using System;public category take a look at one.00m; Console.W.pdf
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using System; public category take a look at one.00m; Console.WriteLine (d); } } When I initial ran the on top of (or one thing similar) I expected it to output simply one (which is what it might are on .NET 1.0) - however in reality, the output was one.00. The decimal sort does not normalize itself - it remembers what percentage decimal digits it\'s (by maintaining the exponent wherever possible) and on info, zero could also be counted as a major digit. i do not understand the precise nature of what exponent is chosen (where there\'s a choice) once 2 totally different decimals ar increased, divided, added etc, however you will realize it attention-grabbing to manipulate with programs like the following: using System; public category take a look at zero.00000000000010000m; whereas (d != 0m) } } Solution using System; public category take a look at one.00m; Console.WriteLine (d); } } When I initial ran the on top of (or one thing similar) I expected it to output simply one (which is what it might are on .NET 1.0) - however in reality, the output was one.00. The decimal sort does not normalize itself - it remembers what percentage decimal digits it\'s (by maintaining the exponent wherever possible) and on info, zero could also be counted as a major digit. i do not understand the precise nature of what exponent is chosen (where there\'s a choice) once 2 totally different decimals ar increased, divided, added etc, however you will realize it attention-grabbing to manipulate with programs like the following: using System; public category take a look at zero.00000000000010000m; whereas (d != 0m) } }.
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To do this work you will need to connect to a teaching server
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facility and entering the web interface address including
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The address of the web interface is http://aineko.eps.surrey.ac.uk:34448
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to install DIGITS yourself on your own machine then you might find those instructions a useful
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1 https://en.wikipedia.org/wiki/Accelerando
You must first create your own work area on the server and download the database.
Remotely login to the teaching server using ssh. In the Linux labs you can open up a terminal
window using Ctrl-Alt-T and then enter the following
ssh aineko.eps.surrey.ac.uk –l your_username
Note that character after the hyphen is a lower case L not a 1! Please substitute username you’re
your actual username. The password is your URN. If you are prompted by an “are you sure…”
prompt just type the word yes. If you can’t log in the we will have to create you an account.
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Problem solving using computers - Chapter 1
Problem solving using computers | Python Basics | College Presentation | Python Programming Language
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higher.
CNNs are state of the art deep neural networks that perform well at machine perception tasks such
as image classification. You will start learning about these in detail within Week 6.
To do this work you will need to connect to a teaching server
called aineko1 on the campus network. The server is not
visible off campus, so you are encouraged to use the lab or
library PCs on campus to do this work. If you want to use
your own laptop then you will be able to connect via the
campus wide “eduroam” wifi network. The campus wide
“The Cloud” network will not work. It is possible to connect
from offcampus using the https://anywhere.surrey.ac.uk
facility and entering the web interface address including
http:// into the text box in the upper-right of the portal.
The address of the web interface is http://aineko.eps.surrey.ac.uk:34448
Verify now that you can connect to the web interface via your browser, or you will not be able to
progress any further.
It is possible to install your own version of DIGITS on your local lab PC (e.g. in the Penguin lab). Since
we have a large class this year this may be useful if the aineko server becomes very busy. You can
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1 https://en.wikipedia.org/wiki/Accelerando
You must first create your own work area on the server and download the database.
Remotely login to the teaching server using ssh. In the Linux labs you can open up a terminal
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ssh aineko.eps.surrey.ac.uk –l your_username
Note that character after the hyphen is a lower case L not a 1! Please substitute username you’re
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1 of 6 LAB 5 IMAGE FILTERING ECE180 Introduction to.docx
1 of 6
LAB 5: IMAGE FILTERING
ECE180: Introduction to Signal Processing
OVERVIEW
You have recently learned about the convolution sum that serves as the basis of the FIR filter difference equation. The filter
coefficient sequence {𝑏𝑘} – equivalent to the filter’s impulse response ℎ[𝑛] – may be viewed as a one-dimensional moving
window that slides over the input signal 𝑥[𝑛] to compute the output signal 𝑦[𝑛] at each time step. Extending the moving
window concept to a 2-D array that slides over an image pixel array provides a useful and popular way to filter an image.
In this lab project you will implement two types of moving-window image filters, one based on convolution and the other
based on the median value of the pixel grayscale values spanned by the window. You will also gain experience with the
built-in image convolution filter imfilter.
OUTLINE
1. Develop and test a 33 median filter
2. Develop and test a 33 convolution filter
3. Evaluate the median and convolution filters to reduce noise while preserving edges
4. Study the behavior of various 33 convolution filter kernels for smoothing, edge detection, and sharpening
5. Learn how to use imfilter to convolution-filter color images, and study the various mechanisms offered by
imfilter to deal with boundary effects
PREPARATION – TO BE COMPLETED BEFORE LAB
Study these tutorial videos:
1. Nested “for” loops -- http://youtu.be/q2xfz8mOuSI?t=1m8s (review this part)
2. Functions -- http://youtu.be/0zTmMIh6I8A (review as needed)
Ensure that you have added the ECE180 DFS folders to your MATLAB path, especially the “images” and “matlab” subfolders.
Follow along with the tutorial video http://youtu.be/MEqUd0dJNBA, if necessary.
LAB ACTIVITIES
1. Develop and test a 33 median filter function:
1.1. Implement the following algorithm as the function med3x3:
TIP: First implement and debug the algorithm as a script and then convert it to a function as a final step. Use any
of the smaller grayscale images from the ECE180 “images” folder as you develop the function, or use the test
image X described in the Step 1.2.
(a) Create the function template and save it to an .m file with the same name as the function,
(b) Accept a grayscale image x as the function input,
http://youtu.be/q2xfz8mOuSI?t=1m8s
http://youtu.be/0zTmMIh6I8A
http://youtu.be/MEqUd0dJNBA
2 of 6
(c) Copy x to the output image y and then initialize y(:) to zero; this technique creates y as the same size and
data type as x,
(d) Determine the number of image rows and columns (see size),
(e) Loop over all pixels in image x (subject to boundary limits):
Extract a 33 neighborhood (subarray) about the current pixel,
Flatten the 2-D array to a 1-D array,
Sort the 1-D array values (see sort),
Assign the middle value of the sorted array to the current output pixel, and
(f) Return the median-filtered image y.
1.2. Enter load lab_5_verify to load the
Develop a system flowchart and then write a menu-driven C++ program .pdf
Develop a system flowchart and then write a menu-driven C++ program that uses user-defined
functions arrays, and a random number generator.
Upon program execution, the screen will be cleared and the menu shown below will appear at
the top of the screen and centered. The menu items are explained below.
Help Smallest Largest Quit
H or h ( for Help ) option will invoke a function named help() which will display a help screen.
The help screen(s) should guide the user how to interact with the program, type of data to be
used, and what results would the program produce. Each help screen should remain on the
monitor until the user strikes any key. Once the user completes reading the help() screens,
striking any key will clear the screen and the menu is displayed again.
The main() function will declare an array of 60 elements. The elements are of type double. The
menu options S and L, will ask the user for the actual number of elements for the array. The
program must verify that the actual number of elements does not exceed 60 and is not a negative
number or 0. Your code must check for these.
S or s ( for Smallest ) option will invoke a function named smallest( ) which will prompt the user
for the number of elements for the array to be examined using the function sizeOfArray( ) which
will read and validate the desired array elements and returns it to the calling function. The
program will then use the returned size of the array to fill the array using the number generator
functions shown at the end of this assignment. Your program will ask the user for the range of
values for the elements of the array and uses these numbers in a random generator function to fill
the array. Once the array is filled, the program will call the function findSmallest( ) which will
find and return the smallest number in the array to the calling function. The program will then
call the function frequency(), that will compute and return the frequency of occurrence of the
smallest number. The program will then display the array elements, the smallest number, and its
frequency of occurrence using the function display( ), in the format shown below. The output
shown is for an array of six elements with an array identifier a.
a[0] = xxxx.xx
a[1] = xxxx.xx
a[2] = xxxx.xx
a[3] = xxxx.xx
a[4] = xxxx.xx
a[5] = xxxx.xx
Smallest no. = xxxx.xx Frequency = xx
Note that for the array elements, two digits after the decimal point is required (i.e., 365.24) and
the frequence of occurrence is of type integer.
The function prototypes to be used are as follows:
void smallest(double s[], int size);
where s is the original array declared in the main() function and size is the maximum array size
specified (60 elements in our case).
//gets the desired array size and returns it
int sizeOfArray(void);
//finds the smallest number in the array and returns it via the return statement.
double findSmallest(double s[], int myS);
//where s is the specified array name and myS is the desired size of the array.
Cu(H2O)42+(aq) + 4CN-(aq) = Cu(CN)42-(aq) +4 H2O(l)Cu+2 is a Lew.pdf
Cu(H2O)42+(aq) + 4CN-(aq) <=> Cu(CN)42-(aq) +4 H2O(l)
Cu+2 is a Lewis acid as it has vacantd-orbitals that can accomodate an electron pair(Electron
pairacceptor)
CN- is a Lewis base as it has lone pair ofelectron that is donated to Cu+2 in the reaction(Electron
pair donor)
H2O is a Lewis base as the Oxygen atom haslone pair of electrons and therefre, it acts as an
electronpair donor.
Solution
Cu(H2O)42+(aq) + 4CN-(aq) <=> Cu(CN)42-(aq) +4 H2O(l)
Cu+2 is a Lewis acid as it has vacantd-orbitals that can accomodate an electron pair(Electron
pairacceptor)
CN- is a Lewis base as it has lone pair ofelectron that is donated to Cu+2 in the reaction(Electron
pair donor)
H2O is a Lewis base as the Oxygen atom haslone pair of electrons and therefre, it acts as an
electronpair donor..
C) CS2 for sureA, B, D and E are all ionic compoundsSolution.pdf
C) CS2 for sure
A, B, D and E are all ionic compounds
Solution
C) CS2 for sure
A, B, D and E are all ionic compounds.
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1 of 6 LAB 5 IMAGE FILTERING ECE180 Introduction to.docx
1 of 6
LAB 5: IMAGE FILTERING
ECE180: Introduction to Signal Processing
OVERVIEW
You have recently learned about the convolution sum that serves as the basis of the FIR filter difference equation. The filter
coefficient sequence {𝑏𝑘} – equivalent to the filter’s impulse response ℎ[𝑛] – may be viewed as a one-dimensional moving
window that slides over the input signal 𝑥[𝑛] to compute the output signal 𝑦[𝑛] at each time step. Extending the moving
window concept to a 2-D array that slides over an image pixel array provides a useful and popular way to filter an image.
In this lab project you will implement two types of moving-window image filters, one based on convolution and the other
based on the median value of the pixel grayscale values spanned by the window. You will also gain experience with the
built-in image convolution filter imfilter.
OUTLINE
1. Develop and test a 33 median filter
2. Develop and test a 33 convolution filter
3. Evaluate the median and convolution filters to reduce noise while preserving edges
4. Study the behavior of various 33 convolution filter kernels for smoothing, edge detection, and sharpening
5. Learn how to use imfilter to convolution-filter color images, and study the various mechanisms offered by
imfilter to deal with boundary effects
PREPARATION – TO BE COMPLETED BEFORE LAB
Study these tutorial videos:
1. Nested “for” loops -- http://youtu.be/q2xfz8mOuSI?t=1m8s (review this part)
2. Functions -- http://youtu.be/0zTmMIh6I8A (review as needed)
Ensure that you have added the ECE180 DFS folders to your MATLAB path, especially the “images” and “matlab” subfolders.
Follow along with the tutorial video http://youtu.be/MEqUd0dJNBA, if necessary.
LAB ACTIVITIES
1. Develop and test a 33 median filter function:
1.1. Implement the following algorithm as the function med3x3:
TIP: First implement and debug the algorithm as a script and then convert it to a function as a final step. Use any
of the smaller grayscale images from the ECE180 “images” folder as you develop the function, or use the test
image X described in the Step 1.2.
(a) Create the function template and save it to an .m file with the same name as the function,
(b) Accept a grayscale image x as the function input,
http://youtu.be/q2xfz8mOuSI?t=1m8s
http://youtu.be/0zTmMIh6I8A
http://youtu.be/MEqUd0dJNBA
2 of 6
(c) Copy x to the output image y and then initialize y(:) to zero; this technique creates y as the same size and
data type as x,
(d) Determine the number of image rows and columns (see size),
(e) Loop over all pixels in image x (subject to boundary limits):
Extract a 33 neighborhood (subarray) about the current pixel,
Flatten the 2-D array to a 1-D array,
Sort the 1-D array values (see sort),
Assign the middle value of the sorted array to the current output pixel, and
(f) Return the median-filtered image y.
1.2. Enter load lab_5_verify to load the
Develop a system flowchart and then write a menu-driven C++ program .pdf
Develop a system flowchart and then write a menu-driven C++ program that uses user-defined
functions arrays, and a random number generator.
Upon program execution, the screen will be cleared and the menu shown below will appear at
the top of the screen and centered. The menu items are explained below.
Help Smallest Largest Quit
H or h ( for Help ) option will invoke a function named help() which will display a help screen.
The help screen(s) should guide the user how to interact with the program, type of data to be
used, and what results would the program produce. Each help screen should remain on the
monitor until the user strikes any key. Once the user completes reading the help() screens,
striking any key will clear the screen and the menu is displayed again.
The main() function will declare an array of 60 elements. The elements are of type double. The
menu options S and L, will ask the user for the actual number of elements for the array. The
program must verify that the actual number of elements does not exceed 60 and is not a negative
number or 0. Your code must check for these.
S or s ( for Smallest ) option will invoke a function named smallest( ) which will prompt the user
for the number of elements for the array to be examined using the function sizeOfArray( ) which
will read and validate the desired array elements and returns it to the calling function. The
program will then use the returned size of the array to fill the array using the number generator
functions shown at the end of this assignment. Your program will ask the user for the range of
values for the elements of the array and uses these numbers in a random generator function to fill
the array. Once the array is filled, the program will call the function findSmallest( ) which will
find and return the smallest number in the array to the calling function. The program will then
call the function frequency(), that will compute and return the frequency of occurrence of the
smallest number. The program will then display the array elements, the smallest number, and its
frequency of occurrence using the function display( ), in the format shown below. The output
shown is for an array of six elements with an array identifier a.
a[0] = xxxx.xx
a[1] = xxxx.xx
a[2] = xxxx.xx
a[3] = xxxx.xx
a[4] = xxxx.xx
a[5] = xxxx.xx
Smallest no. = xxxx.xx Frequency = xx
Note that for the array elements, two digits after the decimal point is required (i.e., 365.24) and
the frequence of occurrence is of type integer.
The function prototypes to be used are as follows:
void smallest(double s[], int size);
where s is the original array declared in the main() function and size is the maximum array size
specified (60 elements in our case).
//gets the desired array size and returns it
int sizeOfArray(void);
//finds the smallest number in the array and returns it via the return statement.
double findSmallest(double s[], int myS);
//where s is the specified array name and myS is the desired size of the array.
Similar to using System;public category take a look at one.00m; Console.W.pdf (20)
1 of 6 LAB 5 IMAGE FILTERING ECE180 Introduction to.docx
1 of 6 LAB 5 IMAGE FILTERING ECE180 Introduction to.docx
Develop a system flowchart and then write a menu-driven C++ program .pdf
Develop a system flowchart and then write a menu-driven C++ program .pdf
More from apnafreez
Cu(H2O)42+(aq) + 4CN-(aq) = Cu(CN)42-(aq) +4 H2O(l)Cu+2 is a Lew.pdf
Cu(H2O)42+(aq) + 4CN-(aq) <=> Cu(CN)42-(aq) +4 H2O(l)
Cu+2 is a Lewis acid as it has vacantd-orbitals that can accomodate an electron pair(Electron
pairacceptor)
CN- is a Lewis base as it has lone pair ofelectron that is donated to Cu+2 in the reaction(Electron
pair donor)
H2O is a Lewis base as the Oxygen atom haslone pair of electrons and therefre, it acts as an
electronpair donor.
Solution
Cu(H2O)42+(aq) + 4CN-(aq) <=> Cu(CN)42-(aq) +4 H2O(l)
Cu+2 is a Lewis acid as it has vacantd-orbitals that can accomodate an electron pair(Electron
pairacceptor)
CN- is a Lewis base as it has lone pair ofelectron that is donated to Cu+2 in the reaction(Electron
pair donor)
H2O is a Lewis base as the Oxygen atom haslone pair of electrons and therefre, it acts as an
electronpair donor..
C) CS2 for sureA, B, D and E are all ionic compoundsSolution.pdf
C) CS2 for sure
A, B, D and E are all ionic compounds
Solution
C) CS2 for sure
A, B, D and E are all ionic compounds.
Before getting public issues he should consider are 1. Increase i.pdf
Before getting public issues he should consider are :
1. Increase in scrutiny by regulatory authorities
2. Become answerable to shareholders.
3. Take consideration on any big decision from new shareholders.
4. Bearing additional cost of filing financial IPO
Solution
Before getting public issues he should consider are :
1. Increase in scrutiny by regulatory authorities
2. Become answerable to shareholders.
3. Take consideration on any big decision from new shareholders.
4. Bearing additional cost of filing financial IPO.
Assets=Liabilities+Stockholders EquityCash given $1000 on the sa.pdf
Assets=Liabilities+Stockholder\'s Equity
Cash given $1000 on the same day and $1000 is Accounts Payable
Cash given $1000 on the same day accounting equation -$1000=Liabilities-$1000
Accounts payable for remaining half of the expenses accounting equation Assets= +$1000-$1000
Solution
Assets=Liabilities+Stockholder\'s Equity
Cash given $1000 on the same day and $1000 is Accounts Payable
Cash given $1000 on the same day accounting equation -$1000=Liabilities-$1000
Accounts payable for remaining half of the expenses accounting equation Assets= +$1000-$1000.
1)Interest per year 8106= 135Simple interest = Amount RateTime.pdf
1)Interest per year 810/6= 135
Simple interest = Amount * Rate*Time
135 = 3750 *r * 1
135 = 3750*r
r = 135/3750
= .036 or 3.6%
2) Quarterly rate = 4.9/4 = 1.225% [4 quarters in a year]
Term (number of quarters) = IN[Future value/present value]/IN(1+r)
= IN[8600/2000]/IN[1+.01225]
= IN 4.3/ IN 1.01225
= 1.45862/ .012176
= 119.79468
Number of years =119.79468 / 4
= 29.95 years
Term : 29.95
Solution
1)Interest per year 810/6= 135
Simple interest = Amount * Rate*Time
135 = 3750 *r * 1
135 = 3750*r
r = 135/3750
= .036 or 3.6%
2) Quarterly rate = 4.9/4 = 1.225% [4 quarters in a year]
Term (number of quarters) = IN[Future value/present value]/IN(1+r)
= IN[8600/2000]/IN[1+.01225]
= IN 4.3/ IN 1.01225
= 1.45862/ .012176
= 119.79468
Number of years =119.79468 / 4
= 29.95 years
Term : 29.95
.
AnswerAnswer1. The body of a fungus is in the form of solid sh.pdf
Answer:
Answer:
1. The body of a fungus is in the form of solid sheets: True
The body of fungus is called mycelium, it is not differentiated into roots, stem or leaf.
2. Yeasts are a kind of fungus: True
Yeast are unicellular fungus
3. False
Spore are haploid, the need to fuse with another to produce diploid, multicellular individual.
4. False
Phloem is alive at maturity because movement of material within the phloem requires energy.
5. True
6. True
7. True
Ferns have motile sperms and non motile egg.
8. False
Most of the plant species are angiosperms.
9. False
Angiosperms are divided into monocots and dicots
10. Charophyte---e. not a plant
11. fern---d. seedless vascular plant
12. pine tree—b. gymnosperm
13. moss—c. non-vascular plant
14. palm tree—a. angiosperm
Solution
Answer:
Answer:
1. The body of a fungus is in the form of solid sheets: True
The body of fungus is called mycelium, it is not differentiated into roots, stem or leaf.
2. Yeasts are a kind of fungus: True
Yeast are unicellular fungus
3. False
Spore are haploid, the need to fuse with another to produce diploid, multicellular individual.
4. False
Phloem is alive at maturity because movement of material within the phloem requires energy.
5. True
6. True
7. True
Ferns have motile sperms and non motile egg.
8. False
Most of the plant species are angiosperms.
9. False
Angiosperms are divided into monocots and dicots
10. Charophyte---e. not a plant
11. fern---d. seedless vascular plant
12. pine tree—b. gymnosperm
13. moss—c. non-vascular plant
14. palm tree—a. angiosperm.
A. This constitutes part of a gene that encodes information for RNA .pdf
A. This constitutes part of a gene that encodes information for RNA Polymerase : Operator.
B. An activator and a repressor are the examples of : Transcription factors.
C. regulation of gene expression leads to changes in cell functions and in multicellular
organisms,can also lead to changes in cellular phenotype.
D. In bacteria, related genes can be organised in units called Operon controlled by a single
Promoter.
E. Lactose is an example of a compound that can induce lac operon.
F.Repressors are proteins which bind to DNA and alter gene expression.
G. The lac operon can be turned ON by the accumulation of the CRP-cAMP complex. Once
tghis complex binds to the DNA , it causes a change in the conformation of the DNA double
helix , making the operator, for the lac operon more attractive to RNA Polymerase.
H. Inhibition of a gene\'s expression often called Negative control , is due to the activity of a
negative regulator (Repressor) which can act to block RNA polymaerase from assessing the
promoter of a gene or inhibit RNA Polymerase from physically moving down the DNA strand.
I. The lac operon is an example of an Inducible operon because it can be induced to turn ON by
the presence of Lactose.
J. activation of a gene\'s expression often called, Positive control is due to the activity of the
Positive regulator (activator), to attract RNA polymerase to the Promoter for the gene.
K. People who cannot digest lactose, because they fail to produce the enzyme lactase, are said to
be lactose intolerant.
Solution
A. This constitutes part of a gene that encodes information for RNA Polymerase : Operator.
B. An activator and a repressor are the examples of : Transcription factors.
C. regulation of gene expression leads to changes in cell functions and in multicellular
organisms,can also lead to changes in cellular phenotype.
D. In bacteria, related genes can be organised in units called Operon controlled by a single
Promoter.
E. Lactose is an example of a compound that can induce lac operon.
F.Repressors are proteins which bind to DNA and alter gene expression.
G. The lac operon can be turned ON by the accumulation of the CRP-cAMP complex. Once
tghis complex binds to the DNA , it causes a change in the conformation of the DNA double
helix , making the operator, for the lac operon more attractive to RNA Polymerase.
H. Inhibition of a gene\'s expression often called Negative control , is due to the activity of a
negative regulator (Repressor) which can act to block RNA polymaerase from assessing the
promoter of a gene or inhibit RNA Polymerase from physically moving down the DNA strand.
I. The lac operon is an example of an Inducible operon because it can be induced to turn ON by
the presence of Lactose.
J. activation of a gene\'s expression often called, Positive control is due to the activity of the
Positive regulator (activator), to attract RNA polymerase to the Promoter for the gene.
K. People who cannot digest lactose, because they fail to produce the en.
A. ANSAutosomal dominant (a trait could passed down from previous.pdf
A. ANS:
Autosomal dominant (a trait could passed down from previous generation), X – linked recessive
(in this disease the dominant gene is not expressed in female, it only expressed in male) and
Autosomal recessive in female (Females are only carriers) effects would be included according
to pedigree.
B. ANS:
Leber hereditary optic atrophy which causes blindness in adulthood predominantly occurs in
male, in this the inheritance a point mutation was arisen in mitochondrial DNA and this affect
causing gene is directly transmitted from mother to offspring. It is mitochondrial inherited
disease in which a retinal ganglion cell was degenerated, and it leads to lose of central vision.
C. ANS:
The individuals II 3 (Male) and II 4 (Female) both are normal, so the offspring always should be
normal (According to the pedigree in offspring 4 (Female) is normal).
D. ANS:
The individuals II 2 (Male) and II 1 (Female) both are not in normal condition. According to the
pedigree male is normal and the female is not normal so the entire offspring is affected in this
disease. (According to the pedigree in offspring 1 (Female) is carrier and 2 & 3 (Males) are
affected).
Solution
A. ANS:
Autosomal dominant (a trait could passed down from previous generation), X – linked recessive
(in this disease the dominant gene is not expressed in female, it only expressed in male) and
Autosomal recessive in female (Females are only carriers) effects would be included according
to pedigree.
B. ANS:
Leber hereditary optic atrophy which causes blindness in adulthood predominantly occurs in
male, in this the inheritance a point mutation was arisen in mitochondrial DNA and this affect
causing gene is directly transmitted from mother to offspring. It is mitochondrial inherited
disease in which a retinal ganglion cell was degenerated, and it leads to lose of central vision.
C. ANS:
The individuals II 3 (Male) and II 4 (Female) both are normal, so the offspring always should be
normal (According to the pedigree in offspring 4 (Female) is normal).
D. ANS:
The individuals II 2 (Male) and II 1 (Female) both are not in normal condition. According to the
pedigree male is normal and the female is not normal so the entire offspring is affected in this
disease. (According to the pedigree in offspring 1 (Female) is carrier and 2 & 3 (Males) are
affected)..
888678123317368915236 #include stdio.h #include stdlib.h.pdf
8886781233
17368915236
#include
#include
int main() {
int i, n, *data , p;
/* get the number of inputs from the user */
printf(\"Enter the number of inputs:\");
scanf(\"%d\", &n);
/* dynamically allocate memory to store i/p values */
data = (int *) malloc(sizeof(int) * n);
/* get the input data from the user */
for (i = 0; i < n; i++) {
printf(\"data[%d]: \", i);
scanf(\"%d\", &data[i]);
}
/* sorts the given numbers */
bubbleSort(data, n ,p );
}
void bubbleSort(int *data, int n , int p) {
int i, temp, k = 1, l,j;
if (n > 0) {
for (i = n; i > 0; i--) {
if (data[i] > data[i - 1])
{
temp = data[i];
data[i] = data[i - 1];
data[i - 1] = temp;
}
if( p = = 0)
{
p++;
l = n;
for ( i = 0 ; i < n ; i++)
{
prinft ( %d, data[i])
}
bubbleSort(data, i - 2 , p );
}
else
{
for ( j = n - 1 ; j < l ; j++)
{
if (data[j] > data[j - 1])
{
temp = data[j];
data[j] = data[j - 1];
data[j - 1] = temp;
}
}
for ( j = 0 ; j < l ; j++)
{
prinft ( %d, data[j])
}
bubbleSort(data, i - 2,p);
}
}
}
return;
}
Solution
8886781233
17368915236
#include
#include
int main() {
int i, n, *data , p;
/* get the number of inputs from the user */
printf(\"Enter the number of inputs:\");
scanf(\"%d\", &n);
/* dynamically allocate memory to store i/p values */
data = (int *) malloc(sizeof(int) * n);
/* get the input data from the user */
for (i = 0; i < n; i++) {
printf(\"data[%d]: \", i);
scanf(\"%d\", &data[i]);
}
/* sorts the given numbers */
bubbleSort(data, n ,p );
}
void bubbleSort(int *data, int n , int p) {
int i, temp, k = 1, l,j;
if (n > 0) {
for (i = n; i > 0; i--) {
if (data[i] > data[i - 1])
{
temp = data[i];
data[i] = data[i - 1];
data[i - 1] = temp;
}
if( p = = 0)
{
p++;
l = n;
for ( i = 0 ; i < n ; i++)
{
prinft ( %d, data[i])
}
bubbleSort(data, i - 2 , p );
}
else
{
for ( j = n - 1 ; j < l ; j++)
{
if (data[j] > data[j - 1])
{
temp = data[j];
data[j] = data[j - 1];
data[j - 1] = temp;
}
}
for ( j = 0 ; j < l ; j++)
{
prinft ( %d, data[j])
}
bubbleSort(data, i - 2,p);
}
}
}
return;
}.
18) One device is attached to a physical interface The switch c.pdf
18) One device is attached to a physical interface
The switch can be remotely managed
The default VIan has been configured
19) Flow control
20) Encapsulation
21) HTTP,TCP,IP,Ethernet
22) They define how messages are exchanged between the source and the destination
Solution
18) One device is attached to a physical interface
The switch can be remotely managed
The default VIan has been configured
19) Flow control
20) Encapsulation
21) HTTP,TCP,IP,Ethernet
22) They define how messages are exchanged between the source and the destination.
1.Joseph could not restrain himself before all those who stood by hi.pdf
1.Joseph could not restrain himself before all those who stood by him, and he cried out, “Make
everyone go out from me!” So no one stood with him while Joseph made himself known to his
brothers. And he wept aloud, and the Egyptians and the house of Pharaoh heard it. Then Joseph
said to his brothers, “I am Joseph; does my father still live?” But his brothers could not answer
him, for they were dismayed in his presence.
a. Joseph could not restrain himself before all those who stood by him: Joseph ordered all the
Egyptians out of the room and was then alone with his brothers. His great emotion showed that
Joseph did not cruelly manipulate his brothers. He was directed by God to make these
arrangements and it hurt him to do it.
b. Joseph made himself known to his brothers: This perhaps means that Joseph told them he was
Joseph and showed his brothers that he was circumcised. Jewish legend says the brothers could
never believe this high Egyptian official was Joseph unless he showed he was circumcised.
c. But his brothers could not answer him, for they were dismayed in his presence: Because of the
punishment they anticipated, the great emotion of Joseph, his manner of revelation, and the total
shock of learning Joseph was not only alive but right in front of them, the brothers were
dismayed.
Pharaoh and Joseph send the brothers home with many gifts.
Joseph gave them carts, according to the command of Pharaoh, and he gave them provisions for
the journey: The sons of Israel received transportation, provision, garments, and riches because
of who their favored brother was. Pharaoh blessed the sons of Jacob for Joseph’s sake.
i. “To return to Canaan with ‘carts from Egypt’ was the cultural equivalent of landing a jumbo jet
among a tribe of isolated savages. It would be the stuff legends are made of.” (Boice)
b. See that you do not become troubled along the way: The idea behind the words “become
troubled” is literally become angry or quarrel. Joseph knew as soon as these men left his
presence they would be tempted to act in selfish, unspiritual ways. They had to anticipate and
guard against this.
Jacob hears the good news – that Joseph lives.
Then they went up out of Egypt, and came to the land of Canaan to Jacob their father. And they
told him, saying, “Joseph is still alive, and he is governor over all the land of Egypt.” And
Jacob’s heart stood still, because he did not believe them. But when they told him all the words
which Joseph had said to them, and when he saw the carts which Joseph had sent to carry him,
the spirit of Jacob their father revived. Then Israel said, “It is enough. Joseph my son is still
alive. I will go and see him before I die.”
a. He did not believe them: Jacob was told Joseph was dead and believed it. Then he was told
Joseph was alive, and he did not believe it until his sons told him the words of Joseph and
showed him the blessings that came to them through Joseph. Then he believed Joseph was alive,
though he had not yet seen.
(a) Total amount of I2 = 501000 x 0.010 = 0.0005 mol = 0.5 mmolLe.pdf
(a) Total amount of I2 = 50/1000 x 0.010 = 0.0005 mol = 0.5 mmol
Let a be the number of millimoles of I2 extracted
Millimoles of I2 left in aqueous phase = 0.5 - a
Distribution coefficient D = [I2](organic)/[I2](aqueous)
= [(mmol of I2)(organic)/V(organic)]/[mmol of I2)(aqueous)/V(aqueous)]
= [(a/50)]/[(0.5 - a)/50] = 85
a/(0.5 - a) = 85
a = 0.4942 mmol
Thus amount of I2 extracted = 0.4942 mmol
(b) Amount of I2 left in aqueous phase
= 0.5 - 0.4942 = 0.0058 mmol = 5.8 x 10-6 mol
Concentration of I2 in aqueous phase = moles/volume (in L)
= 5.8 x 10-6/0.050 = 1.16 x 10-4 M
Solution
(a) Total amount of I2 = 50/1000 x 0.010 = 0.0005 mol = 0.5 mmol
Let a be the number of millimoles of I2 extracted
Millimoles of I2 left in aqueous phase = 0.5 - a
Distribution coefficient D = [I2](organic)/[I2](aqueous)
= [(mmol of I2)(organic)/V(organic)]/[mmol of I2)(aqueous)/V(aqueous)]
= [(a/50)]/[(0.5 - a)/50] = 85
a/(0.5 - a) = 85
a = 0.4942 mmol
Thus amount of I2 extracted = 0.4942 mmol
(b) Amount of I2 left in aqueous phase
= 0.5 - 0.4942 = 0.0058 mmol = 5.8 x 10-6 mol
Concentration of I2 in aqueous phase = moles/volume (in L)
= 5.8 x 10-6/0.050 = 1.16 x 10-4 M.
Nursing is a healthcare profession in which they serve the patients .pdf
Nursing is a healthcare profession in which they serve the patients and ought to have medical
knowledge. Nurses generally are more in contact with the patients and have knowledge of
patient\'s relatives, behaviour, experience in dealing with patients, communication skills. They
are the most important connection link between the doctors and the patient. Nurses also have
knowledge about the medical cases, projects related to community and cultural awareness.Hence,
they form integral part of the medical system. In genetic testing, various people from higher to
lower classes, different caste and creed come for checkup. The nurses are supposed to inform
and consult the patients about the necessities of genetic testing. The informed consent is
necessary to be obtained by explaining the importance of genetic testing. Also, the ethical issues
are to be followed by each and every medical institution. Since the public has trust in nurses,
they are supposed to convince the patients about the genetic testing advantages. They help the
people to understand the basics of genetic testing and the cons of it if not performed before the
delivery. Some set of population think that carrying out these tests are against the will of nature
or sometimes economical issues arise due to higher charges of these tests. Hence it is necessary
to make people aware of their misconceptions and provide them good medical facilities.
Solution
Nursing is a healthcare profession in which they serve the patients and ought to have medical
knowledge. Nurses generally are more in contact with the patients and have knowledge of
patient\'s relatives, behaviour, experience in dealing with patients, communication skills. They
are the most important connection link between the doctors and the patient. Nurses also have
knowledge about the medical cases, projects related to community and cultural awareness.Hence,
they form integral part of the medical system. In genetic testing, various people from higher to
lower classes, different caste and creed come for checkup. The nurses are supposed to inform
and consult the patients about the necessities of genetic testing. The informed consent is
necessary to be obtained by explaining the importance of genetic testing. Also, the ethical issues
are to be followed by each and every medical institution. Since the public has trust in nurses,
they are supposed to convince the patients about the genetic testing advantages. They help the
people to understand the basics of genetic testing and the cons of it if not performed before the
delivery. Some set of population think that carrying out these tests are against the will of nature
or sometimes economical issues arise due to higher charges of these tests. Hence it is necessary
to make people aware of their misconceptions and provide them good medical facilities..
yes.because the deposted money can be multiply with 10 and it beco.pdf
yes.
because the deposted money can be multiply with 10 and it becomes 100*10= $1000
Solution
yes.
because the deposted money can be multiply with 10 and it becomes 100*10= $1000.
When the substituent groups are oriented in the s.pdf
When the substituent groups are oriented in the same direction, the diastereomer is
referred to as cis, whereas, when the substituents are oriented in opposing directions, the
diastereomer is referred to as trans. An example of a small hydrocarbon displaying cis/trans
isomerism is 2-butene. Alicyclic compounds can also display cis/trans isomerism. As an example
of a geometric isomer due to a ring structure, consider 1,2-dichlorocyclohexane: is/trans
isomerism (also known as geometric isomerism, configuration isomerism, or E/Z isomerism) is a
form of stereoisomerism describing the orientation of functional groups within a molecule. In
general, such isomers contain double bonds, which cannot rotate, but they can also arise from
ring structures, wherein the rotation of bonds is greatly restricted. Cis and trans isomers occur
both in organic molecules and in inorganic coordination complexes. The terms cis and trans are
from Latin, in which cis means \"on the same side\" and trans means \"on the other side\" or
\"across\". The term \"geometric isomerism\" is considered an obsolete synonym of \"cis/trans
isomerism\" by IUPAC.[1] It is sometimes used as a synonym for general stereoisomerism (e.g.,
optical isomerism being called geometric isomerism); the correct term for non-optical
stereoisomerism is diastereomerism.
Solution
When the substituent groups are oriented in the same direction, the diastereomer is
referred to as cis, whereas, when the substituents are oriented in opposing directions, the
diastereomer is referred to as trans. An example of a small hydrocarbon displaying cis/trans
isomerism is 2-butene. Alicyclic compounds can also display cis/trans isomerism. As an example
of a geometric isomer due to a ring structure, consider 1,2-dichlorocyclohexane: is/trans
isomerism (also known as geometric isomerism, configuration isomerism, or E/Z isomerism) is a
form of stereoisomerism describing the orientation of functional groups within a molecule. In
general, such isomers contain double bonds, which cannot rotate, but they can also arise from
ring structures, wherein the rotation of bonds is greatly restricted. Cis and trans isomers occur
both in organic molecules and in inorganic coordination complexes. The terms cis and trans are
from Latin, in which cis means \"on the same side\" and trans means \"on the other side\" or
\"across\". The term \"geometric isomerism\" is considered an obsolete synonym of \"cis/trans
isomerism\" by IUPAC.[1] It is sometimes used as a synonym for general stereoisomerism (e.g.,
optical isomerism being called geometric isomerism); the correct term for non-optical
stereoisomerism is diastereomerism..
Tunnel through the IPv4--Internet traffic is expected to be carr.pdf
Tunnel through the IPv4:-
->Internet traffic is expected to be carried via tunnels as the Internet infrastructure migrates from
IPv4, the current version of the Internet protocol, to the long-anticipated upgrade known as IPv6.
->IPv4 uses 32-bit addresses and can support 4.3 billion devices connected directly to the
Internet.
->Internet:-Will probably move to IPv6 “from the edges in”
-> IPv6 will be adopted:-First by smaller organizations with greater flexibility and higher
tolerance for difficulties of pioneering
-> IPv6 packet is formed normally:- Sent to a router capable of encapsulating it in an IPv4
packet.
->Used to allow IPv6 network nodes to send packets over an IPv4 network infrastructure
->Presents a challenge for IPv6 header construction
-> Source node determines which packets must be encapsulated
->Based on the routing information the node maintains in its own routing table.
Types of Tunnels:-
1)RFC 2893 originally specified two different tunneling types
–Configured and automatic
2)RFC 4213, which made RFC 2893 obsolete
– Removed references to automatic tunneling
3)Configured tunnels
–Require that end point addresses be determined in the encapsulator device
4) From configuration data stored for each tunnel.
Solution
Tunnel through the IPv4:-
->Internet traffic is expected to be carried via tunnels as the Internet infrastructure migrates from
IPv4, the current version of the Internet protocol, to the long-anticipated upgrade known as IPv6.
->IPv4 uses 32-bit addresses and can support 4.3 billion devices connected directly to the
Internet.
->Internet:-Will probably move to IPv6 “from the edges in”
-> IPv6 will be adopted:-First by smaller organizations with greater flexibility and higher
tolerance for difficulties of pioneering
-> IPv6 packet is formed normally:- Sent to a router capable of encapsulating it in an IPv4
packet.
->Used to allow IPv6 network nodes to send packets over an IPv4 network infrastructure
->Presents a challenge for IPv6 header construction
-> Source node determines which packets must be encapsulated
->Based on the routing information the node maintains in its own routing table.
Types of Tunnels:-
1)RFC 2893 originally specified two different tunneling types
–Configured and automatic
2)RFC 4213, which made RFC 2893 obsolete
– Removed references to automatic tunneling
3)Configured tunnels
–Require that end point addresses be determined in the encapsulator device
4) From configuration data stored for each tunnel..
The key activity areas for securities firms are i. Investing Se.pdf
The key activity areas for securities firms are :
i. Investing: Security firms act as agents for investors and provide risk diversification and
liquidity services to holders of mutual funds. The major risks are underperforming stock market
whereby investors flee to other avenues of investing.
ii. Investment Banking: These functions relate to underwriting both debt and equity. The major
risk is unsold securities and money on the table due to ineffective pricing.
iii. Market Making: Market making involves the creation of a secondary market in an asset by a
securities firm or investment bank facilitating the smooth transactions of trading in the market. In
case of a crash this can lead to unsold securities and selling at a significant discount thus losing a
lot.
iv. Trading: Securities trader takes an active net position in an underlying instrument or asset.
The risk are snall bid ask spread and high fluctuating prices leading to loss on trading positions
v. Cash Management: Security firms bank deposit-like cash management accounts to individual
investors. Again the proper pricing is needed else there may be a significant loss on the position.
vi. Mergers and Acquisitions: Investment banks provide advice on mergers and acquisitions. The
risk are improper pricing leading to loss or not realizing proper synergies
vi. Other Service Functions: include custody and escrow services, clearance and settlement
services and research and advisory services. They may result in losses if there is a credit risk
present which usually is common.
Solution
The key activity areas for securities firms are :
i. Investing: Security firms act as agents for investors and provide risk diversification and
liquidity services to holders of mutual funds. The major risks are underperforming stock market
whereby investors flee to other avenues of investing.
ii. Investment Banking: These functions relate to underwriting both debt and equity. The major
risk is unsold securities and money on the table due to ineffective pricing.
iii. Market Making: Market making involves the creation of a secondary market in an asset by a
securities firm or investment bank facilitating the smooth transactions of trading in the market. In
case of a crash this can lead to unsold securities and selling at a significant discount thus losing a
lot.
iv. Trading: Securities trader takes an active net position in an underlying instrument or asset.
The risk are snall bid ask spread and high fluctuating prices leading to loss on trading positions
v. Cash Management: Security firms bank deposit-like cash management accounts to individual
investors. Again the proper pricing is needed else there may be a significant loss on the position.
vi. Mergers and Acquisitions: Investment banks provide advice on mergers and acquisitions. The
risk are improper pricing leading to loss or not realizing proper synergies
vi. Other Service Functions: include custody and escrow services, clearance and settlement
services and re.
The finger like projections of intestine known as villi increases su.pdf
The finger like projections of intestine known as villi increases surface area of small intestines.
The surface of small intestine is lined up with epithelial cells containing brush borders or
microvilli that help to absorb nutrients into the blood or lymph, which later moves to the body
cells.
Due to the presence of specialized proteins within the microvilli known as porters, the nutrients
move into the cells.
Human gut flora contains several types of bacterial species in the small intestines that are
involved in the production of vitamin K, B as well as metabolism of bile acid and xenobiotics.
So, correct option is d.transferring nutrients across intestinal membranes
Solution
The finger like projections of intestine known as villi increases surface area of small intestines.
The surface of small intestine is lined up with epithelial cells containing brush borders or
microvilli that help to absorb nutrients into the blood or lymph, which later moves to the body
cells.
Due to the presence of specialized proteins within the microvilli known as porters, the nutrients
move into the cells.
Human gut flora contains several types of bacterial species in the small intestines that are
involved in the production of vitamin K, B as well as metabolism of bile acid and xenobiotics.
So, correct option is d.transferring nutrients across intestinal membranes.
The answer isC. 12In establishing paternity by dna fingerprintin.pdf
The answer is:C. 1/2
In establishing paternity by dna fingerprinting, 50% of microsatellite repeat alleles in a child
come from the father.
Solution
The answer is:C. 1/2
In establishing paternity by dna fingerprinting, 50% of microsatellite repeat alleles in a child
come from the father..
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Cu(H2O)42+(aq) + 4CN-(aq) = Cu(CN)42-(aq) +4 H2O(l)Cu+2 is a Lew.pdf
Cu(H2O)42+(aq) + 4CN-(aq) = Cu(CN)42-(aq) +4 H2O(l)Cu+2 is a Lew.pdf
C) CS2 for sureA, B, D and E are all ionic compoundsSolution.pdf
C) CS2 for sureA, B, D and E are all ionic compoundsSolution.pdf
Before getting public issues he should consider are 1. Increase i.pdf
Before getting public issues he should consider are 1. Increase i.pdf
Assets=Liabilities+Stockholders EquityCash given $1000 on the sa.pdf
Assets=Liabilities+Stockholders EquityCash given $1000 on the sa.pdf
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1)Interest per year 8106= 135Simple interest = Amount RateTime.pdf
AnswerAnswer1. The body of a fungus is in the form of solid sh.pdf
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A. This constitutes part of a gene that encodes information for RNA .pdf
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A. ANSAutosomal dominant (a trait could passed down from previous.pdf
A. ANSAutosomal dominant (a trait could passed down from previous.pdf
888678123317368915236 #include stdio.h #include stdlib.h.pdf
888678123317368915236 #include stdio.h #include stdlib.h.pdf
18) One device is attached to a physical interface The switch c.pdf
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1.Joseph could not restrain himself before all those who stood by hi.pdf
1.Joseph could not restrain himself before all those who stood by hi.pdf
(a) Total amount of I2 = 501000 x 0.010 = 0.0005 mol = 0.5 mmolLe.pdf
(a) Total amount of I2 = 501000 x 0.010 = 0.0005 mol = 0.5 mmolLe.pdf
Nursing is a healthcare profession in which they serve the patients .pdf
Nursing is a healthcare profession in which they serve the patients .pdf
yes.because the deposted money can be multiply with 10 and it beco.pdf
yes.because the deposted money can be multiply with 10 and it beco.pdf
When the substituent groups are oriented in the s.pdf
When the substituent groups are oriented in the s.pdf
Tunnel through the IPv4--Internet traffic is expected to be carr.pdf
Tunnel through the IPv4--Internet traffic is expected to be carr.pdf
The key activity areas for securities firms are i. Investing Se.pdf
The key activity areas for securities firms are i. Investing Se.pdf
The finger like projections of intestine known as villi increases su.pdf
The finger like projections of intestine known as villi increases su.pdf
The answer isC. 12In establishing paternity by dna fingerprintin.pdf
The answer isC. 12In establishing paternity by dna fingerprintin.pdf
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● Why and by when do customers need to upgrade to Java 17?
● Is there any immediate impact to upgrading to Mule Runtime 4.6 and beyond?
● Which MuleSoft products are in scope?
For Upcoming Meetups Join Mysore Meetup Group - https://meetups.mulesoft.com/mysore/
YouTube:- youtube.com/@mulesoftmysore
Mysore WhatsApp group:- https://chat.whatsapp.com/EhqtHtCC75vCAX7gaO842N
Speaker:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Organizers:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Giridhar Meka - https://www.linkedin.com/in/giridharmeka
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Shyam Raj Prasad-
https://www.linkedin.com/in/shyam-raj-prasad/
Honest Reviews of Tim Han LMA Course Program.pptx
Personal development courses are widely available today, with each one promising life-changing outcomes. Tim Han’s Life Mastery Achievers (LMA) Course has drawn a lot of interest. In addition to offering my frank assessment of Success Insider’s LMA Course, this piece examines the course’s effects via a variety of Tim Han LMA course reviews and Success Insider comments.
678020731-Sumas-y-Restas-Para-Colorear.pdf
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How libraries can support authors with open access requirements for UKRI fund...
How libraries can support authors with open access requirements for UKRI funded books
Wednesday 22 May 2024, 14:00-15:00.
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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.
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Polish students mobility to the Czech Republic within eTwinning project "Medieval adventures with Marco Polo"
Francesca Gottschalk - How can education support child empowerment.pptx
Francesca Gottschalk from the OECD’s Centre for Educational Research and Innovation presents at the Ask an Expert Webinar: How can education support child empowerment?
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using System;public category take a look at one.00m; Console.W.pdf
- 1. using System; public category take a look at one.00m; Console.WriteLine (d); } } When I initial ran the on top of (or one thing similar) I expected it to output simply one (which is what it might are on .NET 1.0) - however in reality, the output was one.00. The decimal sort does not normalize itself - it remembers what percentage decimal digits it's (by maintaining the exponent wherever possible) and on info, zero could also be counted as a major digit. i do not understand the precise nature of what exponent is chosen (where there's a choice) once 2 totally different decimals ar increased, divided, added etc, however you will realize it attention-grabbing to manipulate with programs like the following: using System; public category take a look at zero.00000000000010000m; whereas (d != 0m) } } Solution using System; public category take a look at one.00m; Console.WriteLine (d); } } When I initial ran the on top of (or one thing similar) I expected it to output simply one (which is what it might are on .NET 1.0) - however in reality, the output was one.00. The decimal sort does not normalize itself - it remembers what percentage decimal digits it's (by maintaining the exponent wherever possible) and on info, zero could also be counted as a major digit. i do not understand the precise nature of what exponent is chosen (where there's a choice) once 2 totally different decimals ar increased, divided, added etc, however you will realize it attention-grabbing
- 2. to manipulate with programs like the following: using System; public category take a look at zero.00000000000010000m; whereas (d != 0m) } }