14. const unsigned long periodBlinkLed = 1500; 15. const unsigned
long periodThreeLeds 500; break; default: break; ) // State actions 17, const unsigned long
tasksPeriodGCD = 500; return state 19. int TickFct BlinkLed (int state) 20. int TickFct
ThreeLeds (int state) 22. unsigned char process!ngRdyTasks = 0; 23. void TimerISR 98. enum
TL States TL TO, TL T1, TL T2 TL State; 100 99. int TickFct ThreeLeds (int state) ( 101
VARIABLES MUST BE DECLARED STATIC/ unsigned char i; if (processingRdyTasks)
/*e.g., static int x = 0; */ /*variables for state machine go here. printf (\"Period too short \ \"): 103
switch (state) case -1: process!ngRdyTasks = 1; for (1 = 0; 1 = 107 108 109 tasks [i].period) case
TL TO: tasks [1]. state = 1 tasks[i].TickFct (tasks[i].state); state = TL T1; tasks[i].elapsedTime0
break; case TL T1: 112 113 tasks [1].elapsedTime += tasksPeriodGCD; processingRdyTasks - 0;
state = TL T2 115 116 38. int main) break; case TL T2: unsigned char i=0; tasks [1]. state =-1;
tasks [1].period = periodBlinkLed; tasks [1] .elapsedTime tasks [1] ·period; tasks [1].TickFct
6T1ckFct BlinkLed; 118 119 120 121 122 123 1 state = TL T0; break; default: state --1; tasks [1]
, state = -1; tasks [1].period periodThreeLeas; tasks [1].elapsedTime = tasks [1].period; tasks
[1].TICkFct = &TICkfct; ThreeLeas; 125 switch (state) 127 128 129 130 case TL TO: get
Timerset (tasksPeriodGCD); break case TL T1: TimerOn () while (1) Sleep return 0 133 135 137
139 h break case TL T2: 61. enum BL States BL Ledoff, BL Ledon BL State; 62. int TickFct
BlinkLed (int state) *VARIABLES MUST BE DECLARED STATIC break; 64. /*e.g., static Int
X = 0; */ 65. /*variables for state machine go here/ default: break switch (state) case-1: return
state state = BL LedOff break; 146 case BL Ledoff: state = BL Ledon; break; case BL Ledon: 1
Solution
c) Line 25 is correct
Line 25 contains if condition and processingRdyTasks variable is used to detect the unfinished
tasks.
For analysis of Thermodynamic Cycle, we draw T-s or P-h diagram. Why.pdf
For analysis of Thermodynamic Cycle, we draw T-s or P-h diagram. Why don’t we draw T-h or
P-t diagram?
Solution
For analysis of thermodynamics cycle we draw various diagrams, In all these diagram T-s,P-h,T-
h and P-T diagram are some of them. Every diagram give some specific informations about
system working like T-s diagram give information about Heat transfer, P-h diagram use in air
conditioning because air is mixture so pressure affect more and enthalphy is measure of energy.
like these two diagram we use P-t diagram use in pure substance to see the transformation at
triple point. and T-h diagram use for visualise how energy change with temprature..
Fill in the blank. A contingency table relates only continuous rando.pdf
Fill in the blank. A contingency table relates only continuous random variables the difference
in the means of two random variables two categories of data a particular response with order in
which that response should be applied
Solution
A contingency table relates
Two categories of data
Contingency table (cross tabulation or crosstab) is a type of table format that displays the
frequency distribution of the two or more categorical variables..
Discuss the dangers associated with hydrogen sulfide in the petro.pdf
** Discuss the dangers associated with hydrogen sulfide in the petrochemical environment.
*** Your original post should be between 100 - 150 words in length.
Solution
1.on humans:Hydrogen sulfide enters your body primarily through the air you breathe. Much
smaller amounts can enter your body through the skin. Hydrogen sulfide is a gas, so you would
not likely be exposed to it by ingestion. When you breathe air containing hydrogen sulfide or
when hydrogen sulfide comes into contact with skin, it is absorbed into the blood stream and
distributed throughout the body.Exposure to low concentrations of hydrogen sulfide may cause
irritation to the eyes, nose, or throat. It may also cause difficulty in breathing for some
asthmatics.Respiratory distress or arrest has been found in people exposed to very high
concentrations of hydrogen sulfide.
The health effects of hydrogen sulfide depend on several factors such as how much hydrogen
sulfide you are exposed to and the length of that exposure. Studies in workers, communities
living near industrial sources of hydrogen sulfide, and volunteers suggest that the respiratory
tract and nervous system are the most sensitive targets of hydrogen sulfide toxicity.It is not
known whether exposure to hydrogen sulfide causes birth defects in humans. The results of
studies in animals suggest that exposure to low concentrations of hydrogen sulfide during
pregnancy does not cause birth defects.
Exposure to low concentrations of hydrogen sulfide may cause headaches, poor memory,
tiredness, and balance problems.Brief exposures to high concentrations of hydrogen sulfide
(greater than 500 ppm) can cause a loss of consciousness. In most cases, the person appears to
regain consciousness without any other effects. However, in some individuals, there may be
permanent or long-term effects such as headaches, poor attention span, poor memory, and poor
motor function..
Complete the following paragraph to describe the various reproductive.pdf
Complete the following paragraph to describe the various reproductive strategies seen in the
phylum Ascomycota. Some choices may be used more than once.
Solution
1. The Ascomycota contains two main groups; Asexual sac fungi which include bakers yeast and
Sexual sac fungi, which include candida yeast and molds.
2. Most Ascomyta are conidospores and reproduced by meiosis, as seen in candida yeast, or
through the production of fruiting body as seen in molds.
3. In contrast in haploid a(n) diploid containing two nuclei develops within a(n) Conidiospores
.
4. The nuclei fuse creating haploid cell that undergose mitosis to produce diploid cells.
5. Typically at this point, the ascospores swells and burst, releasing the ascus in to the air to
germinate once landed..
CommunicationsSystems have four basic properties (holism, equi-fi.pdf
Communications:
Systems have four basic properties (holism, equi-finality, negative entropy). List the principle
behind each property. What does it mean that “a system is more than the sum of its parts?” How
does this concept apply to organizations?
Solution
Four basic properties of systems are:
Principle of Holism:
It emphasizes that system is more than sum of its part due to interdependence upon each other.
Principle of Equi-finality:
It states that outcome can be achieved through many different paths due to interdependence of
components.
Principle of Negative Entropy:
It is based upon openness of the system that says that system owns the ability to prevent
deterioration and thrive.
Principle of Requisite Variety:
It is also based upon openness of the system and it says that internal complexity should be
maintained to deal with external complexity.
Statement of “a system is more than the sum of its parts” means that system achieves more when
all of its parts complement each other and work together. Each component / part of the system
cannot do much in isolation but they can achieve big goals when grouped together as a team.
It is very important in organizational perspective as it consists of many different functions as
components. When, these functions work together, organizational objective is achieved. If not,
organization falls apart..
Based on your reading of the GCU introduction and the textbooks, wha.pdf
Based on your reading of the GCU introduction and the textbooks, what is the main difference
between the modern western worldview and the postmodern worldview? Which parts, if any, do
you identify with? Explain. Cite references from your reading to support your answer.
Solution
Postmodernism is a response against modernism. According to everyone has a point of view
through which they see the world.
Premodern (the worldview before modernism)- says that natural events are caused by
supernatural causes. God, angels, demons, and the like are a reality to these folks..
Answer using Giardias, Shiga-toxin producing E.coli1.Causative Agn.pdf
Answer using Giardias, Shiga-toxin producing E.coli
1.Causative Agnet of Disease :
2.Type of Agent ( eg. Virus,bacterium) :
3.Further Classification ( eg.Gram type,viral serotype) :
4.How is disease diagnosed?
5.How is the disease treated?
6.How can the disease be prevented?
7.Distinguishing characteristics of organism (what makes the organism unique such as virulence
factors that are required for disease, unique attributes that aid or prohibit the identificaiton of the
organism ):
Solution
1). Giardiasis is an infectious disease caused by a parasite called Giardia lamblia. It is spread
through eating contaminated food, contact with the infected people, contaminated water and
through pets. This organism is generally found thought the worlds, particularly in the developing
countries..
Abraham Shine recently opaned his own accounting fiem on April 1, whi.pdf
Abraham Shine recently opaned his own accounting fiem on April 1, which he operates as a
corporation. The name of the new entity is Abraham Shine CPA Shine xparienced the following
events during the organizing phase of the new business and ies Srst month of operations in 2018
(Cick the icon to view the transactions) Read the teaments Requirement 1. Analyze the effects of
the events on the accounšing equation of Abraham Shine, CPA Anal ze the events chronolo caly
one transaction at a rme egning with the transaction on the 6th caklate the balance in each
account after analyzing the efect of the transaction on the account g equation Aher calculating
the ending balance of each account on the 30m caloulate total assets and totall Babilities and
equity (Complete only the nacessary answer boxes for your transaction ines Do not enter any
zeros for your transaction lines 1 Car down all balances to the al line including zero balance
accounts entering \"Q for any zero balances Enter decrease in ar account with 4 minus sign or
pare theses Abbreviations used Ap Accounts Payable A Accounts Receivable. Com Common
Contr “Conttuted. Div . Dividends. Exp . Expense. Fum . Fumitu, Liab. Liablies Rev-Revenue
Sup . Supplies, Uni-Uilties) ASSETS LIAB+ EQUITY Contr. Retained Earnings ServiceRent
Exp Cap Cash + AR + Ofice + Furn AP + Com Die UtI Sup Stock Rev Exp Enter any inumber
in the edit fhelds and then click Check Answer 13-taring Clear Al Check Answer
Solution
Answer
Assets
=
Liabilities
+
Equity
Contributed Capital
+
Retained earnings
Date
Cash
+
A/R
+
Office Sup
+
Furn
=
A/P
+
Com Stock
-
Div
+
Service Revenue
-
Rent Exp
-
Util Exp
Apr-05
80000
+
+
+
=
+
80000
-
+
-
-
Apr-06
-200
+
+
200
+
=
+
-
+
-
-
Apr-07
+
+
+
7500
=
7500
+
-
+
-
-
Apr-10
3500
+
+
+
=
+
-
+
3500
-
-
Apr-11
-390
+
+
+
=
+
-
+
-
-
390
Apr-12
+
12000
+
+
=
+
-
+
12000
-
-
Apr-18
-1600
+
+
+
=
+
-
+
-
1600
-
Apr-25
12000
+
-12000
+
+
=
+
-
+
-
-
Apr-27
-7500
+
+
+
=
-7500
+
-
+
-
-
Apr-30
-4000
+
+
+
=
+
-
4000
+
-
-
TOTAL
81810
0
200
7500
0
80000
4000
15500
1600
390
Income
Service Revenue
15500
Expenses
Rent Expense
1600
Utilities expense
390
Total expense
1990
Net Income
13510
Opening Bal
0
Add: Net Income
13510
Less: Dividend paid
4000
9510
Closing balance
9510
A. Owner\'s Equity
Common Stock
80000
Retained Earnings
9510
89510
B. Liabilities
Accounts payable
0
Total Owners\' Equity and Liabilities
89510
C. Assets
Cash
81810
Accounts Receivables
0
Office Supplies
200
Furniture
7500
Total Assets
89510
Assets
=
Liabilities
+
Equity
Contributed Capital
+
Retained earnings
Date
Cash
+
A/R
+
Office Sup
+
Furn
=
A/P
+
Com Stock
-
Div
+
Service Revenue
-
Rent Exp
-
Util Exp
Apr-05
80000
+
+
+
=
+
80000
-
+
-
-
Apr-06
-200
+
+
200
+
=
+
-
+
-
-
Apr-07
+
+
+
7500
=
7500
+
-
+
-
-
Apr-10
3500
+
+
+
=
+
-
+
3500
-
-
Apr-11
-390
+
+
+
=
+
-
+
-
-
390
Apr-12
+
12000
+
+
=
+
-
+
12000
-
-
Apr-18
-1600
+
+
+
=
+
-
+
-
1600
-
Apr-25
12000
+
-12000
+
+
=
+
-
+
-
-
Apr-27
-7500
+
+
+
=
-7500
+
-
+
-
-
Apr-3.
29. This figure shows the principal coronary blood vessels. Which one.pdf
29. This figure shows the principal coronary blood vessels. Which one is the left coronary artery
(LCA)? B. 8 C. 5 D. 3 E. 1 will cause a more severe myocardial infarction (MI) 30. Obstruction
of the than the obstruction of any of the others A. left marginal vein B. left coronary artery
(LCA) C. posterior interventricular vein D. anterior interventricular branch E. circumflex branch
Solution
29. D. 3
30. B. left coronary artery
31. A. cardiac muscle fibers have striations
32. C. Sinatrial (SA) node.
33. B. . The tendinous cords.
34. e. They have about the same endurance as skeletal muscle fibers.
35. SA node, AV node, AV bundle, Purkinje fibers, cardiocyte in LV
36. A. Na+ inflow
37. E. slow CA2+ channels.
14. Suppose I have collected the names of people in several separate.pdf
14. Suppose I have collected the names of people in several separate files based on their starting
letter. The names in the first file begin with A through G, the names in the second file begin with
H through N, the names in the third file beginning with O through T, and the names in the fourth
file contain names beginning with U through Z. None of the files are sorted. instead, the names
that appear in each file occur in any order. Describe in detail how I could take advantage of
multithreading to create one sorted list of all the names. Make sure you include details such as
what your Thread class would do and what your main class that uses the Threads would do. (3
points)
Solution
We can slice strings with negative indexes as well. This IDLE transcript should clarify the rules
for slicing where both indexes are specified:
>>> word = \"PYTHONISCOOL\"
>>> print(word[2:7])
THONI
>>> print(word[6:6])
>>> print(word[4:2])
>>> print(word[5:11])
NISCOO
>>> print(word[8:12])
COOL
>>>
>>> print(word[-7:-2])
NISCO
>>> print(word[-9:8])
HONIS
>>> print(word[-3:-7])
>>>
Notice if you attempt to slice a string where the starting index refers to a position that occurs at
or after the ending index, the slice is the empty string, containing no characters.
A string can also be sliced using only one index. If only one index is given, Python must know if
it’s the start or end of the slice. It assumes that the omitted index must refer to the beginning or
end of the string, accordingly. These examples should clarify slicing using only one index:
>>> print(word[8:])
COOL
>>> print(word[-6:])
ISCOOL
>>> print(word[:6])
PYTHON
>>> print(word[:-4])
PYTHONIS.
4. The size of instructions can be fixed or variable. What are advant.pdf
4. The size of instructions can be fixed or variable. What are advantage and disadvantage of
fixed compared to variable? point) eft 2 20 0 C-4131-28 Add suction 131-26 nvaction 125-2 PC
egister daiess mancion 120-16 date 1 regnter 2 nstructien 31-00 eg stcrs Read U ALLu dela tlata
dvts 02 rstend In the above figure (single-cycle implementation), current instruction is SW S1,
4(S2) (SW: Store Word; current addrss for this instruction is 100 in decimal; S1 and $2 are
initially 8). For redundancy, you should write \"X\" instead of 0 or . You should write the reason
for cach question as well. 5. What is the value of RegDst? (1 point) 6. What is the value of
RegWrite? (I point) 7. What is the value after sign-extension? 1 point) 8. What is the value of
MemRead? (1 point) 9. What is the value of MemtoReg? (1 point) In the above figure, current
instruction is ADDI $1. S2, #-1(which is minus l) (ADDI is ADD Immediate; current address for
this instruction s 100 in decimal and S2 are initially1). For redundancy, you should write “X\',
instead of 0 or 1 . You should write the reason for each question as well. 10. What is the value
(immediate value) of instruction [15-0)? Write the value in 6-bit binary (1 point) 11. What is the
value after sign extension? Write the value in 32-bit binary. (I point) 12. What is the value of
Reg Write? (1 point) 13. What is the value of MemtoReg? (1 point) 2/4
Solution
import java.util.AbstractSet; import java.util.Arrays; import java.util.Iterator; import
java.util.NoSuchElementException; /** * Implementation of an abstract set using an array-
based binary tree. This is * used to help teach binary tree\'s and will have more details explained
in * future lectures. * * @author William J. Collins * @author Matthew Hertz * @param
Data type (which must be Comparable) of the elements in this tree. */ public class
ArrayBinaryTree> extends AbstractSet { /** Entry in the data store where the root node can be
found. */ private static final int ROOT = 0; /** Array used to store the nodes which consist of
this binary tree. */ protected Node[] store; /** Number of elements within the tree. */
protected int size; /** * Initializes this ArrayBinaryTree object to be empty. This creates the
array * in which items will be stored. */ @SuppressWarnings(\"unchecked\") public
ArrayBinaryTree() { store = new Node[63]; size = 0; } /** * Initializes this
ArrayBinaryTree object to contain a shallow copy of a * specified ArrayBinaryTree object.
The worstTime(n) is O(n), where n is the * number of elements in the specified
ArrayBinaryTree object. * * @param otherTree The tree which will be copied to create our
new tree, */ @SuppressWarnings(\"unchecked\") public ArrayBinaryTree(ArrayBinaryTree
otherTree) { store = (Node[]) Arrays.copyOf(otherTree.store, otherTree.store.length); size =
otherTree.size; } public int countLeaves() { } /** * Returns the size of this
ArrayBinaryTree object. * * @return the size of this ArrayBinaryTree object. */
@Overrid.
13. What is the difference between post-consumer recycled materials a.pdf
13. What is the difference between post-consumer recycled materials and post- industrial
recycled content materials?
Solution
Postconsumer material means a material or finished product that has served its intended use and
has been diverted or recovered from waste destined for disposal, having completed its life as a
consumer item.
Post-consumer recycled content is defined as waste materials generated by households or by
commercial, industrial and institutional facilities in their role as end-users of a product, which
can no longer be used for its initial purpose..
1a.MinC and MinD promote even bacterial cell division by suppre.pdf
1a.MinC and MinD promote even bacterial cell division by
suppressing FtsZ polymerization at the cell poles
forming a ring at the midline
promoting FtsZ polymerization at the cell poles
localizing to the septum
1b.Biological membranes (phospholipid bilayers)
are impermeable to protons (H+)
are freely premeable to protons (H+)
transduce energy into different forms
both (a) and (c)
1c.The \"super-germ\" Methicillin-resistant Staphylococcus aureus (MRSA) cannot utilize light
as an energy source nor utilize inorganic carbon sources like CO2, making it a
photoheterotroph
chemoheterotroph
photoautotroph
chemoautotroph
Solution
1a. MinC and MinD promote even bacterial cell division by suppressing FtsZ polymerization at
the cell poles.
1b. Biological membranes areimpermeable to protons (H+)and transduceenergy into different
forms. so correct option is both a and c
1c. .The \"super-germ\" Methicillin-resistant Staphylococcus aureus (MRSA) cannot utilize light
as an energy source nor utilize inorganic carbon sources like CO2, making it a chemoheterotroph.
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