11.25In 2010, there were 117,538,000 households in the United Stat.pdf
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11.25 In 2010, there were 117,538,000 households in the United States. There were 78,833,000 family households made up of married couples, single-male households, and single-female households. To determine how many of each type a survey was undertaken. The results were stored using the codes 1=married couples, 2=single male, and 3=single female. Estimate with 95% confidence the total number of American households with married couples. (Adapted from Statistical Abstract of the United States, 2012, Table 59) Solution.
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If the sample means are different from the population mean, what do .pdf
I have a generator and I need a wheel that is going to start the generator. How I know the
diameter of the wheel that I need to start the engine. The require rpm for start of the generator is
3480-3780.
Solution
Basically there are many ways to get the dia of wheel but one method is by using the flywheel of
corresponding power the flwheel size define the wheel size of the used in generator by using the
fluctuation in speed as given.
How would you define the Wilcoxon Signed Rank TestSolutionThe.pdf
Hey guys please I need help with problem 9.87
Thank you, 9.86 Public Policy and Political Science Residential mail boxes in Des Moines, Iowa,
should be installed such that the bottom of the mailbox is 42 inches above the ground. This rule
is designed for safety and to accommodate short mail carriers. A random sample of 75 mailboxes
in the city was selected. The height of each was carefully measured, and the sample mean was
43.22 inches. Assume o- = 7.6 inches and use a = 0.05. Is there any evidence to suggest the true
mean height of mailboxes in Des Moines is different from 42 inches? 9.87 Public Health and
Nutrition Iodine is an important nutri ent for the human body, especially in hormone
development. Milk is a natural source of iodine, and the mean amount in cow?s milk is 88
,Mug/250ml^24 Milk from organic farms is reported to have a lower concentration of elements
such as zinc, iodine, and selenium. A random sample of 35 gallons of organic milk from
Humboldt County, California, was obtained and the iodine concentration was measured in each.
The sample mean was 86.5. Assume sigma = 3.4. Is there any evidence to suggest that the mean
iodine concentration in organic milk is less than 88? Use a = 0.01.
Solution
Formulating the null and alternative hypotheses,
Ho: u >= 88
Ha: u < 88
As we can see, this is a left tailed test.
Thus, getting the critical z, as alpha = 0.01 ,
alpha = 0.01
zcrit = - 2.326347874
Getting the test statistic, as
X = sample mean = 86.5
uo = hypothesized mean = 88
n = sample size = 35
s = standard deviation = 3.4
Thus, z = (X - uo) * sqrt(n) / s = -2.610035198
Also, the p value is
p = 0.004526645
As |z| > 2.326, and P < 0.01, we REJECT THE NULL HYPOTHESIS.
Thus, there is significant evidence that the mean iodine concentration in organic milk is less
than 88. [CONCLUSION].
How could global climate change affect vector-borne or zoonotic di.pdf
Grade
Freshmen
Sophomore
Junior
Senior
A
0
8
9
10
B
0
6
8
11
C
1
7
9
12
D
2
4
1
4
F
0
6
2
1
TOTAL
3
31
29
38
What is the probability that the student is a junior and makes a B in the course?
What is the probability that the student does not make a A in the course, given that the student is
a senior?
What is the probability that the student makes a 0 or F in the course?
Let A be the event that a sophomore is taking the course. Let B be the event that the student
makes a C in the course. Are the events A and B independent? Are the events A and B mutually
exclusive?
Grade
Freshmen
Sophomore
Junior
Senior
A
0
8
9
10
B
0
6
8
11
C
1
7
9
12
D
2
4
1
4
F
0
6
2
1
TOTAL
3
31
29
38
Solution
here total number of students=3+31+29+38=101
##P[student is junior and scores B]=(total juniors who scores B)/total students=8/101=0.0079
[answer]
##P[a student does not make A in the course given that the student is senior]=P[a senior does not
make A]/P[student is a senior]=((38-10)/101)/(38/101)=28/38=0.73684 [answer]
##P[a student makes F]=total students make F/total students=(0+6+2+1)/101=0.08910 [answer]
##A be the event that a sophomore is taking the course. B be the event that the student makes C
in the course.
so P[A]=31/101 P[B]=(1+7+9+12)/101=29/101
now P[AB]=P[a sopomore makes C]=7/101 which is not eual to P[A]*P[B]
so they are not independent
moreover P[AB] is not zero..so not mutually exclusive also..
How Can Sequence Data Be Used to Track Flu Virus Evolution During Pan.pdf
Given the pseudocode below, answer the following questions: Algorithm What (A, n) A
leftarrow 2-dimensional array of n times n integers B leftarrow 1-dimensional array of n integers
for leftarrow n-5 to n-3 do for j leftarrow 0 to n-1 do B[i] leftarrow B[i] + A[i][j] end for end
for What is the Big-O order of the memory required by the algorithm? How many times is the
innermost loop executed, as a function of n? What is the Big-0 order of the operations required
by the algorithm?
Solution
a. Big-O order of memory required = O(n^2) for A and O(n) for B. So result would be O(n^2).
Big-O order of memory required means how much memory is expected to grow as a function of
n. Here, as n increases memory required for A would increase by n^2
b. It is exectued O(2*n) -> O(n)..
Far from being primitive, bacteria have evolved until they are i.pdf
Explain what happens from the time of an interrupt request, (i.e., IRQ line is asserted), through
the start of executing an interrupt service routine, and then how control returns to the interrupted
program. Assume that the interrupt request does not occur at the same time as any other interrupt
request. (Note: There are many steps involved, so take your time and be specific.) in ARM
Assembly.
Solution
Answer
From the time of an interrupt request through the start of executing an interrupt service routine
below steps occur:.
E1-1 Reporting Amounts on the Four Basic Financial Statements [LO 1-2.pdf
During which part of the EKG does the depolarization of the ventricular anterior aspect and
septum occur?
Select one:
a. T wave
b. Q wave
c. R wave
d. S wave
e. P wave
During the plateau of the ventricular muscle action potential, the membrane potential remains
basically unchanged. The movements of which two ions are mainly responsible?
Select one:
a. Sodium influx, calcium efflux
b. Potassium influx, calcium efflux
c. Potassium efflux, calcium efflux
d. Potassium efflux, calcium influx
e. Potassium influx, calcium influx
f. Potassium influx, sodium efflux
g. Sodium influx, potassium efflux
Solution
Ques-1
Answer: b. Q wave
Reason:
QRS complex is also known as initial ventricular complex, which is produced due to the
ventricular depolarization. Q wave is a negative wave that continues as a positive R wave (tallest
wave). R wave is followed by S wave, which is a negative wave. The EKG with Q wave does the
depolarization of the ventricular anterior aspect and septum occur because it is generated via
downward deflection after P wave. In case lead V1, depolarization meticulously associated with
the ventricular anterior aspect and septum & it is not abnormal if it is missing from any
particular lead, it is significant if its higher than 1 mm wide. It also considered as abnormal if it
is 25% - 33% of total amplitude of the R wave..
Do differentiated cells transform into other type of cells How Wha.pdf
discuss the differences between international marketing and global marketing with supporting
details.
Solution
International and global marketing are different in nature
International marketing:imMarketing Decisions are made in the countries and the staff will be
use who got good knowledge about the target markets and positioning strategy
GLOBAL MARKETING: Where as it is treated as whoe world as one market,marketing
activities are co-ordinated and integrated by multiple markets
Common Differntiation.
Define the terms activity, layout, intent, and AVDSolutionACT.pdf
Define the terms: activity, layout, intent, and AVD
Solution
ACTIVITY:
An activity can be defined as the application program component that can be used for the users
for any purpose.
An application might have many number of activities.
Inorder to create an activity we must develop subclass of those activities.
LAYOUT:A layout depicts the graphical view of the user interface for an application.
the layout can be declared in any of the two ways those are:
layout can be declared through xml or it can be through in run time.
INTENT: intent itself can be defined as an Intent object as well as it is a passive data structure
containing an abstract depiction of an operation that has to be done.
AVD:ANDROID VIRTUAL DESIGN.
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If the sample means are different from the population mean, what do .pdf
I have a generator and I need a wheel that is going to start the generator. How I know the
diameter of the wheel that I need to start the engine. The require rpm for start of the generator is
3480-3780.
Solution
Basically there are many ways to get the dia of wheel but one method is by using the flywheel of
corresponding power the flwheel size define the wheel size of the used in generator by using the
fluctuation in speed as given.
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Hey guys please I need help with problem 9.87
Thank you, 9.86 Public Policy and Political Science Residential mail boxes in Des Moines, Iowa,
should be installed such that the bottom of the mailbox is 42 inches above the ground. This rule
is designed for safety and to accommodate short mail carriers. A random sample of 75 mailboxes
in the city was selected. The height of each was carefully measured, and the sample mean was
43.22 inches. Assume o- = 7.6 inches and use a = 0.05. Is there any evidence to suggest the true
mean height of mailboxes in Des Moines is different from 42 inches? 9.87 Public Health and
Nutrition Iodine is an important nutri ent for the human body, especially in hormone
development. Milk is a natural source of iodine, and the mean amount in cow?s milk is 88
,Mug/250ml^24 Milk from organic farms is reported to have a lower concentration of elements
such as zinc, iodine, and selenium. A random sample of 35 gallons of organic milk from
Humboldt County, California, was obtained and the iodine concentration was measured in each.
The sample mean was 86.5. Assume sigma = 3.4. Is there any evidence to suggest that the mean
iodine concentration in organic milk is less than 88? Use a = 0.01.
Solution
Formulating the null and alternative hypotheses,
Ho: u >= 88
Ha: u < 88
As we can see, this is a left tailed test.
Thus, getting the critical z, as alpha = 0.01 ,
alpha = 0.01
zcrit = - 2.326347874
Getting the test statistic, as
X = sample mean = 86.5
uo = hypothesized mean = 88
n = sample size = 35
s = standard deviation = 3.4
Thus, z = (X - uo) * sqrt(n) / s = -2.610035198
Also, the p value is
p = 0.004526645
As |z| > 2.326, and P < 0.01, we REJECT THE NULL HYPOTHESIS.
Thus, there is significant evidence that the mean iodine concentration in organic milk is less
than 88. [CONCLUSION].
How could global climate change affect vector-borne or zoonotic di.pdf
Grade
Freshmen
Sophomore
Junior
Senior
A
0
8
9
10
B
0
6
8
11
C
1
7
9
12
D
2
4
1
4
F
0
6
2
1
TOTAL
3
31
29
38
What is the probability that the student is a junior and makes a B in the course?
What is the probability that the student does not make a A in the course, given that the student is
a senior?
What is the probability that the student makes a 0 or F in the course?
Let A be the event that a sophomore is taking the course. Let B be the event that the student
makes a C in the course. Are the events A and B independent? Are the events A and B mutually
exclusive?
Grade
Freshmen
Sophomore
Junior
Senior
A
0
8
9
10
B
0
6
8
11
C
1
7
9
12
D
2
4
1
4
F
0
6
2
1
TOTAL
3
31
29
38
Solution
here total number of students=3+31+29+38=101
##P[student is junior and scores B]=(total juniors who scores B)/total students=8/101=0.0079
[answer]
##P[a student does not make A in the course given that the student is senior]=P[a senior does not
make A]/P[student is a senior]=((38-10)/101)/(38/101)=28/38=0.73684 [answer]
##P[a student makes F]=total students make F/total students=(0+6+2+1)/101=0.08910 [answer]
##A be the event that a sophomore is taking the course. B be the event that the student makes C
in the course.
so P[A]=31/101 P[B]=(1+7+9+12)/101=29/101
now P[AB]=P[a sopomore makes C]=7/101 which is not eual to P[A]*P[B]
so they are not independent
moreover P[AB] is not zero..so not mutually exclusive also..
How Can Sequence Data Be Used to Track Flu Virus Evolution During Pan.pdf
Given the pseudocode below, answer the following questions: Algorithm What (A, n) A
leftarrow 2-dimensional array of n times n integers B leftarrow 1-dimensional array of n integers
for leftarrow n-5 to n-3 do for j leftarrow 0 to n-1 do B[i] leftarrow B[i] + A[i][j] end for end
for What is the Big-O order of the memory required by the algorithm? How many times is the
innermost loop executed, as a function of n? What is the Big-0 order of the operations required
by the algorithm?
Solution
a. Big-O order of memory required = O(n^2) for A and O(n) for B. So result would be O(n^2).
Big-O order of memory required means how much memory is expected to grow as a function of
n. Here, as n increases memory required for A would increase by n^2
b. It is exectued O(2*n) -> O(n)..
Far from being primitive, bacteria have evolved until they are i.pdf
Explain what happens from the time of an interrupt request, (i.e., IRQ line is asserted), through
the start of executing an interrupt service routine, and then how control returns to the interrupted
program. Assume that the interrupt request does not occur at the same time as any other interrupt
request. (Note: There are many steps involved, so take your time and be specific.) in ARM
Assembly.
Solution
Answer
From the time of an interrupt request through the start of executing an interrupt service routine
below steps occur:.
E1-1 Reporting Amounts on the Four Basic Financial Statements [LO 1-2.pdf
During which part of the EKG does the depolarization of the ventricular anterior aspect and
septum occur?
Select one:
a. T wave
b. Q wave
c. R wave
d. S wave
e. P wave
During the plateau of the ventricular muscle action potential, the membrane potential remains
basically unchanged. The movements of which two ions are mainly responsible?
Select one:
a. Sodium influx, calcium efflux
b. Potassium influx, calcium efflux
c. Potassium efflux, calcium efflux
d. Potassium efflux, calcium influx
e. Potassium influx, calcium influx
f. Potassium influx, sodium efflux
g. Sodium influx, potassium efflux
Solution
Ques-1
Answer: b. Q wave
Reason:
QRS complex is also known as initial ventricular complex, which is produced due to the
ventricular depolarization. Q wave is a negative wave that continues as a positive R wave (tallest
wave). R wave is followed by S wave, which is a negative wave. The EKG with Q wave does the
depolarization of the ventricular anterior aspect and septum occur because it is generated via
downward deflection after P wave. In case lead V1, depolarization meticulously associated with
the ventricular anterior aspect and septum & it is not abnormal if it is missing from any
particular lead, it is significant if its higher than 1 mm wide. It also considered as abnormal if it
is 25% - 33% of total amplitude of the R wave..
Do differentiated cells transform into other type of cells How Wha.pdf
discuss the differences between international marketing and global marketing with supporting
details.
Solution
International and global marketing are different in nature
International marketing:imMarketing Decisions are made in the countries and the staff will be
use who got good knowledge about the target markets and positioning strategy
GLOBAL MARKETING: Where as it is treated as whoe world as one market,marketing
activities are co-ordinated and integrated by multiple markets
Common Differntiation.
Define the terms activity, layout, intent, and AVDSolutionACT.pdf
Define the terms: activity, layout, intent, and AVD
Solution
ACTIVITY:
An activity can be defined as the application program component that can be used for the users
for any purpose.
An application might have many number of activities.
Inorder to create an activity we must develop subclass of those activities.
LAYOUT:A layout depicts the graphical view of the user interface for an application.
the layout can be declared in any of the two ways those are:
layout can be declared through xml or it can be through in run time.
INTENT: intent itself can be defined as an Intent object as well as it is a passive data structure
containing an abstract depiction of an operation that has to be done.
AVD:ANDROID VIRTUAL DESIGN.
Describe the similarities and differences between IPv4 & IPv6.So.pdf
Describe how you would look up the critical z-value for a 72% on a normal curve table, what is
the critical z-value for a 72% confidence interval?
Solution
Za/2 = Z-table value = 1.08
Method 1:
I will use the below formula NORMINV(1-0.28/2,0,1)
Method 2:
To do this, if X ~ N(?, ?^2), then
STEP 1:
Z= X- ? / ? ~ N(0,1)
STEP 2:
A table of standardized normal values, can then be used to obtain an answer in terms of the
converted problem.
STEP 3:
Find x=Z ?+ ?
STEP 4:
Use below method
http://www.mathsisfun.com/data/standard-normal-distribution-table.html
Eg:
TYPE 1: (Z ? 1.65) = F(1.65) = 0.95.
Compare different sexual and asexual life cycles noting their adapti.pdf
Compare different sexual and asexual life cycles noting their adaptive advantages.
Solution
Asexual reproduction sexual reproduction Number of organisms involved One parent
needed Two parents are required to mate Cell division Cells divide by Fission, budding , or
regeneration Cells divide by Meiosis Types Budding, vegetative reproduction, fragmentation,
spore formation Syngamy and conjugation Advantages Time Efficient; no need to search for
mate, requires less energy Variation, Unique., organism is more protected Disadvantages No
variation - if the parent has a genetic disease, offspring does too. Requires two organisms,
requires more energy Evolution There is very little chance of variation with asexual
reproduction. Mutations in DNA can still occur but not nearly as frequently as in sexual
reproduction. Sexual reproduction leads to genetic variation in new generations of offspring.
This is fundamental to evolution. Involvement of sex cells No formation or fusion of gametes
(sex cell) Formation and fusion of gametes (sex cell) occurs Found in Lower organisms Higher
invertebrates and all vertebrates Unit of reproduction May be whole parent body or a bud or a
fragment or a single somatic cell Gamete Time taken Asexual reproduction is completed in a
very short period of time. Sexual reproduction can take several months to complete. Number of
offspring Two or more One or more.
BackgroundIn many applications, the composition of a collection o.pdf
Background:
In many applications, the composition of a collection of data items changes over time. Not only
are new data items added and existing ones removed, but data items may be duplicated. A list
data structure is a member of the general category of abstract data types called containers, whose
purpose is to hold other objects. In C++, lists are provided in the Standard Template Library.
However, for this assignment you will design and write your own linked list implementation to
support the ADT operations specified below.
Objective:
Design and implement the specifications for a List Abstract Data Type where the items in the list
are unsorted.
Requirements:
Define a list and develop a set of operations for creating and manipulating a list that satisfies the
list ADT specification.
List ADT Specification
Structure: The list elements are of ItemType. The list has a property called the current position
which designates the position of the last element accessed by GetNextItem during an iteration
through the list. Only ResetList and GetNextItem alter the current position.
Definitions (provided by the user):
MAX_ITEMS: A constant specifying the maximum capacity of items allowed on the list
Item Type: Class encapsulating the type of items in the list
RelationType: An enumeration type that consists of LESS, GREATER, EQUAL
Member function of ItemType that must be included:
RelationType ComparedTo(ItemType Item)
Function: Determines the ordering of two ItemType objects based on their keys
Precondition: Self and item have their key members initialized
Postcondition:
Function value = LESS if the key of self is less than the key of item
= GREATER if the key of self is greater than the key of item
= EQUAL if the keys are equal
Operations (provided by Unsorted List ADT)
Make Empty
Function: Initializes list to empty state
Preconditions: None
Postcondition: List is empty
Boolean IsFull
Function: Determines whether list is full
Preconditions: List has been initialized
Postcondition: Function value = (list is full)
int GetLength
Function: Determines the number of elements in list
Preconditions: List has been initialized
Postcondition: Function value = number of elements in list
ItemType GetItem(Item Typeitem, Boolean& found)
Function: Get list element whose key matches item’s key (if present)
Preconditions: List has been initialized
Key member of item is initialized
Postcondition: If there is an element someItem whose keymatches item’s key, then found =
true and copy of someItemis returned; otherwise found = false and item is returned
List is unchanged
PutItem(ItemType item)
Function: Puts item into list
Preconditions: List has been initialized
List is not full
Item is not in list
Postcondition: Item is in the list
DeleteItem(ItemType item)
Function: Deletes the element whose key matches item’s key
Preconditions: List has been initialized
Postcondition: One and only one element in list has a key matching item’s key
ResetList
Function: Initializes current p.
Which methodology may be used for detection of incorporated BrdUA.pdf
What is the first enzymatic activity in DNA synthesis?
Endonuclease
Primase
Exonuclease
DNA polymerase
LigaseA.
EndonucleaseB.
PrimaseC.
ExonucleaseD.
DNA polymeraseE.
Ligase
Solution
Ans. Correct option. B. Primase.
The first enzymatic activity in DNA synthesis (among the available options) is Primase.
DNA primase (or, primase) binds to ssDNA strands during initiation period. It synthesizes small
RNA primer to both the strands. DNA polymerase holoenzyme binds during elongation of DNA
synthesis after primase has synthesized RNA primers. It extends the RNA primers with DNA
nucleotide polymerization on it in 5’-3’ direction.
The exonuclease activity of DNA polymerase comes into action during repair or nick translation
during elongation. Other than that, no other endonuclease or exonuclease participate DNA
synthesis.
Ligase (DNA ligase) joins DNA fragments during nick translation much later than DNA pol
begins to synthesize DNA..
Which of the following is a difference between lightweight and heavy.pdf
What is the probability that a five card poker hand contains the jack of diamonds
and the two of spades?
Solution
Jack of diamonds and the two of spades:
No of ways to get 1 Jack of diamonds: = 1
No of ways to get 2 spades: 13C2 = 78
Total No of ways to contains the jack of diamonds and the two of spades = 1*78 = 78
and No of ways to five card poker hand = 52C5 = 2,598,960
Then Probability = 78/2,598,960 = 0.00003001.
What are the common conditions resulting in thrombocytosis Explain .pdf
Use the following information to answer questions 7-11 The following information applies to
the questions displayed below. The following information is available for Lock-Tite Company,
which produces special-order security products and uses a job order costing system. April 30
May 31 Inventories Raw materials Work in process Finlshed goods $39,000 $30,000 9,100
20,500 69,000 34,500 Activities and information for May Raw materials purchases (paid with
cash) Factory payroll (paid with cash) Factory overhead 186,000 300,000 Indirect materials
Indirect labor Other overhead costs 7,000 71,000 111,000 1,100,000 Sales (received in cash)
Predetermined overhead rate based on direct labor cost 55%
Solution
Answer - Exercise 15-7 T-accounts Factory Overheads Cost of material used
Raw Materials Raw Material 7000 125950 Work in process Raw Materials April-30 $39,000
RM-April 30 39000 7000 Factory Overhead Factory Payroll 71000 63050 Cost of goods sold
Add:Raw Material Purchases $186,000 RM Purchases 186000 188000 Work in process Other
Overheads 111000 Less : Raw Materials May-31 $30,000 Cost of material used
$195,000 less : Indirect Material used $7,000 RM May 31 30000 Direct Material
Used $188,000 Work in process Cost of direct labour used = $3,00,000 WIP-
April 30 9100 531550 Finished Goods Raw Material 188000 Cost of goods
Manufactured Factory Payroll 229000 Cost of Materials used $188,000 Factory
Overhead 125950 Cost of direct labour used $229,000 WIP-May 31 20500 Factory
Overhead applied [55% of $2,29,000] $125,950 Cost of Manufacturing for the period
$542,950 Factory Payroll Add: Work in process , April-30 $9,100 Cash 300000 71000
Factory Overheads Less : Work in process,May-31 $20,500 229000 Work in process
Cost of goods Manufactured $531,550 Cost of goods sold Cost of goods
Manufactured $531,550 Add : Finished Goods , April-30 $69,000 Finished Goods
Less : Finished Goods,May-31 $34,500 FG-April 30 69000 Cost of goods sold $566,050
Work in process 531550 566050 Cost of goods sold Gross Profit Sales
$1,100,000 FG May 31 34500 Less : Cost of goods sold (unadjusted) $566,050
Gross Profit $533,950 Overapplied or underapplied overheads = Actual
overheads - applied overheads Actual Overheads = Indirect Material + Indirect Labour +
other overhead costs = $7000 + $71000 + $111000 = $1,89,000 Underapplied overheads =
$189000 -$125950 = $63,050 Journal Entry to allocate underapplied
overheads Account Titles Debit Credit Cost of goods sold $63,050 Factory
Overheads $63,050 Income Statement (Partial) Sales $1,100,000
Cost of goods sold $566,050 Gross Profit $533,950.
What are some differences between the xylem and the phloem Solu.pdf
Use Simplex method and solve the following problem:
Maximize p = x+y subject to x+y <= 2, 3x+y >= 4
Which of teh following is the last matrix table?
A.
x y s1 s2 | b
0 1 3/2 1/2 1
1 0 1/2 -1/2 1
0 0 1 0 2
B.
x y s1 s2 | b
0 1 3/2 1/2 1
1 0 -1/2 1/2 1
0 0 1 0 2
C.
x y s1 s2 | b
0 1 3/2 1/2 1
1 0 -1/2 -1/2 1
0 0 1 0 2
D.
x y s1 s2 | b
0 1 3/2 -1/2 1
1 0 -1/2 -1/2 1
0 0 1 0 2
x y s1 s2 | b
0 1 3/2 1/2 1
1 0 1/2 -1/2 1
0 0 1 0 2
B.
x y s1 s2 | b
0 1 3/2 1/2 1
1 0 -1/2 1/2 1
0 0 1 0 2
C.
x y s1 s2 | b
0 1 3/2 1/2 1
1 0 -1/2 -1/2 1
0 0 1 0 2
D.
x y s1 s2 | b
0 1 3/2 -1/2 1
1 0 -1/2 -1/2 1
0 0 1 0 2
Solution
The step by step solution of the problem is given below -
Tableau #1
x y s1 s2 p
1 1 1 0 0 2
3 1 0 -1 0 4
-1 -1 0 0 1 0
Tableau #2
x y s1 s2 p
0 2/3 1 1/3 0 2/3
1 1/3 0 -1/3 0 4/3
0 -2/3 0 -1/3 1 4/3
Tableau #3
x y s1 s2 p
0 1 3/2 1/2 0 1
1 0 -1/2 -1/2 0 1
0 0 1 0 1 2
So, the correct answer is option (c)..
Two families of sloths exist commonly in Central and South America (t.pdf
There are a number of highly touted search engines for finding things of interest on the Internet.
Recently, a consumer rating system ranked two search engines ahead of the others. Now, a
computer user\'s magazine wishes to make the final determination regarding which one is
actually better at finding particular information. To do this, each search engine was used in an
attempt to locate specific information using specified keywords. Both search engines were
subjected to 100 queries. Search engine 1 successfully located the information 88 times and
search engine 2 located the information 80 times. Using a significance level equal to 0.05, what
is the critical value for the hypothesis test?
z =
z =
Solution
It is a two-tailed test.
Given a=0.05, the critical values are Z(0.025) = -1.96 or 1.96 (from standard normal table)
Answer: z =.
Two populations of rats are discovered on nearby islands. Describe t.pdf
think like an You have three resistors of resistance 30 ohm [Ohm]. There are__configurations
possible, with__unique combinations, resulting in___different resistance values. A
\"configuration\" is a way of combining the resistors. For example, two resistors in parallel is one
configuration; two resistors in series is a second configuration. A \"combination\" is the specific
way of combining given resistors to form an effective resistance. For example, combining
Resistor #1 and #2 in parallel is one combination; combining Resistor #1 and #3 in parallel is a
second combination. These combinations may or may not result in a unique effective resistance.
What is the greatest resistance that can be made from a combination of resistors, and what is the
effective resistance of this combination? What is the least resistance that can be made from a
combination of resistors, and what is the effective resistance of this combination?
Solution
b) GREATEST RESISTANCE CAN BE ACHIEVED THROUGH SERIES CONNECTION OF
THREE RESISTORS.
EFFECTIVE RESISTANCE=30+30+30=90 OHMS
C) LEAST RESISTANCE CAN BE ACHIEVED THROUGH PARALLEL CONNECTION OF
THREE RESISTORS.
EFFECTIVE RESISTANCE=(30*30)/(30+30)=15 OHMS
(15*30)/(15+30)=450/45=10 OHMS.
30-32 Nuclear pores permit the passage of chromosomes outward. glu.pdf
30-32 Nuclear pores permit the passage of chromosomes outward. glucose molecules outward.
assembled DNA molecules outward. RNA outward. sodium ions inward, potassium ions
outward. The cell walls of fungi are made from peptidoglycan. pseudomurein. cellulose.
chitin. Starch. Flattened sacks of membranes involved in the packaging and distribution of
molecules synthesized in the cell are known as Golgi bodies. microbodies. pinocytic vesicles.
vacuoles. chromosomes.
Solution
30). RNA outwarrd
This transport includes RNA and ribosomal proteins moving from nucleus to the cytoplasm.
31). Cellulose and chitin
32). Golgi bodies.
Review Questions for Exam 10182016 1. public class .pdf
Q3 GIVEN THE TREND EQUATION NO OF STUDENTS-3000+200 YEAR BASE: 2003-0
1.INTERPRET EQUATION RESULTS 2.FORECAST FOR 2020 3.WHEN WILL THE NO OF
STUDENTS REACH 6000
Solution
x no of years from 2003 y no of students y = 3000+200x 1) The regression
equation has a positive slope of 200 y intercept is 3000 i.e. no of students in the year 2000
Thus for every year 200 students increase 2) For 2020, x = 17, y = 3000+200(17) = 6400
3) When y =6000, 6000 = 3000+200x or x = 15 Hence in year 2018 students reach 6000
B Regression C D. All the above.
B-tree insertion (Draw node) Show the results of inserting the keys .pdf
B-tree insertion (Draw node) Show the results of inserting the keys 23, 98, 55, 37, 30, 47, 35,
15, 10, 75, 3, 53, 1, 33, 5, 28, 25, 90, 95, 56, 8, and 9 in order into an empty B-tree of order 5.
That is, each internal node in this B-tree other than the root must have at least 3 and at most 5
children (or at least 2 and at most 4 keys).
Solution
/*
* C++ Program to Implement B-Tree
*/
#include
#include
#include
using namespace std;
struct BTreeNode
{
int *data;
BTreeNode **child_ptr;
bool leaf;
int n;
}*root = NULL, *np = NULL, *x = NULL;
BTreeNode * init()
{
int i;
np = new BTreeNode;
np->data = new int[5];
np->child_ptr = new BTreeNode *[6];
np->leaf = true;
np->n = 0;
for (i = 0; i < 6; i++)
{
np->child_ptr[i] = NULL;
}
return np;
}
void traverse(BTreeNode *p)
{
cout<n; i++)
{
if (p->leaf == false)
{
traverse(p->child_ptr[i]);
}
cout << \" \" << p->data[i];
}
if (p->leaf == false)
{
traverse(p->child_ptr[i]);
}
cout< p[j])
{
temp = p[i];
p[i] = p[j];
p[j] = temp;
}
}
}
}
int split_child(BTreeNode *x, int i)
{
int j, mid;
BTreeNode *np1, *np3, *y;
np3 = init();
np3->leaf = true;
if (i == -1)
{
mid = x->data[2];
x->data[2] = 0;
x->n--;
np1 = init();
np1->leaf = false;
x->leaf = true;
for (j = 3; j < 5; j++)
{
np3->data[j - 3] = x->data[j];
np3->child_ptr[j - 3] = x->child_ptr[j];
np3->n++;
x->data[j] = 0;
x->n--;
}
for (j = 0; j < 6; j++)
{
x->child_ptr[j] = NULL;
}
np1->data[0] = mid;
np1->child_ptr[np1->n] = x;
np1->child_ptr[np1->n + 1] = np3;
np1->n++;
root = np1;
}
else
{
y = x->child_ptr[i];
mid = y->data[2];
y->data[2] = 0;
y->n--;
for (j = 3; j < 5; j++)
{
np3->data[j - 3] = y->data[j];
np3->n++;
y->data[j] = 0;
y->n--;
}
x->child_ptr[i + 1] = y;
x->child_ptr[i + 1] = np3;
}
return mid;
}
void insert(int a)
{
int i, temp;
x = root;
if (x == NULL)
{
root = init();
x = root;
}
else
{
if (x->leaf == true && x->n == 5)
{
temp = split_child(x, -1);
x = root;
for (i = 0; i < (x->n); i++)
{
if ((a > x->data[i]) && (a < x->data[i + 1]))
{
i++;
break;
}
else if (a < x->data[0])
{
break;
}
else
{
continue;
}
}
x = x->child_ptr[i];
}
else
{
while (x->leaf == false)
{
for (i = 0; i < (x->n); i++)
{
if ((a > x->data[i]) && (a < x->data[i + 1]))
{
i++;
break;
}
else if (a < x->data[0])
{
break;
}
else
{
continue;
}
}
if ((x->child_ptr[i])->n == 5)
{
temp = split_child(x, i);
x->data[x->n] = temp;
x->n++;
continue;
}
else
{
x = x->child_ptr[i];
}
}
}
}
x->data[x->n] = a;
sort(x->data, x->n);
x->n++;
}
int main()
{
int i, n, t;
cout<<\"enter the no of elements to be inserted\ \";
cin>>n;
for(i = 0; i < n; i++)
{
cout<<\"enter the element\ \";
cin>>t;
insert(t);
}
cout<<\"traversal of constructed tree\ \";
traverse(root);
getch();
}.
Please help me find where I do something wrong with my programI am.pdf
OPERATING SYSTEMS: DISK SCHEDULING
PLEASE SHOW YOUR WORK! I\'M TRYING TO UNDERSTAND HOW TO ACTUALLY
DO THE PROBLEM.
2. (3 points) Suppose that a disk drive has 3000 cylinders, numbered 0 to 2999. The drive is
currently serving a request at cylinder 140, and the previous request was at cylinder 125. The
queue of pending requests, in FIFO order, is
66, 147, 913, 289, 202, 1750, 15
Starting from the current head position, what is the total distance (in cylinders) that the disk arm
moves to satisfy all the pending requests, for each of the following disk-scheduling algorithms?
a. FCFS
b. SSTF
c. SCAN
d. LOOK
e. C-SCAN
3. Requests are not usually uniformly distributed. For example, a cylinder containing the file
system FAT or inodes can be expected to be accessed more frequently than a cylinder that only
contains files. Suppose you know that 50 percent of the requests are for a small, fixed number of
cylinders.
Propose a disk-scheduling algorithm (one that we have discussed) that gives even better
performance by taking advantage of this “hot spot” on the disk. Explain your answer and discuss
any problems and how you might mitigate them.
Solution
The drive is currently serving a request at cylinder 140
request queue -- 66,147,913,289,202,1750,15
a) FCFS
66<--- 140
66---->147
147 ----> 913
289 <----- 913
202<---- 289
202 ------>1750
15 <------ 1750
total time = |66-140|+|66-147|+|147-913|+|289-913|+||202-289|+|202-1750|+|15-1750|
=74+81+766+624+87+1548+1735
=4915
b) SSTF
140 ----> 147
147 ------>202
202----->289
66<----289
15<----66
15----->913
913----->1750
total time=7+55+87+223+51+898+837
=2158
c) SCAN
140--->147---->202---->289----->913----->1750----->2999---
0<-----15<------66<-----------------------------------------------
total time = |140-147|+|147-202|+|202-289|+|289-913|+|913-1750|+|1750-2999|+|2999-66|+|66-
15|+|15-0|
=7+55+87+624+837+1249+2933+51+15
=5858
d)LOOK
140--->147---->202---->289----->913----->1750---
15<------66<-----------------------------------------------
total time = |140-147|+|147-202|+|202-289|+|289-913|+|913-1750|+|1750-66|+|66-15|
=7+55+87+624+837+1684+51
=3345
e)C-SCAN
140--->147---->202---->289----->913----->1750----->2999---
66<-----15<------0<-----------------------------------------------
total time = |140-147|+|147-202|+|202-289|+|289-913|+|913-1750|+|1750-2999|+|2999-0|+|15-
0|+|66-15|
=7+55+87+624+837+1249+2999+15+51
=5924
SSTF can be the best disk scheduling algorithm when data is close to each other.
We can have hot spot in the middle of the disk . Aging can be done on older request by
increasing their priority.
List the transformations to the basic graph and then graph the funct.pdf
Joe\'s Coffee Shoppe orders fresh doughnuts each morning. Joe pays $3.20 per dozen doughnuts
and sell doughnuts for $0.4 each. The leftover doughnuts are salvaged at half the selling price.
Suppose that the demand for doughnuts is uniformly distributed from 16 dozens to 30 dozens.
How many doughnuts (in dozens) should Joe stock each day? A detailed analysis of past data
shows that the number of doughnuts sold per day is better described by a normal distribution,
with a mean 23 dozens and standard deviation 10 dozens. Now, how many doughnuts should Joe
stock each day? The expected demand was the same in parts (a) and (b), but the optimal order
quantities should have been different. What accounted for this different. What accounted for this
difference?
Solution.
In your own words discuss the following. 1. Briefly describe the p.pdf
In a survey conducted by a business management company, 51% of the U.S. workers siad they
have negotiated a pay raise at least once inthier lives. The sample size used in the study was not
disclosed.
A) Suppose that the survey had a sample size of n=500. Construct a condifidence interval
estimate for the proportion of all United States workers who have negotiated a pay raise.
Solution
Answer to the question)
P = 0.51
n = 500
Standard error SE = srt(0.51 *0.49/500)
SE = 0.0224
.
Let the confidence level = 95% , so the Z critical value = 1.96
.
The formula of cofndience interval is:
p - z* SE , p + z * SE
.
On plugging the values we get
0.51 - 1.96 *0.0224 , 0.51 +1.96 *0.0224
0.04661 , 0.5539.
Types of Herbal Cosmetics its standardization.
Physiology and chemistry of skin and pigmentation, hairs, scalp, lips and nail, Cleansing cream, Lotions, Face powders, Face packs, Lipsticks, Bath products, soaps and baby product,
Preparation and standardization of the following : Tonic, Bleaches, Dentifrices and Mouth washes & Tooth Pastes, Cosmetics for Nails.
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Hindi varnamala | hindi alphabet PPT.pdf
हिंदी वर्णमाला पीपीटी, hindi alphabet PPT presentation, hindi varnamala PPT, Hindi Varnamala pdf, हिंदी स्वर, हिंदी व्यंजन, sikhiye hindi varnmala, dr. mulla adam ali, hindi language and literature, hindi alphabet with drawing, hindi alphabet pdf, hindi varnamala for childrens, hindi language, hindi varnamala practice for kids, https://www.drmullaadamali.com
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Describe the similarities and differences between IPv4 & IPv6.So.pdf
Describe how you would look up the critical z-value for a 72% on a normal curve table, what is
the critical z-value for a 72% confidence interval?
Solution
Za/2 = Z-table value = 1.08
Method 1:
I will use the below formula NORMINV(1-0.28/2,0,1)
Method 2:
To do this, if X ~ N(?, ?^2), then
STEP 1:
Z= X- ? / ? ~ N(0,1)
STEP 2:
A table of standardized normal values, can then be used to obtain an answer in terms of the
converted problem.
STEP 3:
Find x=Z ?+ ?
STEP 4:
Use below method
http://www.mathsisfun.com/data/standard-normal-distribution-table.html
Eg:
TYPE 1: (Z ? 1.65) = F(1.65) = 0.95.
Compare different sexual and asexual life cycles noting their adapti.pdf
Compare different sexual and asexual life cycles noting their adaptive advantages.
Solution
Asexual reproduction sexual reproduction Number of organisms involved One parent
needed Two parents are required to mate Cell division Cells divide by Fission, budding , or
regeneration Cells divide by Meiosis Types Budding, vegetative reproduction, fragmentation,
spore formation Syngamy and conjugation Advantages Time Efficient; no need to search for
mate, requires less energy Variation, Unique., organism is more protected Disadvantages No
variation - if the parent has a genetic disease, offspring does too. Requires two organisms,
requires more energy Evolution There is very little chance of variation with asexual
reproduction. Mutations in DNA can still occur but not nearly as frequently as in sexual
reproduction. Sexual reproduction leads to genetic variation in new generations of offspring.
This is fundamental to evolution. Involvement of sex cells No formation or fusion of gametes
(sex cell) Formation and fusion of gametes (sex cell) occurs Found in Lower organisms Higher
invertebrates and all vertebrates Unit of reproduction May be whole parent body or a bud or a
fragment or a single somatic cell Gamete Time taken Asexual reproduction is completed in a
very short period of time. Sexual reproduction can take several months to complete. Number of
offspring Two or more One or more.
BackgroundIn many applications, the composition of a collection o.pdf
Background:
In many applications, the composition of a collection of data items changes over time. Not only
are new data items added and existing ones removed, but data items may be duplicated. A list
data structure is a member of the general category of abstract data types called containers, whose
purpose is to hold other objects. In C++, lists are provided in the Standard Template Library.
However, for this assignment you will design and write your own linked list implementation to
support the ADT operations specified below.
Objective:
Design and implement the specifications for a List Abstract Data Type where the items in the list
are unsorted.
Requirements:
Define a list and develop a set of operations for creating and manipulating a list that satisfies the
list ADT specification.
List ADT Specification
Structure: The list elements are of ItemType. The list has a property called the current position
which designates the position of the last element accessed by GetNextItem during an iteration
through the list. Only ResetList and GetNextItem alter the current position.
Definitions (provided by the user):
MAX_ITEMS: A constant specifying the maximum capacity of items allowed on the list
Item Type: Class encapsulating the type of items in the list
RelationType: An enumeration type that consists of LESS, GREATER, EQUAL
Member function of ItemType that must be included:
RelationType ComparedTo(ItemType Item)
Function: Determines the ordering of two ItemType objects based on their keys
Precondition: Self and item have their key members initialized
Postcondition:
Function value = LESS if the key of self is less than the key of item
= GREATER if the key of self is greater than the key of item
= EQUAL if the keys are equal
Operations (provided by Unsorted List ADT)
Make Empty
Function: Initializes list to empty state
Preconditions: None
Postcondition: List is empty
Boolean IsFull
Function: Determines whether list is full
Preconditions: List has been initialized
Postcondition: Function value = (list is full)
int GetLength
Function: Determines the number of elements in list
Preconditions: List has been initialized
Postcondition: Function value = number of elements in list
ItemType GetItem(Item Typeitem, Boolean& found)
Function: Get list element whose key matches item’s key (if present)
Preconditions: List has been initialized
Key member of item is initialized
Postcondition: If there is an element someItem whose keymatches item’s key, then found =
true and copy of someItemis returned; otherwise found = false and item is returned
List is unchanged
PutItem(ItemType item)
Function: Puts item into list
Preconditions: List has been initialized
List is not full
Item is not in list
Postcondition: Item is in the list
DeleteItem(ItemType item)
Function: Deletes the element whose key matches item’s key
Preconditions: List has been initialized
Postcondition: One and only one element in list has a key matching item’s key
ResetList
Function: Initializes current p.
Which methodology may be used for detection of incorporated BrdUA.pdf
What is the first enzymatic activity in DNA synthesis?
Endonuclease
Primase
Exonuclease
DNA polymerase
LigaseA.
EndonucleaseB.
PrimaseC.
ExonucleaseD.
DNA polymeraseE.
Ligase
Solution
Ans. Correct option. B. Primase.
The first enzymatic activity in DNA synthesis (among the available options) is Primase.
DNA primase (or, primase) binds to ssDNA strands during initiation period. It synthesizes small
RNA primer to both the strands. DNA polymerase holoenzyme binds during elongation of DNA
synthesis after primase has synthesized RNA primers. It extends the RNA primers with DNA
nucleotide polymerization on it in 5’-3’ direction.
The exonuclease activity of DNA polymerase comes into action during repair or nick translation
during elongation. Other than that, no other endonuclease or exonuclease participate DNA
synthesis.
Ligase (DNA ligase) joins DNA fragments during nick translation much later than DNA pol
begins to synthesize DNA..
Which of the following is a difference between lightweight and heavy.pdf
What is the probability that a five card poker hand contains the jack of diamonds
and the two of spades?
Solution
Jack of diamonds and the two of spades:
No of ways to get 1 Jack of diamonds: = 1
No of ways to get 2 spades: 13C2 = 78
Total No of ways to contains the jack of diamonds and the two of spades = 1*78 = 78
and No of ways to five card poker hand = 52C5 = 2,598,960
Then Probability = 78/2,598,960 = 0.00003001.
What are the common conditions resulting in thrombocytosis Explain .pdf
Use the following information to answer questions 7-11 The following information applies to
the questions displayed below. The following information is available for Lock-Tite Company,
which produces special-order security products and uses a job order costing system. April 30
May 31 Inventories Raw materials Work in process Finlshed goods $39,000 $30,000 9,100
20,500 69,000 34,500 Activities and information for May Raw materials purchases (paid with
cash) Factory payroll (paid with cash) Factory overhead 186,000 300,000 Indirect materials
Indirect labor Other overhead costs 7,000 71,000 111,000 1,100,000 Sales (received in cash)
Predetermined overhead rate based on direct labor cost 55%
Solution
Answer - Exercise 15-7 T-accounts Factory Overheads Cost of material used
Raw Materials Raw Material 7000 125950 Work in process Raw Materials April-30 $39,000
RM-April 30 39000 7000 Factory Overhead Factory Payroll 71000 63050 Cost of goods sold
Add:Raw Material Purchases $186,000 RM Purchases 186000 188000 Work in process Other
Overheads 111000 Less : Raw Materials May-31 $30,000 Cost of material used
$195,000 less : Indirect Material used $7,000 RM May 31 30000 Direct Material
Used $188,000 Work in process Cost of direct labour used = $3,00,000 WIP-
April 30 9100 531550 Finished Goods Raw Material 188000 Cost of goods
Manufactured Factory Payroll 229000 Cost of Materials used $188,000 Factory
Overhead 125950 Cost of direct labour used $229,000 WIP-May 31 20500 Factory
Overhead applied [55% of $2,29,000] $125,950 Cost of Manufacturing for the period
$542,950 Factory Payroll Add: Work in process , April-30 $9,100 Cash 300000 71000
Factory Overheads Less : Work in process,May-31 $20,500 229000 Work in process
Cost of goods Manufactured $531,550 Cost of goods sold Cost of goods
Manufactured $531,550 Add : Finished Goods , April-30 $69,000 Finished Goods
Less : Finished Goods,May-31 $34,500 FG-April 30 69000 Cost of goods sold $566,050
Work in process 531550 566050 Cost of goods sold Gross Profit Sales
$1,100,000 FG May 31 34500 Less : Cost of goods sold (unadjusted) $566,050
Gross Profit $533,950 Overapplied or underapplied overheads = Actual
overheads - applied overheads Actual Overheads = Indirect Material + Indirect Labour +
other overhead costs = $7000 + $71000 + $111000 = $1,89,000 Underapplied overheads =
$189000 -$125950 = $63,050 Journal Entry to allocate underapplied
overheads Account Titles Debit Credit Cost of goods sold $63,050 Factory
Overheads $63,050 Income Statement (Partial) Sales $1,100,000
Cost of goods sold $566,050 Gross Profit $533,950.
What are some differences between the xylem and the phloem Solu.pdf
Use Simplex method and solve the following problem:
Maximize p = x+y subject to x+y <= 2, 3x+y >= 4
Which of teh following is the last matrix table?
A.
x y s1 s2 | b
0 1 3/2 1/2 1
1 0 1/2 -1/2 1
0 0 1 0 2
B.
x y s1 s2 | b
0 1 3/2 1/2 1
1 0 -1/2 1/2 1
0 0 1 0 2
C.
x y s1 s2 | b
0 1 3/2 1/2 1
1 0 -1/2 -1/2 1
0 0 1 0 2
D.
x y s1 s2 | b
0 1 3/2 -1/2 1
1 0 -1/2 -1/2 1
0 0 1 0 2
x y s1 s2 | b
0 1 3/2 1/2 1
1 0 1/2 -1/2 1
0 0 1 0 2
B.
x y s1 s2 | b
0 1 3/2 1/2 1
1 0 -1/2 1/2 1
0 0 1 0 2
C.
x y s1 s2 | b
0 1 3/2 1/2 1
1 0 -1/2 -1/2 1
0 0 1 0 2
D.
x y s1 s2 | b
0 1 3/2 -1/2 1
1 0 -1/2 -1/2 1
0 0 1 0 2
Solution
The step by step solution of the problem is given below -
Tableau #1
x y s1 s2 p
1 1 1 0 0 2
3 1 0 -1 0 4
-1 -1 0 0 1 0
Tableau #2
x y s1 s2 p
0 2/3 1 1/3 0 2/3
1 1/3 0 -1/3 0 4/3
0 -2/3 0 -1/3 1 4/3
Tableau #3
x y s1 s2 p
0 1 3/2 1/2 0 1
1 0 -1/2 -1/2 0 1
0 0 1 0 1 2
So, the correct answer is option (c)..
Two families of sloths exist commonly in Central and South America (t.pdf
There are a number of highly touted search engines for finding things of interest on the Internet.
Recently, a consumer rating system ranked two search engines ahead of the others. Now, a
computer user\'s magazine wishes to make the final determination regarding which one is
actually better at finding particular information. To do this, each search engine was used in an
attempt to locate specific information using specified keywords. Both search engines were
subjected to 100 queries. Search engine 1 successfully located the information 88 times and
search engine 2 located the information 80 times. Using a significance level equal to 0.05, what
is the critical value for the hypothesis test?
z =
z =
Solution
It is a two-tailed test.
Given a=0.05, the critical values are Z(0.025) = -1.96 or 1.96 (from standard normal table)
Answer: z =.
Two populations of rats are discovered on nearby islands. Describe t.pdf
think like an You have three resistors of resistance 30 ohm [Ohm]. There are__configurations
possible, with__unique combinations, resulting in___different resistance values. A
\"configuration\" is a way of combining the resistors. For example, two resistors in parallel is one
configuration; two resistors in series is a second configuration. A \"combination\" is the specific
way of combining given resistors to form an effective resistance. For example, combining
Resistor #1 and #2 in parallel is one combination; combining Resistor #1 and #3 in parallel is a
second combination. These combinations may or may not result in a unique effective resistance.
What is the greatest resistance that can be made from a combination of resistors, and what is the
effective resistance of this combination? What is the least resistance that can be made from a
combination of resistors, and what is the effective resistance of this combination?
Solution
b) GREATEST RESISTANCE CAN BE ACHIEVED THROUGH SERIES CONNECTION OF
THREE RESISTORS.
EFFECTIVE RESISTANCE=30+30+30=90 OHMS
C) LEAST RESISTANCE CAN BE ACHIEVED THROUGH PARALLEL CONNECTION OF
THREE RESISTORS.
EFFECTIVE RESISTANCE=(30*30)/(30+30)=15 OHMS
(15*30)/(15+30)=450/45=10 OHMS.
30-32 Nuclear pores permit the passage of chromosomes outward. glu.pdf
30-32 Nuclear pores permit the passage of chromosomes outward. glucose molecules outward.
assembled DNA molecules outward. RNA outward. sodium ions inward, potassium ions
outward. The cell walls of fungi are made from peptidoglycan. pseudomurein. cellulose.
chitin. Starch. Flattened sacks of membranes involved in the packaging and distribution of
molecules synthesized in the cell are known as Golgi bodies. microbodies. pinocytic vesicles.
vacuoles. chromosomes.
Solution
30). RNA outwarrd
This transport includes RNA and ribosomal proteins moving from nucleus to the cytoplasm.
31). Cellulose and chitin
32). Golgi bodies.
Review Questions for Exam 10182016 1. public class .pdf
Q3 GIVEN THE TREND EQUATION NO OF STUDENTS-3000+200 YEAR BASE: 2003-0
1.INTERPRET EQUATION RESULTS 2.FORECAST FOR 2020 3.WHEN WILL THE NO OF
STUDENTS REACH 6000
Solution
x no of years from 2003 y no of students y = 3000+200x 1) The regression
equation has a positive slope of 200 y intercept is 3000 i.e. no of students in the year 2000
Thus for every year 200 students increase 2) For 2020, x = 17, y = 3000+200(17) = 6400
3) When y =6000, 6000 = 3000+200x or x = 15 Hence in year 2018 students reach 6000
B Regression C D. All the above.
B-tree insertion (Draw node) Show the results of inserting the keys .pdf
B-tree insertion (Draw node) Show the results of inserting the keys 23, 98, 55, 37, 30, 47, 35,
15, 10, 75, 3, 53, 1, 33, 5, 28, 25, 90, 95, 56, 8, and 9 in order into an empty B-tree of order 5.
That is, each internal node in this B-tree other than the root must have at least 3 and at most 5
children (or at least 2 and at most 4 keys).
Solution
/*
* C++ Program to Implement B-Tree
*/
#include
#include
#include
using namespace std;
struct BTreeNode
{
int *data;
BTreeNode **child_ptr;
bool leaf;
int n;
}*root = NULL, *np = NULL, *x = NULL;
BTreeNode * init()
{
int i;
np = new BTreeNode;
np->data = new int[5];
np->child_ptr = new BTreeNode *[6];
np->leaf = true;
np->n = 0;
for (i = 0; i < 6; i++)
{
np->child_ptr[i] = NULL;
}
return np;
}
void traverse(BTreeNode *p)
{
cout<n; i++)
{
if (p->leaf == false)
{
traverse(p->child_ptr[i]);
}
cout << \" \" << p->data[i];
}
if (p->leaf == false)
{
traverse(p->child_ptr[i]);
}
cout< p[j])
{
temp = p[i];
p[i] = p[j];
p[j] = temp;
}
}
}
}
int split_child(BTreeNode *x, int i)
{
int j, mid;
BTreeNode *np1, *np3, *y;
np3 = init();
np3->leaf = true;
if (i == -1)
{
mid = x->data[2];
x->data[2] = 0;
x->n--;
np1 = init();
np1->leaf = false;
x->leaf = true;
for (j = 3; j < 5; j++)
{
np3->data[j - 3] = x->data[j];
np3->child_ptr[j - 3] = x->child_ptr[j];
np3->n++;
x->data[j] = 0;
x->n--;
}
for (j = 0; j < 6; j++)
{
x->child_ptr[j] = NULL;
}
np1->data[0] = mid;
np1->child_ptr[np1->n] = x;
np1->child_ptr[np1->n + 1] = np3;
np1->n++;
root = np1;
}
else
{
y = x->child_ptr[i];
mid = y->data[2];
y->data[2] = 0;
y->n--;
for (j = 3; j < 5; j++)
{
np3->data[j - 3] = y->data[j];
np3->n++;
y->data[j] = 0;
y->n--;
}
x->child_ptr[i + 1] = y;
x->child_ptr[i + 1] = np3;
}
return mid;
}
void insert(int a)
{
int i, temp;
x = root;
if (x == NULL)
{
root = init();
x = root;
}
else
{
if (x->leaf == true && x->n == 5)
{
temp = split_child(x, -1);
x = root;
for (i = 0; i < (x->n); i++)
{
if ((a > x->data[i]) && (a < x->data[i + 1]))
{
i++;
break;
}
else if (a < x->data[0])
{
break;
}
else
{
continue;
}
}
x = x->child_ptr[i];
}
else
{
while (x->leaf == false)
{
for (i = 0; i < (x->n); i++)
{
if ((a > x->data[i]) && (a < x->data[i + 1]))
{
i++;
break;
}
else if (a < x->data[0])
{
break;
}
else
{
continue;
}
}
if ((x->child_ptr[i])->n == 5)
{
temp = split_child(x, i);
x->data[x->n] = temp;
x->n++;
continue;
}
else
{
x = x->child_ptr[i];
}
}
}
}
x->data[x->n] = a;
sort(x->data, x->n);
x->n++;
}
int main()
{
int i, n, t;
cout<<\"enter the no of elements to be inserted\ \";
cin>>n;
for(i = 0; i < n; i++)
{
cout<<\"enter the element\ \";
cin>>t;
insert(t);
}
cout<<\"traversal of constructed tree\ \";
traverse(root);
getch();
}.
Please help me find where I do something wrong with my programI am.pdf
OPERATING SYSTEMS: DISK SCHEDULING
PLEASE SHOW YOUR WORK! I\'M TRYING TO UNDERSTAND HOW TO ACTUALLY
DO THE PROBLEM.
2. (3 points) Suppose that a disk drive has 3000 cylinders, numbered 0 to 2999. The drive is
currently serving a request at cylinder 140, and the previous request was at cylinder 125. The
queue of pending requests, in FIFO order, is
66, 147, 913, 289, 202, 1750, 15
Starting from the current head position, what is the total distance (in cylinders) that the disk arm
moves to satisfy all the pending requests, for each of the following disk-scheduling algorithms?
a. FCFS
b. SSTF
c. SCAN
d. LOOK
e. C-SCAN
3. Requests are not usually uniformly distributed. For example, a cylinder containing the file
system FAT or inodes can be expected to be accessed more frequently than a cylinder that only
contains files. Suppose you know that 50 percent of the requests are for a small, fixed number of
cylinders.
Propose a disk-scheduling algorithm (one that we have discussed) that gives even better
performance by taking advantage of this “hot spot” on the disk. Explain your answer and discuss
any problems and how you might mitigate them.
Solution
The drive is currently serving a request at cylinder 140
request queue -- 66,147,913,289,202,1750,15
a) FCFS
66<--- 140
66---->147
147 ----> 913
289 <----- 913
202<---- 289
202 ------>1750
15 <------ 1750
total time = |66-140|+|66-147|+|147-913|+|289-913|+||202-289|+|202-1750|+|15-1750|
=74+81+766+624+87+1548+1735
=4915
b) SSTF
140 ----> 147
147 ------>202
202----->289
66<----289
15<----66
15----->913
913----->1750
total time=7+55+87+223+51+898+837
=2158
c) SCAN
140--->147---->202---->289----->913----->1750----->2999---
0<-----15<------66<-----------------------------------------------
total time = |140-147|+|147-202|+|202-289|+|289-913|+|913-1750|+|1750-2999|+|2999-66|+|66-
15|+|15-0|
=7+55+87+624+837+1249+2933+51+15
=5858
d)LOOK
140--->147---->202---->289----->913----->1750---
15<------66<-----------------------------------------------
total time = |140-147|+|147-202|+|202-289|+|289-913|+|913-1750|+|1750-66|+|66-15|
=7+55+87+624+837+1684+51
=3345
e)C-SCAN
140--->147---->202---->289----->913----->1750----->2999---
66<-----15<------0<-----------------------------------------------
total time = |140-147|+|147-202|+|202-289|+|289-913|+|913-1750|+|1750-2999|+|2999-0|+|15-
0|+|66-15|
=7+55+87+624+837+1249+2999+15+51
=5924
SSTF can be the best disk scheduling algorithm when data is close to each other.
We can have hot spot in the middle of the disk . Aging can be done on older request by
increasing their priority.
List the transformations to the basic graph and then graph the funct.pdf
Joe\'s Coffee Shoppe orders fresh doughnuts each morning. Joe pays $3.20 per dozen doughnuts
and sell doughnuts for $0.4 each. The leftover doughnuts are salvaged at half the selling price.
Suppose that the demand for doughnuts is uniformly distributed from 16 dozens to 30 dozens.
How many doughnuts (in dozens) should Joe stock each day? A detailed analysis of past data
shows that the number of doughnuts sold per day is better described by a normal distribution,
with a mean 23 dozens and standard deviation 10 dozens. Now, how many doughnuts should Joe
stock each day? The expected demand was the same in parts (a) and (b), but the optimal order
quantities should have been different. What accounted for this different. What accounted for this
difference?
Solution.
In your own words discuss the following. 1. Briefly describe the p.pdf
In a survey conducted by a business management company, 51% of the U.S. workers siad they
have negotiated a pay raise at least once inthier lives. The sample size used in the study was not
disclosed.
A) Suppose that the survey had a sample size of n=500. Construct a condifidence interval
estimate for the proportion of all United States workers who have negotiated a pay raise.
Solution
Answer to the question)
P = 0.51
n = 500
Standard error SE = srt(0.51 *0.49/500)
SE = 0.0224
.
Let the confidence level = 95% , so the Z critical value = 1.96
.
The formula of cofndience interval is:
p - z* SE , p + z * SE
.
On plugging the values we get
0.51 - 1.96 *0.0224 , 0.51 +1.96 *0.0224
0.04661 , 0.5539.
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