Tree: inin A recursave tree data structure. Note the retationship between this class and
Recursivelist; the only major difference is that rest has been replaced by _ subtrees to handte
multiple recursive sub-parts. Hen \# wa Private Attributes =m= \# The item stored at this tree's
root, or None if the tree is enpty. root: optional [Anyl \# The list of alt subtrees of this tree. -
subtrees: List [Tree] \# mum Representation Invariants === \# - If self. root is None then self.
subtrees is an enpty list. This setting of attributes represents an enpty Tree. Note: self. _ subtrees
may be empty when selt._root is not None. This setting of attributes represents a tree consisting
of just one node. def swap_down(setf) None: "*=Swap the root of this Tree with the frargest of
its chitdren until the originat root's value is at a position where it's larger than or equal to all of its
children. In the case of a tie, swap with the one that cones first in subtrees. >t= Tree(1,
[1)tsswapdownt) print(t) \& No swaps are made 1 t._subtrees =[Tree(2,[1]. Tree (3, [1)] print(t)
1 3233> t. 5wapdown() ss print (t) in 1 swapped with 3 3 2 1 t-large
=Tree(3,ITree(5,ITree(7,tTree(2,11), Tree(1, [1]) 1), 1) Tree (4, [1] 1]> print (t __targe) 3 7 2 1
mos targe-swap_down( ) print(t_large) =3 swapped with 5 , and then with 7 .
\# The beton are methods given to you. Do NoT change these. def init_(self, root: Any, subtrees:
List (Treel) None: wanitiatize a new Tree with the given root vatue and subtrees. If eroots is
None, the tree is eapty. Preconditiont if sroot> is None, then ksubtreess is enpty. self, _root =root
self.-subtrees = subtrees def is empty(self boolt nHinReturn True if this tree is enty. t1=Trec(
None, [1)t1.15 _etopty() True t2=Tree(3,11) t2.1s_empty () Fatise winl return self,_root is None
det _str (sett) str:
\# The below are nethods given to you. Do NoT change these. def _init (self, root: Any, subtrees:
List(TreeJ) Wonet "anitialize a new Tree with the given root value and subtrees. If eroot> is
None, the tree is empty. Precondition: if is None, then esubtrees> is empty. ren self root = root
self._subtrees: = subtrees def is empty (self) bool: wawpeturn True if this tree is empty. tI= Tree
(None, []) t1,is_empty () True >t2=Tree(3, [1) > t2.is_empty ( ) False a return self. root is None
def _str_ ( self f str: m*Return a string representation of this: tree. For each node, its iten is
printed before any of its descendants' items. The output is nicely indented. You may find this
method helpful for debugging. return self._str_indented () , steip( ) def str_indented(setf, depth:
int =0) str: in*peturn an indented string representation of this-tree. The indentation level is:
specified by the : return e tiset 5=1+ depth + str (set( root )+ ' ln3 for subtree in self . subtrees: *
Note that the "depth' argument to the recursive catl is \# modified. 5+= subtree.
_str_indented(depth + 1) returnis if nane _ingort doctest doctest. testmod (1) import python_ta.
class Tree:
"""A recursive tree data structure.
Note the relationship between this class and RecursiveList; the only major
difference is that _rest has been replaced by _subtrees to handle multiple
recursive sub-parts.
"""
# === Private Attributes ===
# The item stored at this tree's root, or None if the tree is empty.
_root: Optional[Any]
# The list of all subtrees of this tree.
_subtrees: List[Tree]
# === Representation Invariants ===
# - If self._root is None then self._subtrees is an empty list.
# This setting of attributes represents an empty Tree.
#
# Note: self._subtrees may be empty when self._root is not None.
# This setting of attributes represents a tree consisting of just one
# node.
def swap_down(self) -> None:
"""Swap the root of this Tree with the largest of its children until
the original root's value is at a position where it's larger than or
equal to all of its children.
In the case of a tie, swap with the one that comes first in _subtrees.
>>> t = Tree(1, [])
>>> t.swap_down()
>>> print(t) # No swaps are made
1
>>> t._subtrees = [Tree(2, []), Tree(3, [])]
>>> print(t)
1
2
3
>>> t.swap_down()
>>> print(t) # 1 swapped with 3
3
2
1
>>> t_large = Tree(3, [Tree(5, [Tree(7, [Tree(2, []), Tree(1, [])])]), \
Tree(4, [])])
>>> print(t_large)
3
5
7
2
1
4
>>> t_large.swap_down()
>>> print(t_large) # 3 swapped with 5, and then with 7
5
7
3
2
1
4
"""
############################################################################(cid:13)(cid:10)
# The below are methods given to you. Do NOT change these.
############################################################################(cid:13)(cid:10)
def __init__(self, root: Any, subtrees: List[Tree]) -> None:
"""Initialize a new Tree with the given root value and subtrees.
If is None, the tree is empty.
Precondition: if is None, then is empty.
"""
self._root = root
self._subtrees = subtrees
def is_empty(self) -> bool:
"""Return True if this tree is empty.
>>> t1 = Tree(None, [])
>>> t1.is_empty()
True
>>> t2 = Tree(3, [])
>>> t2.is_empty()
False
"""
return self._root is None
def __str__(self) -> str:
"""Return a string representation of this tree.
For each node, its item is printed before any of its
descendants' items. The output is nicely indented.
You may find this method helpful for debugging.
"""
return self._str_indented().strip()
def _str_indented(self, depth: int = 0) -> str:
"""Return an indented string representation of this tree.
The indentation level is specified by the parameter.
"""
if self.is_empty():
return ''
else:
s = ' ' * depth + str(self._root) + '\n'
for subtree in self._subtrees:
# Note that the 'depth' argument to the recursive call is
# modified.
s += subtree._str_indented(depth + 1)
return s
* The below are methods given to you. Do NOT change these. def __init__(self, root: Any,
subtrees: List[Tree]) None: "n"Initialize a new Tree with the given root value and subtrees. If
root> is None, the tree is empty. Precondition: if is None, then is empty. nin self._.root = root
s.
Tree and Binary search tree in data structure.
The complete explanation of working of trees and Binary Search Tree is given. It is discussed such a way that everyone can easily understand it. Trees have great role in the data structures.
class Tree:
"""A recursive tree data structure.
Note the relationship between this class and RecursiveList; the only major
difference is that _rest has been replaced by _subtrees to handle multiple
recursive sub-parts.
"""
# === Private Attributes ===
# The item stored at this tree's root, or None if the tree is empty.
_root: Optional[Any]
# The list of all subtrees of this tree.
_subtrees: List[Tree]
# === Representation Invariants ===
# - If self._root is None then self._subtrees is an empty list.
# This setting of attributes represents an empty Tree.
#
# Note: self._subtrees may be empty when self._root is not None.
# This setting of attributes represents a tree consisting of just one
# node.
def swap_down(self) -> None:
"""Swap the root of this Tree with the largest of its children until
the original root's value is at a position where it's larger than or
equal to all of its children.
In the case of a tie, swap with the one that comes first in _subtrees.
>>> t = Tree(1, [])
>>> t.swap_down()
>>> print(t) # No swaps are made
1
>>> t._subtrees = [Tree(2, []), Tree(3, [])]
>>> print(t)
1
2
3
>>> t.swap_down()
>>> print(t) # 1 swapped with 3
3
2
1
>>> t_large = Tree(3, [Tree(5, [Tree(7, [Tree(2, []), Tree(1, [])])]), \
Tree(4, [])])
>>> print(t_large)
3
5
7
2
1
4
>>> t_large.swap_down()
>>> print(t_large) # 3 swapped with 5, and then with 7
5
7
3
2
1
4
"""
############################################################################(cid:13)(cid:10)
# The below are methods given to you. Do NOT change these.
############################################################################(cid:13)(cid:10)
def __init__(self, root: Any, subtrees: List[Tree]) -> None:
"""Initialize a new Tree with the given root value and subtrees.
If is None, the tree is empty.
Precondition: if is None, then is empty.
"""
self._root = root
self._subtrees = subtrees
def is_empty(self) -> bool:
"""Return True if this tree is empty.
>>> t1 = Tree(None, [])
>>> t1.is_empty()
True
>>> t2 = Tree(3, [])
>>> t2.is_empty()
False
"""
return self._root is None
def __str__(self) -> str:
"""Return a string representation of this tree.
For each node, its item is printed before any of its
descendants' items. The output is nicely indented.
You may find this method helpful for debugging.
"""
return self._str_indented().strip()
def _str_indented(self, depth: int = 0) -> str:
"""Return an indented string representation of this tree.
The indentation level is specified by the parameter.
"""
if self.is_empty():
return ''
else:
s = ' ' * depth + str(self._root) + '\n'
for subtree in self._subtrees:
# Note that the 'depth' argument to the recursive call is
# modified.
s += subtree._str_indented(depth + 1)
return s
* The below are methods given to you. Do NOT change these. def __init__(self, root: Any,
subtrees: List[Tree]) None: "n"Initialize a new Tree with the given root value and subtrees. If
root> is None, the tree is empty. Precondition: if is None, then is empty. nin self._.root = root
s.
Tree and Binary search tree in data structure.
The complete explanation of working of trees and Binary Search Tree is given. It is discussed such a way that everyone can easily understand it. Trees have great role in the data structures.
Short overview of balance tree: Data Structures, Binary search tree, BST Problem, Self Balancing BST, Usage, Red Black Trees, Red Black Insertion, AVL Tree, Rotations, B-Tree
date structure have a great application .Trees are basic important part of this structure . all basic discreption is provided in this presentation . data structure and trees theory .
types of binary search,representation of trees , implementation of trees , important algorithm , extended binary tree , linked list representation of trees. orders in binary trees .
Once you have all the structures working as intended- it is time to co.docxfarrahkur54
Once you have all the structures working as intended, it is time to compare the
performances of the 2. Use runBenchmark() to start.
a. First, generate an increasing number (N) of random integers (1000, 5000, 10000,
50000, 75000, 100000, 500000 or as far as your computing power will let you)
i. Time and print how long it takes to insert the random integers into an initially
empty BST. Do not print the tree.
You can get a random list using the java class Random, located in
java.util.Random. To test the speed of the insertion algorithm, you should use
the System.currentTimeMillis() method, which returns a long that contains the
current time (in milliseconds). Call System.currentTimeMillis() before and after
the algorithm runs and subtract the two times. (Instant.now() is an alternative
way of recording the time.)
ii. Time and print how long it takes to insert the same random integers into an
initially empty AVL tree. Do not print the tree.
iii. If T(N) is the time function, how does the growth of TBST(N) compare with the
growth of TAVL(N)?
Plot a simple graph to of time against N for the two types of BSTs to visualize
your results.
b. Second, generate a list of k random integers. k is also some large value.
i. Time how long it takes to search your various N-node BSTs for all k random
integers. It does not matter whether the search succeeds.
ii. Time how long it takes to search your N-node AVL trees for the same k random
integers.
iii. Compare the growth rates of these two search time-functions with a graph the
same way you did in part a.
public class BinarySearchTree<AnyType extends Comparable<? super AnyType>> {
protected BinaryNode<AnyType> root;
public BinarySearchTree() {
root = null;
}
/**
* Insert into the tree; duplicates are ignored.
*
* @param x the item to insert.
* @param root
* @return
*/
protected BinaryNode<AnyType> insert(AnyType x, BinaryNode<AnyType> root) {
// If the root is null, we've reached an empty leaf node, so we create a new node
// with the value x and return it
if (root == null) {
return new BinaryNode<>(x, null, null);
}
// Compare the value of x to the value stored in the root node
int compareResult = x.compareTo(root.element);
// If x is less than the value stored in the root node, we insert it into the left subtree
if (compareResult < 0) {
root.left = insert(x, root.left);
}
// If x is greater than the value stored in the root node, we insert it into the right subtree
else if (compareResult > 0) {
root.right = insert(x, root.right);
}
// If x is equal to the value stored in the root node, we ignore it since the tree does not allow duplicates
return root;
}
/**
* Counts the number of leaf nodes in this tree.
*
* @param t The root of the tree.
* @return
*/
private int countLeafNodes(BinaryNode<AnyType> root) {
// If the root is null, it means the tree is empty, so we return 0
if (root == null) {
return 0;
}
// If the root has no children, it means it is a leaf node, so we return 1
if (root.left == .
Two babies in a maternity ward have lost their identity bands and oth.pdfamazingkidstoys
Two babies in a maternity ward have lost their identity bands and other paperwork was lost.
Baby \#1 is type A and baby \#2 is type B. Betty is one of the mothers and she has type O. Which
of the following responses will you give to Betty based only on the blood types? The type B
baby belongs to Betty Either baby could belong to Betty The type A baby belongs to Betty.
Neither baby could be hers. Incorrect. Remember, her genotype is "ii".
Twenty-two true-false statements appear below. You must answer all of.pdfamazingkidstoys
Twenty-two true-false statements appear below. You must answer all of them. For each
statements you answer, insert "T" or "F" in the appropriate place on the separate answer sheet. 1.
Jury duty pay is excluded from gross income. 2. A item of small value, a Christmas ham for
example, given by an employer to an employee must be reported as gross income by the
recipient employee. 3. A compensatory damages payment received by a taxpayer as a lump sum
on account of sickness will be included in the taxpayer's gross income. 4. A disabled spouse, age
27 , is eligible to be a qualifying person under an employer maintained dependent care assistance
program. 5. If a dependent child receives Social Security benefits, the benefits, if taxable, will be
taxable to the taxpayer(s) claiming the dependency deduction. 6. Payments received under
workers' compensation acts are generally included in gross income. 7. Per the constructive
receipt doctrine, a cash basis taxpayer receives income when it is credited to a taxpayer's
account, or set apart for him/her, or otherwise made available for him/her to draw upon at any
time. 8. A tenant renting business-use property who reports on the cash basis of accounting is not
allowed to claim a deduction for a payment of rent in advance, but must wait until the year in
which the payment was due to claim the deduction. 9. Security deposits received by a landlord
from tenants are not picked up in the landlord's gross income unless they are to be returned to the
tenant upon termination of the lease. 10. An amount paid by the Veterans Administration to a
family member of a military disabled veteran as a subsistence allowance is excluded from gross
income. 11. An employee achievement award (under a nonqualified plan) in the amount of
$1,600 to employee Anne Able would not represent taxable gross income to Anne, 12. Gross
income may be received via cash, receivables, property, and land, but not any any other form of
economic benefit. 13. Payments received by a landlord from a tenant in exchange for a
modification of the lease are not included in the gross income of the landlord..
ts.) Recall the case study entitled Not Exactly... The Complexity o.pdfamazingkidstoys
ts.) Recall the case study entitled "Not Exactly..." The Complexity of a Human Genetic Disease.
Explain the meaning of the title of the case. Be specific. Use this case to explain how a gene can
have many alleles within a population, not just a single dominant and recessive allele as is often
modeled (such as the C and c that would be the novice way of explaining the inheritance of
cystic fibrosis). Return to the allelic variant section of the OMIM (omim.org) record for cystic
fibrosis. Choose three specific mutations (allelic variants) that were not listed in the case study.
Provide the abbreviation for the mutations as listed in the database and explain specifically the
effect of the mutation. Review question 8 in your case study for guidance but do not use those
mutations as responses. c. In addition to a gene having multiple alleles, there are many other
extensions to Mendelian inheritance. Describe the following extensions in your own words and
provide a specific example to support your response. i. incomplete dominance ii. codominance d.
Explain in your own words what it means that recessive disorders are typically loss of function
mutations, but dominant mutations are typically gain of function mutations..
Tsunami Ltd. wanted to acquire WiseCom, and tried to ensure their cor.pdfamazingkidstoys
Tsunami Ltd. wanted to acquire WiseCom, and tried to ensure their corporate cultures were
compatible. WiseCom executives welcomed the acquisition and eagerly described their
perceptions of the company's culture. Tsunami also relied heavily on WiseCom's website and
public relations statements to decipher WiseCom's culture. The information gathered by Tsunami
is most likely measuring WiseCom's: Enacted values Assumptions Counterculture Socialization
process Espoused values.
True or false Opisthokonts are a monophyletic group (a clade) True F.pdfamazingkidstoys
True or false: Opisthokonts are a monophyletic group (a clade) True False Question 46 Why is
Chemoheterotrophy NOT a good synapomorphy of Metazoa (Animalia)? Most animals are
actually autotrophs Choanoflageilates that are NOT animals also are chemoheterotrophis The
most recent (last) common ancestor of all Metazoa was a photoautotroph It is a good
synapomorphy, as it is something that all animals have.
More Related Content
Similar to Tree inin A recursave tree data structure. Note the retationship bet.pdf
Short overview of balance tree: Data Structures, Binary search tree, BST Problem, Self Balancing BST, Usage, Red Black Trees, Red Black Insertion, AVL Tree, Rotations, B-Tree
date structure have a great application .Trees are basic important part of this structure . all basic discreption is provided in this presentation . data structure and trees theory .
types of binary search,representation of trees , implementation of trees , important algorithm , extended binary tree , linked list representation of trees. orders in binary trees .
Once you have all the structures working as intended- it is time to co.docxfarrahkur54
Once you have all the structures working as intended, it is time to compare the
performances of the 2. Use runBenchmark() to start.
a. First, generate an increasing number (N) of random integers (1000, 5000, 10000,
50000, 75000, 100000, 500000 or as far as your computing power will let you)
i. Time and print how long it takes to insert the random integers into an initially
empty BST. Do not print the tree.
You can get a random list using the java class Random, located in
java.util.Random. To test the speed of the insertion algorithm, you should use
the System.currentTimeMillis() method, which returns a long that contains the
current time (in milliseconds). Call System.currentTimeMillis() before and after
the algorithm runs and subtract the two times. (Instant.now() is an alternative
way of recording the time.)
ii. Time and print how long it takes to insert the same random integers into an
initially empty AVL tree. Do not print the tree.
iii. If T(N) is the time function, how does the growth of TBST(N) compare with the
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Plot a simple graph to of time against N for the two types of BSTs to visualize
your results.
b. Second, generate a list of k random integers. k is also some large value.
i. Time how long it takes to search your various N-node BSTs for all k random
integers. It does not matter whether the search succeeds.
ii. Time how long it takes to search your N-node AVL trees for the same k random
integers.
iii. Compare the growth rates of these two search time-functions with a graph the
same way you did in part a.
public class BinarySearchTree<AnyType extends Comparable<? super AnyType>> {
protected BinaryNode<AnyType> root;
public BinarySearchTree() {
root = null;
}
/**
* Insert into the tree; duplicates are ignored.
*
* @param x the item to insert.
* @param root
* @return
*/
protected BinaryNode<AnyType> insert(AnyType x, BinaryNode<AnyType> root) {
// If the root is null, we've reached an empty leaf node, so we create a new node
// with the value x and return it
if (root == null) {
return new BinaryNode<>(x, null, null);
}
// Compare the value of x to the value stored in the root node
int compareResult = x.compareTo(root.element);
// If x is less than the value stored in the root node, we insert it into the left subtree
if (compareResult < 0) {
root.left = insert(x, root.left);
}
// If x is greater than the value stored in the root node, we insert it into the right subtree
else if (compareResult > 0) {
root.right = insert(x, root.right);
}
// If x is equal to the value stored in the root node, we ignore it since the tree does not allow duplicates
return root;
}
/**
* Counts the number of leaf nodes in this tree.
*
* @param t The root of the tree.
* @return
*/
private int countLeafNodes(BinaryNode<AnyType> root) {
// If the root is null, it means the tree is empty, so we return 0
if (root == null) {
return 0;
}
// If the root has no children, it means it is a leaf node, so we return 1
if (root.left == .
Two babies in a maternity ward have lost their identity bands and oth.pdfamazingkidstoys
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example, given by an employer to an employee must be reported as gross income by the
recipient employee. 3. A compensatory damages payment received by a taxpayer as a lump sum
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27 , is eligible to be a qualifying person under an employer maintained dependent care assistance
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taxable to the taxpayer(s) claiming the dependency deduction. 6. Payments received under
workers' compensation acts are generally included in gross income. 7. Per the constructive
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time. 8. A tenant renting business-use property who reports on the cash basis of accounting is not
allowed to claim a deduction for a payment of rent in advance, but must wait until the year in
which the payment was due to claim the deduction. 9. Security deposits received by a landlord
from tenants are not picked up in the landlord's gross income unless they are to be returned to the
tenant upon termination of the lease. 10. An amount paid by the Veterans Administration to a
family member of a military disabled veteran as a subsistence allowance is excluded from gross
income. 11. An employee achievement award (under a nonqualified plan) in the amount of
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Explain the meaning of the title of the case. Be specific. Use this case to explain how a gene can
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modeled (such as the C and c that would be the novice way of explaining the inheritance of
cystic fibrosis). Return to the allelic variant section of the OMIM (omim.org) record for cystic
fibrosis. Choose three specific mutations (allelic variants) that were not listed in the case study.
Provide the abbreviation for the mutations as listed in the database and explain specifically the
effect of the mutation. Review question 8 in your case study for guidance but do not use those
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extensions to Mendelian inheritance. Describe the following extensions in your own words and
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Tsunami Ltd. wanted to acquire WiseCom, and tried to ensure their cor.pdfamazingkidstoys
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compatible. WiseCom executives welcomed the acquisition and eagerly described their
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public relations statements to decipher WiseCom's culture. The information gathered by Tsunami
is most likely measuring WiseCom's: Enacted values Assumptions Counterculture Socialization
process Espoused values.
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is usually double-stranded, while RNA is usually (but not always) single-stranded. RNA uses
uracil (U) in place of DNA's thymine (T). Only DNA is capable of complementary base-pairing.
RNA nucleotides contain ribose, while DNA nucleotides contain deoxyribose. RNA-directed
RNA polymerase copies viral RNA into DNA and simultaneously inserts it into the host cell's
genome..
Trace the output of the following short program. If the program crash.pdfamazingkidstoys
Trace the output of the following short program. If the program crashes write the word 'error'
without delimiters. If the program generates no output write the word 'no output' without delimit
deffun5():print(apple,end=*) def fun 4(): print('tuna', end =) def f un 1(): print('mouse', end = ' *')
fun3() fun 4() fun 1().
Treasury spot interest rates are as follows Part 1 Q. Attempt 110 f.pdfamazingkidstoys
Treasury spot interest rates are as follows: Part 1 Q. Attempt 1/10 for 10 pts. What is the fair
price of a Treasury note with 2 years to maturity, a $1,000 face value, and a coupon rate of 5.2%,
paid annually (in $ )? Correct dP=1+r1C+(1+r2)2C+F=1+0.02152+(1+0.028)21,052=1,046.4
Part 2 A Attempt 4/10 for 7 pts. Suppose this Treasury note is currently priced in the market fair
and square (no mispricing), what is the yield to maturity?.
Total magnification of a microscope is determined by Multiple C.pdfamazingkidstoys
Total magnification of a microscope is determined by Multiple Choice dividing the objective
magnification value by the ocular value. adding all objective magnification values together. none
of the above. multiplying all objective magnification values together.
Bacteria divide by Multiple Choice regular time intervals, regardless of the presence of nutrients.
mitosis. meiosis. binary fission..
Use of Computer Services This case pertains to determining the approp.pdfamazingkidstoys
Use of Computer Services This case pertains to determining the appropriate use of compuier
time, which is a matter of both aceess by an individual and the availability of quality service to
others. The individual in question has been granted permission to the the computing facilities for
a specific purpose. Many companies have an implicit code of conduct that regulates the
behaviour of individuals who have suthorized access to a computing system. The ethical
considerations involved in this case can provide insight into this unwritten standard. The Case
Dave is employed as a programmer in a large software company, where he is respoasible for
developing and testing utility programs, such as compilers. The company operates two
computing shifts; during the day, program development and enline applieations are run, while at
night, bateh production jobs are completed. Dave has access to workload data, which indicates
that the evening bateh runs are complementary to the daytime programuine tasks, meaning that
additional programming work dering the night shift would not negatively impact the computer's
performance for other users. However, Dave returns after nermal working bours to work on a
personal project, which invelves developing a program to manage his own stock portfolio.
Although his use of the system is minimal, and be consunes very few expendable supplies, sueh
as printer paper, the question arises as to whether his behaviour is ethical. Some of the ethical
principles involved in this case are listed below: - Ownershp of resources. - Effect on others. -
Universalism principle. - Possibility of detection, punishment. Analyse the given case and would
it affect the ethics of the situation if any of the following actions or characteristics were
considered? - Dave initiated a business venture whereby be managed stock portfolios on behalf
of numerous individuals for a fec. - Despite his qualifications, Dave's income was substandard,
implying that he regarded access to the company's computing resourees as a fringe benefit. -
Dave's employer was aware of other employees who were engaged in similar activities and
implicitly sanctioned them by refraining from interveniing. - Since Dave worked for a
government agency instead of a private enterprise, he reasoned that the computer system.
Travis Company purchased merchandise on account from a supplier for $.pdfamazingkidstoys
Travis Company purchased merchandise on account from a supplier for $11,100, terms 2/10, net
30 . Travis Company paid for the merchandise within the discount period. Under a perpetual
inventory system, journalize these transactions. If an amount box does not require an entry, leave
it blank..
Use numList file which got created above. Write a program that will.pdfamazingkidstoys
Use 'numList' file which got created above. Write a program that will prompt the user to enter
the name of the file lie. numlistl. and display the first 15 lines of the file's content. Use try/except
construct to capture any file opening related errors such as invalid file name ctc. Hints: Define
main 0 Declare variables: line and counter, and initialize them Prompt for fite name: use input
function Open the fle in read mode. Read the lines in the file with readine function:
'inputfile,readline0' and in 'tialize counter to 1 Use a while loop to read and display the first 15
lines Use istripi) method to remove ' in' characters at the end of the line Print the line Update the
counter when the line is read Close fite.
urrent Attempt in Progress At December 31, 2021, Martinez Imports rep.pdfamazingkidstoys
urrent Attempt in Progress At December 31, 2021, Martinez Imports reported this information
on its balance sheet. During 2022, the company had the following transactions related to
receivables.
Prepare a tabular summary that includes the following accounts: Cash, Accounts Receivable,
Allowance for Doubtful Accounts, Revenue, and Expense. Enter the January 1, 2022, balances in
Accounts Receivable and Allowance for Doubtful Accounts. (Enter negative amounts using
either a negative sign preceding the number e.g. -45 or parentheses e.g. (45).)
Prepare a tabular summary that includes the following accounts: Cash, Accounts Receivable,
Allowance for Doubtful Accounts, Revenue, and Expense. Enter the January 1, 2022, balances in
Accounts Receivable and Allowance for Doubtful Accounts. (Enter negative amounts using
either a negative sign preceding the number e.g. -45 or parentheses e.g. (45).).
Use Figure 12.2 to answer the following questio.pdfamazingkidstoys
Use Figure 12.2 to answer the following questions about cellular respiration: Does the equation
for cellular respiration describe everything that happens in the process? Explain your answer..
Use Excels Analysis ToolPak, with a seed of 1 , to generate 1,000 ra.pdfamazingkidstoys
Use Excel's Analysis ToolPak, with a seed of 1 , to generate 1,000 random observations based
on a Poisson distribution with =4.25. What are the mean and standard deviation of the 1,000
observations? Note: Round your answers to 2 decimal places..
Under the heterogenous treatment effect setup and the usual LATE (Loc.pdfamazingkidstoys
Under the heterogenous treatment effect setup and the usual LATE (Local Average Treatment
Effect) assumptions, the usual Instrumental Variable (IV) estimator consistently estimate the
average treatment effect for compliers. Complier is a subgroup of people who would always
participate if assigned to the control group or treatment group. would not participate if assigned
to the treatment group; and they would participate if assigned to the control group. would not
participate if assigned to the control group; and they would participate if assigned to the
treatment group. would always not participate if assigned to the control group or treatment
group..
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
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.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
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
Introduction to AI for Nonprofits with Tapp NetworkTechSoup
Dive into the world of AI! Experts Jon Hill and Tareq Monaur will guide you through AI's role in enhancing nonprofit websites and basic marketing strategies, making it easy to understand and apply.
Acetabularia Information For Class 9 .docxvaibhavrinwa19
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
Tree inin A recursave tree data structure. Note the retationship bet.pdf
1. Tree: inin A recursave tree data structure. Note the retationship between this class and
Recursivelist; the only major difference is that rest has been replaced by _ subtrees to handte
multiple recursive sub-parts. Hen # wa Private Attributes =m= # The item stored at this tree's
root, or None if the tree is enpty. root: optional [Anyl # The list of alt subtrees of this tree. -
subtrees: List [Tree] # mum Representation Invariants === # - If self. root is None then self.
subtrees is an enpty list. This setting of attributes represents an enpty Tree. Note: self. _ subtrees
may be empty when selt._root is not None. This setting of attributes represents a tree consisting
of just one node. def swap_down(setf) None: "*=Swap the root of this Tree with the frargest of
its chitdren until the originat root's value is at a position where it's larger than or equal to all of its
children. In the case of a tie, swap with the one that cones first in subtrees. >t= Tree(1,
[1)tsswapdownt) print(t) & No swaps are made 1 t._subtrees =[Tree(2,[1]. Tree (3, [1)] print(t)
1 3233> t. 5wapdown() ss print (t) in 1 swapped with 3 3 2 1 t-large
=Tree(3,ITree(5,ITree(7,tTree(2,11), Tree(1, [1]) 1), 1) Tree (4, [1] 1]> print (t __targe) 3 7 2 1
mos targe-swap_down( ) print(t_large) =3 swapped with 5 , and then with 7 .
# The beton are methods given to you. Do NoT change these. def init_(self, root: Any, subtrees:
List (Treel) None: wanitiatize a new Tree with the given root vatue and subtrees. If eroots is
None, the tree is eapty. Preconditiont if sroot> is None, then ksubtreess is enpty. self, _root =root
self.-subtrees = subtrees def is empty(self boolt nHinReturn True if this tree is enty. t1=Trec(
None, [1)t1.15 _etopty() True t2=Tree(3,11) t2.1s_empty () Fatise winl return self,_root is None
det _str (sett) str:
# The below are nethods given to you. Do NoT change these. def _init (self, root: Any, subtrees:
List(TreeJ) Wonet "anitialize a new Tree with the given root value and subtrees. If eroot> is
None, the tree is empty. Precondition: if is None, then esubtrees> is empty. ren self root = root
self._subtrees: = subtrees def is empty (self) bool: wawpeturn True if this tree is empty. tI= Tree
(None, []) t1,is_empty () True >t2=Tree(3, [1) > t2.is_empty ( ) False a return self. root is None
def _str_ ( self f str: m*Return a string representation of this: tree. For each node, its iten is
printed before any of its descendants' items. The output is nicely indented. You may find this
method helpful for debugging. return self._str_indented () , steip( ) def str_indented(setf, depth:
int =0) str: in*peturn an indented string representation of this-tree. The indentation level is:
specified by the : return e tiset 5=1+ depth + str (set( root )+ ' ln3 for subtree in self . subtrees: *
Note that the "depth' argument to the recursive catl is # modified. 5+= subtree.
_str_indented(depth + 1) returnis if nane _ingort doctest doctest. testmod (1) import python_ta
![Tree: inin A recursave tree data structure. Note the retationship between this class and
Recursivelist; the only major difference is that rest has been replaced by _ subtrees to handte
multiple recursive sub-parts. Hen # wa Private Attributes =m= # The item stored at this tree's
root, or None if the tree is enpty. root: optional [Anyl # The list of alt subtrees of this tree. -
subtrees: List [Tree] # mum Representation Invariants === # - If self. root is None then self.
subtrees is an enpty list. This setting of attributes represents an enpty Tree. Note: self. _ subtrees
may be empty when selt._root is not None. This setting of attributes represents a tree consisting
of just one node. def swap_down(setf) None: "*=Swap the root of this Tree with the frargest of
its chitdren until the originat root's value is at a position where it's larger than or equal to all of its
children. In the case of a tie, swap with the one that cones first in subtrees. >t= Tree(1,
[1)tsswapdownt) print(t) & No swaps are made 1 t._subtrees =[Tree(2,[1]. Tree (3, [1)] print(t)
1 3233> t. 5wapdown() ss print (t) in 1 swapped with 3 3 2 1 t-large
=Tree(3,ITree(5,ITree(7,tTree(2,11), Tree(1, [1]) 1), 1) Tree (4, [1] 1]> print (t __targe) 3 7 2 1
mos targe-swap_down( ) print(t_large) =3 swapped with 5 , and then with 7 .
# The beton are methods given to you. Do NoT change these. def init_(self, root: Any, subtrees:
List (Treel) None: wanitiatize a new Tree with the given root vatue and subtrees. If eroots is
None, the tree is eapty. Preconditiont if sroot> is None, then ksubtreess is enpty. self, _root =root
self.-subtrees = subtrees def is empty(self boolt nHinReturn True if this tree is enty. t1=Trec(
None, [1)t1.15 _etopty() True t2=Tree(3,11) t2.1s_empty () Fatise winl return self,_root is None
det _str (sett) str:
# The below are nethods given to you. Do NoT change these. def _init (self, root: Any, subtrees:
List(TreeJ) Wonet "anitialize a new Tree with the given root value and subtrees. If eroot> is
None, the tree is empty. Precondition: if is None, then esubtrees> is empty. ren self root = root
self._subtrees: = subtrees def is empty (self) bool: wawpeturn True if this tree is empty. tI= Tree
(None, []) t1,is_empty () True >t2=Tree(3, [1) > t2.is_empty ( ) False a return self. root is None
def _str_ ( self f str: m*Return a string representation of this: tree. For each node, its iten is
printed before any of its descendants' items. The output is nicely indented. You may find this
method helpful for debugging. return self._str_indented () , steip( ) def str_indented(setf, depth:
int =0) str: in*peturn an indented string representation of this-tree. The indentation level is:
specified by the : return e tiset 5=1+ depth + str (set( root )+ ' ln3 for subtree in self . subtrees: *
Note that the "depth' argument to the recursive catl is # modified. 5+= subtree.
_str_indented(depth + 1) returnis if nane _ingort doctest doctest. testmod (1) import python_ta](https://image.slidesharecdn.com/treeininarecursavetreedatastructure-230331034025-e63d539b/85/Tree-inin-A-recursave-tree-data-structure-Note-the-retationship-bet-pdf-1-320.jpg)
