The following diagram shows nondisjunction In either Melosis I or Mel.pdf
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The following diagram shows nondisjunction In either Melosis I or Melosis II. The cell has only one pair homologous. chromosomes. If fertilized with a normal healthy gamete, which cell(s) will most likely give normal.offspring?. Select one: only c A, C and D A and D A, B and D Solution The gametes produced in \"C\", will only give rise to normal offspring. because it is a haploid gamete..
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How are the epimysium and tendons relatedA. The epimysium surroun.pdf
How are the epimysium and tendons related?
A. The epimysium surrounds the tendon.
B. Tendons surround the epimysium.
C. Tendons and epimysium both surround the end of the muscle.
D. The tendon is a continuation of the epimysium.
Solution
Answer:
D. The tendon is a continuation of the epimysium.
Explanation:
At the ends of a muscle, the fibers of the Epimysium may form a cable-like tendon or a sheet-
like aponeurosis..
How does DNA differ from RNAA. RNA uses only purinesB. RNA uses.pdf
How does DNA differ from RNA?
A. RNA uses only purines
B. RNA uses a different five-carbon sugar
C. RNA has a contains different sized phosphate groups
D. RNA has multiple bases attached to the sugar
Solution
B. RNA uses a different five-carbon sugar. RNA has the ribose pentose; whereas in DNA, the
pentose is deoxyribose..
Gerard throws a discus distances of 10 meters, 14.5 meters, 14.8 met.pdf
Gerard throws a discus distances of 10 meters, 14.5 meters, 14.8 meters, 16.7 meters, and 15.4
meters. On his last attempt he faults, which counts as 0 meters. What is his average throwing
distance?
A. 11.9 meters
B. 12.8 meters
C. 13.78 meters
D. 14.28 meters
Solution
A. To find the average, add the numbers and divide by the number of numbers. In this problem
add 10, 14.5, 14.8, 16.7, 15.4, and 0 to get 71.4. Divide this by 6.
71.4/6 = 11.9.
For Printing and personal use ONLY. DO NOT Hand in this copy of the a.pdf
For Printing and personal use ONLY. DO NOT Hand in this copy of the assignment. Only
responses submitted to the corresponding D2L quiz will be graded. The D2L version is exactly
like this document. Membrane proteins are proteins that are suspended within the lipid bilayer of
a membrane. They are held in place because some R-groups of the protein interact with the
phosphate heads of the phospholipids of the membrane, while other R-groups of the protein
interact with the hydrocarbon tails of the phospholipids of the membrane. From the amino acids
shown below, which one likely has an R-group that interacts with the phopholipid tails? If a cell
has some salt inside it but lacks ion channels, and the cell is put into a beaker of pure water, what
will happen?
Solution
Leucine (green color) is most likely to be part of the protein that interact with the hydrocarbon
chain of phospholipid in the membrane (since it is hydrophobic amino acid).
Find and provide a link to a multi-year capital plan from any Illino.pdf
Find and provide a link to a multi-year capital plan from any Illinois municipality. Discuss the
presentation of the plan (i.e. - is there detail about specific projects, funding sources, etc) and
how you think it could be improved.
Solution
The Link can\'t be provided so you can find it in the document centre of the village of the round
lake beach and the multi year capital plan is for the years from 2015- 2024.
The plan is presented in detail proving the names of the specific projects that would be
undertaken by the village and all the ways in which they will be collecting the funds needed. The
plan talks about the following specific projects:
The funding is alot dependent upon the taxes from the people inhabiting the place and a very
small amount has been estimated as allocations from outside. The plan is highly unrealistic as
there may be instances where the people residing the place may not choose to pay so much or
simply not buy the bonds they are expecting to issue..
A geneticist is working with a new bacteriophage called phage Y3 that.pdf
A geneticist is working with a new bacteriophage called phage Y3 that infects E. coli. He has
isolated eight mutant phages that fail to produce plaques when grown on E. coli strain K. To
determine whether these mutations occur at the same functional gene, he simultaneously infects
E. coli K with paired combination of the mutants and looks to see whether plaques are formed.
He obtains the following results. (A plus sign means that plaques were formed on E. coll K; a
minus sign means that no plaques were formed on E. coli K.) To how many functional genes
(cistrons) do these mutations belong? 5 1 8 2 7 3
Solution
Answer: Correct answer: 3
Reason: The geneticist has conducted the transformation experiment in which “trans” portion of
the cis–trans test to determine the complementation between the different phage mutants infected
with E. coli K cells. Finally, these cells were looked whether the plaques are generated in the
plates. It can be seen in the above results table, in which plus sign means that plaques were
formed due to complimentaion in 3 mutants (5, 6 & 8) have meticulously specific defects in the
same functional so that on E. coli K; a minus sign means that no plaques. Therefore, it can be
concluded that complementation can happen when the mutations of the different phages come
together in the sample as part of transformation.
Assume that the four living species in the figure below evolved from.pdf
Assume that the four living species in the figure below evolved from their common ancestor
without convergence or reversal.
1. Reconstruct their evolutionary relationships and label the character transitions. Show diagram.
2. Which evolved first: red eyes (living species 3) or fin stripes? Explain your answer.
***I have the answer but I don\'t know how we got it. I\'ve seen some incorrect/incomplete
answers here and that wasn\'t helpful. Please explain you answer for BOTH parts and show
diagram. If you\'re not sure don\'t answer! Thanks!*** Common ancestor Living species
Solution
a)the transition is as follows
common ancestor->Panderichthys,->Tiktaalik with the fins->tetrapods with legs and wings-
>amphibian
b)The eyes came first because the fish first has a body with eyes and then it has fins to swim.The
fins stripes increase with type into long wings and it can swim more and more.The eye is to see
the direction to swim.
Declining BalanceThe cost of the asset $10,000.00 The salvage v.pdf
Declining Balance
The cost of the asset: $10,000.00
The salvage value: $800.00
The life of the asset: 4 years
Calculate the Depreciation Expense to the Nearest One Dollar at End of Year 2.
Solution
Since No depreciation rate is given it is assumed theat depreciation is calclucated based on
double decling balance method
Hence depreciation Rate =1/4*2 =50%
Depreciation Exxpense at year 2 is $3750Carrying Value
at the beginningDepreciationCarrying value at the EndYear 11000025007500Year
2750037503750.
Recommended
How are the epimysium and tendons relatedA. The epimysium surroun.pdf
How are the epimysium and tendons related?
A. The epimysium surrounds the tendon.
B. Tendons surround the epimysium.
C. Tendons and epimysium both surround the end of the muscle.
D. The tendon is a continuation of the epimysium.
Solution
Answer:
D. The tendon is a continuation of the epimysium.
Explanation:
At the ends of a muscle, the fibers of the Epimysium may form a cable-like tendon or a sheet-
like aponeurosis..
How does DNA differ from RNAA. RNA uses only purinesB. RNA uses.pdf
How does DNA differ from RNA?
A. RNA uses only purines
B. RNA uses a different five-carbon sugar
C. RNA has a contains different sized phosphate groups
D. RNA has multiple bases attached to the sugar
Solution
B. RNA uses a different five-carbon sugar. RNA has the ribose pentose; whereas in DNA, the
pentose is deoxyribose..
Gerard throws a discus distances of 10 meters, 14.5 meters, 14.8 met.pdf
Gerard throws a discus distances of 10 meters, 14.5 meters, 14.8 meters, 16.7 meters, and 15.4
meters. On his last attempt he faults, which counts as 0 meters. What is his average throwing
distance?
A. 11.9 meters
B. 12.8 meters
C. 13.78 meters
D. 14.28 meters
Solution
A. To find the average, add the numbers and divide by the number of numbers. In this problem
add 10, 14.5, 14.8, 16.7, 15.4, and 0 to get 71.4. Divide this by 6.
71.4/6 = 11.9.
For Printing and personal use ONLY. DO NOT Hand in this copy of the a.pdf
For Printing and personal use ONLY. DO NOT Hand in this copy of the assignment. Only
responses submitted to the corresponding D2L quiz will be graded. The D2L version is exactly
like this document. Membrane proteins are proteins that are suspended within the lipid bilayer of
a membrane. They are held in place because some R-groups of the protein interact with the
phosphate heads of the phospholipids of the membrane, while other R-groups of the protein
interact with the hydrocarbon tails of the phospholipids of the membrane. From the amino acids
shown below, which one likely has an R-group that interacts with the phopholipid tails? If a cell
has some salt inside it but lacks ion channels, and the cell is put into a beaker of pure water, what
will happen?
Solution
Leucine (green color) is most likely to be part of the protein that interact with the hydrocarbon
chain of phospholipid in the membrane (since it is hydrophobic amino acid).
Find and provide a link to a multi-year capital plan from any Illino.pdf
Find and provide a link to a multi-year capital plan from any Illinois municipality. Discuss the
presentation of the plan (i.e. - is there detail about specific projects, funding sources, etc) and
how you think it could be improved.
Solution
The Link can\'t be provided so you can find it in the document centre of the village of the round
lake beach and the multi year capital plan is for the years from 2015- 2024.
The plan is presented in detail proving the names of the specific projects that would be
undertaken by the village and all the ways in which they will be collecting the funds needed. The
plan talks about the following specific projects:
The funding is alot dependent upon the taxes from the people inhabiting the place and a very
small amount has been estimated as allocations from outside. The plan is highly unrealistic as
there may be instances where the people residing the place may not choose to pay so much or
simply not buy the bonds they are expecting to issue..
A geneticist is working with a new bacteriophage called phage Y3 that.pdf
A geneticist is working with a new bacteriophage called phage Y3 that infects E. coli. He has
isolated eight mutant phages that fail to produce plaques when grown on E. coli strain K. To
determine whether these mutations occur at the same functional gene, he simultaneously infects
E. coli K with paired combination of the mutants and looks to see whether plaques are formed.
He obtains the following results. (A plus sign means that plaques were formed on E. coll K; a
minus sign means that no plaques were formed on E. coli K.) To how many functional genes
(cistrons) do these mutations belong? 5 1 8 2 7 3
Solution
Answer: Correct answer: 3
Reason: The geneticist has conducted the transformation experiment in which “trans” portion of
the cis–trans test to determine the complementation between the different phage mutants infected
with E. coli K cells. Finally, these cells were looked whether the plaques are generated in the
plates. It can be seen in the above results table, in which plus sign means that plaques were
formed due to complimentaion in 3 mutants (5, 6 & 8) have meticulously specific defects in the
same functional so that on E. coli K; a minus sign means that no plaques. Therefore, it can be
concluded that complementation can happen when the mutations of the different phages come
together in the sample as part of transformation.
Assume that the four living species in the figure below evolved from.pdf
Assume that the four living species in the figure below evolved from their common ancestor
without convergence or reversal.
1. Reconstruct their evolutionary relationships and label the character transitions. Show diagram.
2. Which evolved first: red eyes (living species 3) or fin stripes? Explain your answer.
***I have the answer but I don\'t know how we got it. I\'ve seen some incorrect/incomplete
answers here and that wasn\'t helpful. Please explain you answer for BOTH parts and show
diagram. If you\'re not sure don\'t answer! Thanks!*** Common ancestor Living species
Solution
a)the transition is as follows
common ancestor->Panderichthys,->Tiktaalik with the fins->tetrapods with legs and wings-
>amphibian
b)The eyes came first because the fish first has a body with eyes and then it has fins to swim.The
fins stripes increase with type into long wings and it can swim more and more.The eye is to see
the direction to swim.
Declining BalanceThe cost of the asset $10,000.00 The salvage v.pdf
Declining Balance
The cost of the asset: $10,000.00
The salvage value: $800.00
The life of the asset: 4 years
Calculate the Depreciation Expense to the Nearest One Dollar at End of Year 2.
Solution
Since No depreciation rate is given it is assumed theat depreciation is calclucated based on
double decling balance method
Hence depreciation Rate =1/4*2 =50%
Depreciation Exxpense at year 2 is $3750Carrying Value
at the beginningDepreciationCarrying value at the EndYear 11000025007500Year
2750037503750.
Create a java project that - Draw a circle with three random init.pdf
Create a java project that:
- Draw a circle with three random initial points on the circle.
- Link the points to form a triangle.
- Print the angles values in the triangle.
- Use the mouse to drag a point along the perimeter of the circle. As you drag it, the triangle and
angles are redisplayed dynamically.
Here is the formula to compute angles:
A = acos((a * a - b * b - c * c) / (-2 * b * c))
B = acos((b * b - a * a - c * c) / (-2 * a * c))
C = acos((c * c - b * b - a * a) / (-2 * a * b))
Solution
@Suppresswarnings(\"WeakerAccess\")
public class DragPoints extends Application {
@Override
public void start(Stage primaryStage) {
final PointPane pane = new PointPane(640, 480);
pane.setStyle(\"-fx-background-color: wheat;\");
Label label = new Label(\"Click and drag the points.\");
BorderPane borderPane = new BorderPane(pane);
BorderPane.setAlignment(label, Pos.CENTER);
label.setPadding(new Insets(5));
borderPane.setBottom(label);
Scene scene = new Scene(borderPane);
primaryStage.setTitle(\"Exercise15_21\");
primaryStage.setScene(scene);
primaryStage.show();
}
private class PointPane extends Pane {
final Circle circle = new Circle();
final Vertex[] v = new Vertex[3];
final int strokeWidth = 2;
final Color circleStroke = Color.GRAY, legStroke = Color.BLACK;
@SuppressWarnings(\"SameParameterValue\")
PointPane(double w, double h) {
this.setPrefSize(w, h);
this.setWidth(w);
this.setHeight(h);
circle.setStroke(circleStroke);
circle.setFill(Color.TRANSPARENT);
circle.setStrokeWidth(strokeWidth);
circle.radiusProperty().bind(this.heightProperty().multiply(0.4));
circle.centerXProperty().bind(this.widthProperty().divide(2));
circle.centerYProperty().bind(this.heightProperty().divide(2));
this.getChildren().add(circle);
for (int i = 0; i < v.length; i++) {
v[i] = new Vertex(circle, 2 * Math.PI / v.length * (i + Math.random()));
v[i].radiusProperty().bind(circle.radiusProperty().divide(10));
v[i].setPosition();
v[i].setStroke(new Color(i == 0 ? 1 : 0, i == 1 ? 1 : 0, i == 2 ? 1 : 0, 1));
v[i].setFill(Color.TRANSPARENT);
v[i].setStrokeWidth(strokeWidth);
this.getChildren().add(v[i]);
v[i].setOnMouseDragged(new EventHandler() {
@Override
public void handle(MouseEvent event) {
int i;
for (i = 0; i < v.length; i++)
if (v[i] == event.getSource())
break;
v[i].setAngle(event.getX(), event.getY());
moveUpdate((Vertex) event.getSource());
}
});
}
for (int i = 0; i < v.length; i++) {
int j = i + 1 < v.length ? i + 1 : 0;
int k = j + 1 < v.length ? j + 1 : 0;
v[i].bindLeg(v[j], v[k]);
v[i].leg.setStroke(legStroke);
v[i].leg.setStrokeWidth(strokeWidth);
this.getChildren().add(v[i].leg);
this.getChildren().add(v[i].text);
}
for(DoubleProperty p: new DoubleProperty[]
{circle.radiusProperty(), circle.centerXProperty(), circle.centerYProperty()})
p.addListener(new ResizeListener());
moveUpdate(v[0]);
}
void moveUpdate(Vertex vert) {
vert.setPosition();
double[] legLength = new double[3];
for (int i = 0; i < v.length; i++)
legLength[i] = v[i].getLegLength();
for (int i = .
aphase The spindle contracts and the sister chromatids are separated.pdf
aphase: The spindle contracts and the sister chromatids are separated at the centromere Once the
sister for each replicated chromosome have separated, each is now called in the cell an
unreplicated chromosome. It is at this point t chromosome num Telophase: Telophase reverse.
List the events below that happen in is essentially prophase in telophase SIMULATE CELL
DIVISION MITOSIS Using pop beads and magnetic centromeres, create 4 chromosomes as
follows: Red 6 pop beads on each side of the centromere Yellow: Same as Red. Blue 5 pop beads
on one sided of the centromere, 3 pop beads on the other side Green: Same as Blue. Work
through the phases of cell division, beginning at G1, with 4 unreplicated chromosomes
Remember that we can not see chromosomes during interphase, but this can not be modeled in
this lab. It is important to note what happens during the S phase when you do the simulation
Chromosome count Use the adjective \"unreplicated\" or \"replicated\" with your count G1: rom
DS ones S: osomes As best as you are able, and without looking at the descriptions above or
your text, describe what is happening in each phase of mitosis as you simulate the process. Avoid
using pronouns (it and they). Speak the language of genetics as you simulate mitosis. Help your
lab partner if she and provide feedback with the language. Chromosome counts and description
(replicated or unreplicated) in each phase of mitosis: Prophase: mm Somes Metaphase 1,-L
Chromosome Anaphase Telophase: What event occurs after Telophase to complete the process
of mitosis? What C4 Kenesis mitosis has been completed? is the chromosome count in each cell
after Summarizing Mitosis 1 parent cell produces 2 essentially identical daughter cells. Mitosis
separates sister chromatids.
Solution
G1: 4 unreplicated chromosomes (singlet chromosomes)
S: 4 replicated chromosomes (doublet chromosomes)
G2: 4 replicated chromosomes (doublet chromosomes)
Prophase: 4 replicated chromosomes
Metaphase: 4 replicated chromosomes arranged on metaphasic plate.
Anaphase: 8 unreplicated (singlet) chromosomes (4 singlet chromosomes move toward one end,
while the other 4 singlet (unreplicated) chromosomes move to the opposite end.)
Telophase: 8 unreplicated chromosomes (4 unreplicated chromosomes in each of the two
dividing daughter cells)
cytokinesis occurs after telophase that completes mitosis
after mitosis, 2 daughter cells are formed. each daughter cell contains 4 unreplicated
chromosomes. (similar to parent cell at G1 phase).
A population of wild deer mice includes individuals with long or sho.pdf
A population of wild deer mice includes individuals with long or short hair. When two short
haird indviduals breed, one-third of the pups have long hair. How do you explain the unusual
phenotype ratio?
Solution
it can be autosomal inheritance where homozygous dominant is lethal.
consider short hair to be dominant S, long hair to be recessive s
so cross between two short haired can be
Ss X Ss
S s
S SS Ss
s Ss ss
1:2:1 ratio for SS:Ss:ss
but we considered homozygous dominant to be lethal, so SS are dead, therefore Ss:ss is 2:1 and
1/3rd population have long hair..
Who is it important that the lymphatic system operate separately f.pdf
Who is it important that the lymphatic system operate separately from the cardiovascular
system? Explain
Describe the relationship between the lymphatic vessels and blood vessels.
Who is it important that the lymphatic system operate separately from the cardiovascular
system? Explain
Describe the relationship between the lymphatic vessels and blood vessels.
Describe the relationship between the lymphatic vessels and blood vessels.
Solution
It is important that the lymphatic system operate separately from the cardiovascular system
becuase lymphatic system is a closed circulatory system that consists of lymph capillaries, lymph
vessels, and lymph nodes different to blood vessels. The lymph present in capillaries flows into
lymphatic vessels, which are presented deep or superficial but blood vessels are deep innervating
various parts of body to maintain pheripheral resistance & mean arterial blood pressure. Blood
flows through three types of blood vessels (arteries, capillaries and veins) distributed throughout
the body. Capillaries are the fine endings of blood vessels. These capillaries are responsible for
the exchange of respiratory gases, nutrients and waste products. The molecules move across the
capillaries through a process called “diffusion.” These lymph vessels combine to form lymphatic
trunks that empty into the large collecting vessels of lymphatic system called “thoracic duct and
right lymphatic duct.”
The lymphoid tissue is mostly composed of a reticular connective tissue (a type of loose
connective tissue). The fibres of the reticular network contain macrophages. A number of
lymphocytes from the postcapillary venule walls occupy this network and leave to the body
again. This ensures the lymphocytes to identify the infected sites at a faster rate.
Lymphatic system is a closed circulatory system that consists of lymph capillaries, lymph
vessels, and lymph nodes. Lymph is a tissue fluid, which is formed by the diffusion of
substances from blood capillaries into tissue spaces. The lymph present in capillaries flows into
lymphatic vessels, which are presented deep or superficial. These lymph vessels combine to form
lymphatic trunks that empty into the large collecting vessels of lymphatic system called “thoracic
duct and right lymphatic duct.”
It is important that the lymphatic system operate separately from the cardiovascular system due
to the follwoing reasons and exerts differentail effects of innate immune system. The lymph is
generated to produce defensive mechanisms via innate immunity include physical barriers,
phagocytes, immune surveillance,interferons, complement system, inflammation and fever.
1. Fluid recovery: Filtration of the fluid into the tissue spaces from the blood capillaries will
occur continuously. 85% of it will be absorbed by the blood capillaries. About 15% will be
remained unabsorbed. This unabsorbed portion of the fluid is about 2-4 litres of water and 1/4th
to 1/2th of plasma protein. If this water protein an.
Provide a full explanation to the below question.1. Summarize the .pdf
Provide a full explanation to the below question.
1. Summarize the 802.11 standard and describe the various flavors of the standard, its
architecture and how the standard is contributing to easing of congestion in cellular networks.
Solution
IEEE 802.11 is a standard that defines the physical and MAC layers of Wireless Local Area
Network (WLAN). Under this standard, the Mobile terminals (MTs) can communicate with a
Access Point (AP) in two modes. First mode is the Infrastructure Mode, in which the MTs
communicate with the APs which forward their data to the WLAN. Second mode is the Adhoc
mode, in which the MTs communicate directly with each other without the use of a AP. The
IEEE 802.11 has a very robust Medium Access Control (MAC) mechanism that helps in
alleviating te congestion in the LANs. The inherent working of MAC protocol employed by
802.11 standard is same as that of Carrier Sense Multiple Access/ Collision Aviodance
(CSMA/CA). However, in IEEE 802.11 the protocol is implemented in two different ways. First
way is called as the Distributed Coordination Function based Wireless MAC (DCFWMAC), and
the second way is Point Coordination Function based Wireless MAC (PCFWMAC).
In the DCFWMAC, every node tries to access the medium based on some fixed duration of time
defined in the standard. Normally there are three such time intervals defined based on which a
node attempts to access the channel and transmit its packets. The first interval is the Short Inter
frame Spacing (SIFS). It is the smallest duration of time between two frames and gives the node
higher priority for sending its data. Such a interval is employed only when the node has correctly
sent its data through the channel and the receiver node needs to send a Acknowlegement (ACK)
after waiting for SIFS period of time. The second most priority interval is PCF Inter Frame
Spacing (PIFS) used during Polling mechanism by the AP. Its duration is between SIFS and
DCF Inter frame Spacing (DIFS). The DIFS is the longest time duration and hence of least
priority, in which a AP has to wait between two successive channel accesses for the given
duration. Once the channel is sensed idle, the AP waits for DIFS period of time to transmit data
through the channel. If the channel is sensed busy, then it backs off for a period of time based on
the minimum and maximum value of a contention window. During the back off period, if the
node senses the channel as busy, then it freezes the backoff counter and starts the backoff(with
the remaining time left) once the channel is idle. In this manner, the nodes which have waited
longer get higher priority over others which accessing the channel. Once the node sends the data
after accessing the channel, the receiver waits for SIFS duration of time to finally send the
acknowledgement to the sender. In order to tackle the hidden terminal problem, the DCFWMAC
employs an additional mechanism called as the RTS-CTS mechanism, in which any node that
wants to transmit.
Write a program to generate the entire calendar for one year. The pr.pdf
Write a program to generate the entire calendar for one year. The program must get two values
from the user: (1) the year and (2) the day of the week for January 1st of that year. The year,
which should be positive, is needed to check for and handle leap years1. The day of the week for
January 1st is needed so that you know where to start the calendar. The user should enter 0 for
Sunday, 1 for Monday, ... or 6 for Saturday. As always, you need to validate the user\'s input. To
actually print the calendar, you must use a single method that prints out the calendar for one
month and then call this function 12 times from main(), once for each month in the year. To
check for a leap year you will need to write another method that takes the year as a parameter
and returns true if it’s a leap year, or false otherwise. Stubs (i.e. method signatures without any
code) for both of these methods have been provided for you.
The calendar should be printed in the form below. In this example, January starts on a Saturday
(day 6). Note that February starts on a Tuesday in this example because January ended on a
Monday.
January
1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 2829
30 31
February
1 2 3 4 5
6 7 8 9 10 11 12 ...
The format he gave us is as follows:
public class PA6a {
/**
* Error to output if year is not positive
*/
static final String E_YEAR = \"The year must be positive!\";
/**
* Error to output if the day is not between 0 and 6
*/
static final String E_DAY = \"The day of January 1st must be between 0 and 6!\";
/**
* Determines if an input is a leap year
*
* @param year year in question
* @ereturn true if a leap year
*/
public static boolean isLeapYear(int year) {
return false; // TODO: replace with your code
}
/**
* Outputs a month to the console
*
* @param month title
* @param startDay 0=Sunday ... 6=Saturday
* @param numDays number of days in the month
* @return day of the week of the last day of the month
*/
public static int printMonth(String month, int startDay, int numDays) {
return 0; // TODO: replace with your code
}
/**
* Program execution point:
* input year, day of the week (0-6) of january 1
* output calendar for that year
*
* @param args command-line arguments (ignored)
*/
public static void main(String[] args) {
// TODO: write your code here
}
I am confused especially with the print month method, help is greatly appreciated.
Solution
package org.students;
import java.util.Scanner;
import com.sun.org.apache.bcel.internal.generic.IUSHR;
public class Calander {
//declaring static variables
static final String E_YEAR = \"The year must be positive!\";
static final String E_DAY = \"The day of January 1st must be between 0 and 6!\";
public static void main(String[] args) {
//Declaring variables
int getFirstDay;
int year;
Scanner scanner = new Scanner(System.in);
//This while loop continue to execute until user enters valid year
while (true) {
//Getting the year entered by the user
System.out.print(\"Enter the year:.
Write the interval notation for the set of numbers graphed. Solu.pdf
Write the interval notation for the set of numbers graphed.
Solution
The open circle represents that the value is not included in the interval while a solid (filled) circle
represents that the value is included in the interval.
So, we can see that there is open circle for (-2) and closed for (5).
Hence the required interval is (-2,5]..
Hyenas are diploid organisms, and their gametes contain 20 chromosome.pdf
Hyenas are diploid organisms, and their gametes contain 20 chromosomes. Fill in the blank in
each sentences using the terms below. Each term may be used more than once or not at all.
When the somatic cells of a hyena are in interphase, the ploidy level is denoted as The ploidy
level of the hyena\'s gametes is denoted as The 2n number of a hyena is 40 20 1n 80 2n
Solution
When the somatic cells of hyena in interphase = diploid [2n] because they are not divided and
tend to divide by preparing enzymes and multiplying DNA.
The ploidy level of hyena is haploid or 1n because they are developed through meiosis process.
The 2n number of chromosomes of hyena is 40 because gametes are haploid so n=20 [as given]
and the 2n = 2x 20 = 40.
Why doesnt hemoglobin have any intermediate conformations between .pdf
Why doesn\'t hemoglobin have any intermediate conformations between R-state and T-state?
Solution
Answer:
X-ray crystallogarphy has determined two major conformations of hemoglobin- the T state
which is stabilized in the absence of oxygen and the R state which is relatively more stable in the
presence of oxygen.
The pair of identical dimers of Hemoglobin are linked by several ion pairs at the interface that
stabilize it in its deoxy-state. Many of these interactions are disrupted upon binding to oxygen
and new ones are formed.
There are intermediate conformations between T and R states, but the intermediate
conformations have intermediate binding affinities. Also, these transition states are
thermodynamically unfavourable and short lived. These states have been utilized in studying the
allosteric pathways of hemoglobin. The binding of the ligand destabilizes the T state and makes
it more likely to switch to the high-affinity oxygenated (oxy)-R state..
4. What is gyanandromorphy How does this happenSolutionAnswe.pdf
4. What is gyanandromorphy? How does this happen?
Solution
Answer : Gynandromorphy is a phenamenan in which there is an occurence of both male and
female charectorstics in a single individual (part of the body expresses as male charectors, where
as the other part express female charectors. Individuals with this phenotype are called
gynandromorphs.
It results due to the loss of an X chromosome in a particular cell during development through
mitosis.For instance, a cell with XY through mitosis produces XXYY upon duplication which in
turn divides to form XY cells. But in gynandromorphs, this cell divides into an X cell and an
XYY cells. Due to this the organism develops with both male and female charectors..
Which of the following requires the mitochondria to create contact po.pdf
Which of the following requires the mitochondria to create contact points with the ER? mt
fission mt fusion mt biogenesis of more mitochondria autophagy of mitochondria (Mitophagy)
apoptosis
Solution
Answer= The mitochondria requires the following to create a contact with ER is
mt fission - mitochondrial fission
Explaination= The mitochondria & RE are tubular organelles with a characteristic network
structure that facilitates the formation of interorganeller connections
The mito.& ER joins together at multiple contact sites to form specific domains called as
mitochondria ER associated membranes.
why nitrogen fixation is so importantSolutionNitrogen fixation.pdf
why nitrogen fixation is so important
Solution
Nitrogen fixation
It is characteristic and manufactured, is basic for all types of life since nitrogen is required to
biosynthesize fundamental building pieces of plants, creatures and other living things, e.g.,
nucleotides for DNA and RNA, the coenzyme nicotinamide adenine dinucleotide for its part in
digestion system (exchanging electrons amongst atoms), and amino acids for proteins. Along
these lines, as a major aspect of the nitrogen cycle, it is fundamental for horticulture and the
fabricate of compost. It is additionally, in a roundabout way, important to the make of every
concoction intensify that contain nitrogen, which incorporates explosives, most pharmaceuticals,
colors, and so on.
Nitrogen fixation : it happens normally in the dirt by nitrogen settling microscopic organisms
(for instance, Azotobacter partnered with a few plants, particularly vegetables). Some nitrogen-
settling microbes have cozy associations with plants, alluded to as harmonious nitrogen
obsession. Looser connections between nitrogen-settling microorganisms and plants are regularly
alluded to as cooperative or non-advantageous, as found in nitrogen obsession happening on rice
roots. It additionally happens actually noticeable all around by method for lightning.
All natural nitrogen fixation is finished by method for metalloenzymes called nitrogenases.
These catalysts contain press, frequently with a moment metal, generally molybdenum yet here
and there vanadium. Microorganisms that can settle nitrogen are prokaryotes (both microscopic
organisms and archaea, conveyed all through their particular Domains) called diazotrophs. Some
higher plants, and a few creatures (termites), have shaped affiliations (advantageous interaction)
with diazotrophs..
what will happen to the photon if two photons enter the same atom si.pdf
what will happen to the photon if two photons enter the same atom simultaneously? when their
energies are combined they have exactly the right amount of energy to cause an electron to jump
to an excited state.
emission
absorption
nothing
Solution
absorption
There are certain energy levels in atom.An electron is a part of atom moving in its particular
level.So when an electron absorbs energy,it jumps to higher energy level from its
groundstate.And when it emits energy it comes back to its original state.
Which of the following is not a function of the skeletonA. Storag.pdf
Which of the following is not a function of the skeleton?
A. Storage of calcium and phosphate ions
B. Storage of red blood cells
C. Providing support for most muscles
D. Protecting the brain
E. Protecting the spinal cord
Solution
B. Storage of red blood cells ... FACT BASED QUESTION ALL OTHERS ARE FUNCTION
OF SKELTON....bones store calcium....and provide support skelton... cranium protects the brain
and vertebral column protects the spinal cord ....
Hope it helps ... Regards.
3. Which of the following will completely dissolve in water (More t.pdf
3. Which of the following will completely dissolve in water? (More than one answer may be
correct)
1. A large polar molecule
2. An ionically bonded salt
3. A small polar molecule
4. A small non-polar molecule
5. A large non-polar molecule
Solution
3. Which of the following will completely dissolve in water?
1. A large polar molecule - will completely dissolve in water (some under Heat conditions ) So
True
Lets take starch as Example for large polar molecule.Starch becomes soluble in water when
heated through a process is called starch gelatinization.
2. An ionically bonded salt - will completely dissolve in water So True
Lets take Nacl as Example
3. A small polar molecule - will completely dissolve in water So True
Ex: Glucose or Urea
4. A small non-polar molecule - No - Due to the presence of hydrophobic groups these are not
soluble in water
5. A large non-polar molecule- No - Due to the presence of hydrophobic groups these are not
soluble in water. But at the same time they have hydrophilic groups too. So they dissolve
partially..
True or False-A correlational study is used to examine the relatio.pdf
True or False-
A correlational study is used to examine the relationship between two variables but cannot
determine wheather it is a cause-and-effect relationship.
Solution
True
Correlation only tells whether the two variables are associated, i.e, if there is a linear relationship
between the two variables. But correlation does not imply causation.
what are some good current research proposal topic in psychology..pdf
what are some good current research proposal topic in psychology.
Solution
Current research topics of psychology are as follows
Is Child Obesity a Reflection of Parental Negligence?
Narcissistic Mother\'s Influence on Child\'s Mental Health
Schizophrenia in Young Women and Men
Social Group Dynamics in Infancy: Understanding the factors that contribute and/or influence
social interaction which shape infants\' development.
Cortical auditory evoked potentials and behavioural measures of auditory processing in children
with unilateral hearing loss due to congenital atresia
The Involvement of Psychologists in Military Interrogations
Effects of Postpartum Depression on the Mother and Child
The Psychological Profile of a Terrorist
The Impact of Hate Crimes on Victims and Communities
Efficacy of memory improvement in healthy older adults and adults with Mild Cognitive
Impairment with Memory Tune.
these are some topics that are good and interestong to work on.
Hope you find it useful.
Save Write a program to implement Binary search using recursive algo.pdf
Save Write a program to implement Binary search using recursive algorithm. Submit a
screenshot of your Java source codes and IDE/Comment line results in PDF file.
Solution
BinarySearchTest.java
import java.util.Scanner;
public class BinarySearchTest {
public static void main(String[] args) {
int[] arr1 = { 1,2,3,4,5,6,7,8,9,10};
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter the key: \");
int key = scan.nextInt();
int index = binarySearch(arr1,0,arr1.length,key);
System.out.println(\"Index is \"+index);
}
public static int binarySearch(int[] array, int a, int e, int key) {
if (a < e) {
int mid = a + (e - a) / 2;
if (key < array[mid]) {
return binarySearch(array, a, mid, key);
} else if (key > array[mid]) {
return binarySearch(array, mid+1, e , key);
} else {
return mid;
}
}
return -1;
}
}
Output:
Enter the key:
4
Index is 3.
1. Which of the following is responsible for problem-solving skills .pdf
1. Which of the following is responsible for problem-solving skills? A. central nervous system
B. peripheral nervous system C, somatic nervous system D. autonomic nervous system E. None
of these choices is correct 2. The sensory (afferent) division of the peripheral nervous system A.
transmits action potentials to sensory organs. B. conveys action potentials to the CNS. C.
stimulates glands to release hormones D. stimulates muscle contractions. E. does not involve
sensory receptors. 3. The autonomic nervous system A. stimulates skeletal muscle contractions,
B. has two sets ofneurons in a series. C. is involved in problem solving. D. is under voluntary
control. E. does not include the central nervous system 4. Neurons that have a single axon and a
single dendrite are A. unipolar, B. bipolar. C. multipolar. D. pseudopolar. E. monopolar 5. Which
of the following statements regarding neuroglia is false? A. Neuroglia serve as the major
supporting tissue in the CNS B. Neuroglia form part of the blood-brain barrier. C. Neuroglia
produce action potentials for skeletal muscles. D. Neuroglia form myelin sheaths around some
axons. E. Neuroglia produce cerebrospinal fluid.
Solution
1. The answer is: A. Central nervous system.
The central nervous system is responsible for problem solving skills of the brain.
2. The answer is: B. Conveys action potential to the CNS.
The afferent nerves conveys action potential to the CNS.
3. The answer is: B. has two sets of neurons in a series.
The autonomic nervous system has two sets of neurons in a series.
4. The answer is: B. Bipolar.
Neurons that have a single axon and a single dendrite are bipolar..
Some scientist and science fiction writers have envisioned life on ot.pdf
Some scientist and science fiction writers have envisioned life on other Wanes based on silicon
not carbon. Based on your understanding of the chemical nature of organic molecules why is
silicon a logical substitute for carbon as basis of biomolecules?
Solution
although silicon based life is not well accepted theory of life, I will try to explain why silicon has
been chosen as an alternative to carbon.
initially, the search for carbon initiated to find life or aliens on any other planets, since carbon is
the major component of biomolecules in all forms of life on Earth.
However, it may not be true that life forms in other planets too would contain carbon as its major
component. there might be some other component as well.
then research began to find an alternative for carbon search. some scientists believe that there is a
possibility of siliicon based life. on an extreme world like Titan, the atmosphere contains no
oxygen, all the water had been frozen to solid. But silicon is not oxidized to form inert rock.Titan
has liquid methane and ethane on its surface. methane acts a good solvent for silicon. silicon
molecules mimics organisc chemistry on Earth. They are stable and may be a start of alien
biochemistry..
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Aberdeen
Temple of Asclepius in Thrace. Excavation results
The temple and the sanctuary around were dedicated to Asklepios Zmidrenus. This name has been known since 1875 when an inscription dedicated to him was discovered in Rome. The inscription is dated in 227 AD and was left by soldiers originating from the city of Philippopolis (modern Plovdiv).
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Create a java project that - Draw a circle with three random init.pdf
Create a java project that:
- Draw a circle with three random initial points on the circle.
- Link the points to form a triangle.
- Print the angles values in the triangle.
- Use the mouse to drag a point along the perimeter of the circle. As you drag it, the triangle and
angles are redisplayed dynamically.
Here is the formula to compute angles:
A = acos((a * a - b * b - c * c) / (-2 * b * c))
B = acos((b * b - a * a - c * c) / (-2 * a * c))
C = acos((c * c - b * b - a * a) / (-2 * a * b))
Solution
@Suppresswarnings(\"WeakerAccess\")
public class DragPoints extends Application {
@Override
public void start(Stage primaryStage) {
final PointPane pane = new PointPane(640, 480);
pane.setStyle(\"-fx-background-color: wheat;\");
Label label = new Label(\"Click and drag the points.\");
BorderPane borderPane = new BorderPane(pane);
BorderPane.setAlignment(label, Pos.CENTER);
label.setPadding(new Insets(5));
borderPane.setBottom(label);
Scene scene = new Scene(borderPane);
primaryStage.setTitle(\"Exercise15_21\");
primaryStage.setScene(scene);
primaryStage.show();
}
private class PointPane extends Pane {
final Circle circle = new Circle();
final Vertex[] v = new Vertex[3];
final int strokeWidth = 2;
final Color circleStroke = Color.GRAY, legStroke = Color.BLACK;
@SuppressWarnings(\"SameParameterValue\")
PointPane(double w, double h) {
this.setPrefSize(w, h);
this.setWidth(w);
this.setHeight(h);
circle.setStroke(circleStroke);
circle.setFill(Color.TRANSPARENT);
circle.setStrokeWidth(strokeWidth);
circle.radiusProperty().bind(this.heightProperty().multiply(0.4));
circle.centerXProperty().bind(this.widthProperty().divide(2));
circle.centerYProperty().bind(this.heightProperty().divide(2));
this.getChildren().add(circle);
for (int i = 0; i < v.length; i++) {
v[i] = new Vertex(circle, 2 * Math.PI / v.length * (i + Math.random()));
v[i].radiusProperty().bind(circle.radiusProperty().divide(10));
v[i].setPosition();
v[i].setStroke(new Color(i == 0 ? 1 : 0, i == 1 ? 1 : 0, i == 2 ? 1 : 0, 1));
v[i].setFill(Color.TRANSPARENT);
v[i].setStrokeWidth(strokeWidth);
this.getChildren().add(v[i]);
v[i].setOnMouseDragged(new EventHandler() {
@Override
public void handle(MouseEvent event) {
int i;
for (i = 0; i < v.length; i++)
if (v[i] == event.getSource())
break;
v[i].setAngle(event.getX(), event.getY());
moveUpdate((Vertex) event.getSource());
}
});
}
for (int i = 0; i < v.length; i++) {
int j = i + 1 < v.length ? i + 1 : 0;
int k = j + 1 < v.length ? j + 1 : 0;
v[i].bindLeg(v[j], v[k]);
v[i].leg.setStroke(legStroke);
v[i].leg.setStrokeWidth(strokeWidth);
this.getChildren().add(v[i].leg);
this.getChildren().add(v[i].text);
}
for(DoubleProperty p: new DoubleProperty[]
{circle.radiusProperty(), circle.centerXProperty(), circle.centerYProperty()})
p.addListener(new ResizeListener());
moveUpdate(v[0]);
}
void moveUpdate(Vertex vert) {
vert.setPosition();
double[] legLength = new double[3];
for (int i = 0; i < v.length; i++)
legLength[i] = v[i].getLegLength();
for (int i = .
aphase The spindle contracts and the sister chromatids are separated.pdf
aphase: The spindle contracts and the sister chromatids are separated at the centromere Once the
sister for each replicated chromosome have separated, each is now called in the cell an
unreplicated chromosome. It is at this point t chromosome num Telophase: Telophase reverse.
List the events below that happen in is essentially prophase in telophase SIMULATE CELL
DIVISION MITOSIS Using pop beads and magnetic centromeres, create 4 chromosomes as
follows: Red 6 pop beads on each side of the centromere Yellow: Same as Red. Blue 5 pop beads
on one sided of the centromere, 3 pop beads on the other side Green: Same as Blue. Work
through the phases of cell division, beginning at G1, with 4 unreplicated chromosomes
Remember that we can not see chromosomes during interphase, but this can not be modeled in
this lab. It is important to note what happens during the S phase when you do the simulation
Chromosome count Use the adjective \"unreplicated\" or \"replicated\" with your count G1: rom
DS ones S: osomes As best as you are able, and without looking at the descriptions above or
your text, describe what is happening in each phase of mitosis as you simulate the process. Avoid
using pronouns (it and they). Speak the language of genetics as you simulate mitosis. Help your
lab partner if she and provide feedback with the language. Chromosome counts and description
(replicated or unreplicated) in each phase of mitosis: Prophase: mm Somes Metaphase 1,-L
Chromosome Anaphase Telophase: What event occurs after Telophase to complete the process
of mitosis? What C4 Kenesis mitosis has been completed? is the chromosome count in each cell
after Summarizing Mitosis 1 parent cell produces 2 essentially identical daughter cells. Mitosis
separates sister chromatids.
Solution
G1: 4 unreplicated chromosomes (singlet chromosomes)
S: 4 replicated chromosomes (doublet chromosomes)
G2: 4 replicated chromosomes (doublet chromosomes)
Prophase: 4 replicated chromosomes
Metaphase: 4 replicated chromosomes arranged on metaphasic plate.
Anaphase: 8 unreplicated (singlet) chromosomes (4 singlet chromosomes move toward one end,
while the other 4 singlet (unreplicated) chromosomes move to the opposite end.)
Telophase: 8 unreplicated chromosomes (4 unreplicated chromosomes in each of the two
dividing daughter cells)
cytokinesis occurs after telophase that completes mitosis
after mitosis, 2 daughter cells are formed. each daughter cell contains 4 unreplicated
chromosomes. (similar to parent cell at G1 phase).
A population of wild deer mice includes individuals with long or sho.pdf
A population of wild deer mice includes individuals with long or short hair. When two short
haird indviduals breed, one-third of the pups have long hair. How do you explain the unusual
phenotype ratio?
Solution
it can be autosomal inheritance where homozygous dominant is lethal.
consider short hair to be dominant S, long hair to be recessive s
so cross between two short haired can be
Ss X Ss
S s
S SS Ss
s Ss ss
1:2:1 ratio for SS:Ss:ss
but we considered homozygous dominant to be lethal, so SS are dead, therefore Ss:ss is 2:1 and
1/3rd population have long hair..
Who is it important that the lymphatic system operate separately f.pdf
Who is it important that the lymphatic system operate separately from the cardiovascular
system? Explain
Describe the relationship between the lymphatic vessels and blood vessels.
Who is it important that the lymphatic system operate separately from the cardiovascular
system? Explain
Describe the relationship between the lymphatic vessels and blood vessels.
Describe the relationship between the lymphatic vessels and blood vessels.
Solution
It is important that the lymphatic system operate separately from the cardiovascular system
becuase lymphatic system is a closed circulatory system that consists of lymph capillaries, lymph
vessels, and lymph nodes different to blood vessels. The lymph present in capillaries flows into
lymphatic vessels, which are presented deep or superficial but blood vessels are deep innervating
various parts of body to maintain pheripheral resistance & mean arterial blood pressure. Blood
flows through three types of blood vessels (arteries, capillaries and veins) distributed throughout
the body. Capillaries are the fine endings of blood vessels. These capillaries are responsible for
the exchange of respiratory gases, nutrients and waste products. The molecules move across the
capillaries through a process called “diffusion.” These lymph vessels combine to form lymphatic
trunks that empty into the large collecting vessels of lymphatic system called “thoracic duct and
right lymphatic duct.”
The lymphoid tissue is mostly composed of a reticular connective tissue (a type of loose
connective tissue). The fibres of the reticular network contain macrophages. A number of
lymphocytes from the postcapillary venule walls occupy this network and leave to the body
again. This ensures the lymphocytes to identify the infected sites at a faster rate.
Lymphatic system is a closed circulatory system that consists of lymph capillaries, lymph
vessels, and lymph nodes. Lymph is a tissue fluid, which is formed by the diffusion of
substances from blood capillaries into tissue spaces. The lymph present in capillaries flows into
lymphatic vessels, which are presented deep or superficial. These lymph vessels combine to form
lymphatic trunks that empty into the large collecting vessels of lymphatic system called “thoracic
duct and right lymphatic duct.”
It is important that the lymphatic system operate separately from the cardiovascular system due
to the follwoing reasons and exerts differentail effects of innate immune system. The lymph is
generated to produce defensive mechanisms via innate immunity include physical barriers,
phagocytes, immune surveillance,interferons, complement system, inflammation and fever.
1. Fluid recovery: Filtration of the fluid into the tissue spaces from the blood capillaries will
occur continuously. 85% of it will be absorbed by the blood capillaries. About 15% will be
remained unabsorbed. This unabsorbed portion of the fluid is about 2-4 litres of water and 1/4th
to 1/2th of plasma protein. If this water protein an.
Provide a full explanation to the below question.1. Summarize the .pdf
Provide a full explanation to the below question.
1. Summarize the 802.11 standard and describe the various flavors of the standard, its
architecture and how the standard is contributing to easing of congestion in cellular networks.
Solution
IEEE 802.11 is a standard that defines the physical and MAC layers of Wireless Local Area
Network (WLAN). Under this standard, the Mobile terminals (MTs) can communicate with a
Access Point (AP) in two modes. First mode is the Infrastructure Mode, in which the MTs
communicate with the APs which forward their data to the WLAN. Second mode is the Adhoc
mode, in which the MTs communicate directly with each other without the use of a AP. The
IEEE 802.11 has a very robust Medium Access Control (MAC) mechanism that helps in
alleviating te congestion in the LANs. The inherent working of MAC protocol employed by
802.11 standard is same as that of Carrier Sense Multiple Access/ Collision Aviodance
(CSMA/CA). However, in IEEE 802.11 the protocol is implemented in two different ways. First
way is called as the Distributed Coordination Function based Wireless MAC (DCFWMAC), and
the second way is Point Coordination Function based Wireless MAC (PCFWMAC).
In the DCFWMAC, every node tries to access the medium based on some fixed duration of time
defined in the standard. Normally there are three such time intervals defined based on which a
node attempts to access the channel and transmit its packets. The first interval is the Short Inter
frame Spacing (SIFS). It is the smallest duration of time between two frames and gives the node
higher priority for sending its data. Such a interval is employed only when the node has correctly
sent its data through the channel and the receiver node needs to send a Acknowlegement (ACK)
after waiting for SIFS period of time. The second most priority interval is PCF Inter Frame
Spacing (PIFS) used during Polling mechanism by the AP. Its duration is between SIFS and
DCF Inter frame Spacing (DIFS). The DIFS is the longest time duration and hence of least
priority, in which a AP has to wait between two successive channel accesses for the given
duration. Once the channel is sensed idle, the AP waits for DIFS period of time to transmit data
through the channel. If the channel is sensed busy, then it backs off for a period of time based on
the minimum and maximum value of a contention window. During the back off period, if the
node senses the channel as busy, then it freezes the backoff counter and starts the backoff(with
the remaining time left) once the channel is idle. In this manner, the nodes which have waited
longer get higher priority over others which accessing the channel. Once the node sends the data
after accessing the channel, the receiver waits for SIFS duration of time to finally send the
acknowledgement to the sender. In order to tackle the hidden terminal problem, the DCFWMAC
employs an additional mechanism called as the RTS-CTS mechanism, in which any node that
wants to transmit.
Write a program to generate the entire calendar for one year. The pr.pdf
Write a program to generate the entire calendar for one year. The program must get two values
from the user: (1) the year and (2) the day of the week for January 1st of that year. The year,
which should be positive, is needed to check for and handle leap years1. The day of the week for
January 1st is needed so that you know where to start the calendar. The user should enter 0 for
Sunday, 1 for Monday, ... or 6 for Saturday. As always, you need to validate the user\'s input. To
actually print the calendar, you must use a single method that prints out the calendar for one
month and then call this function 12 times from main(), once for each month in the year. To
check for a leap year you will need to write another method that takes the year as a parameter
and returns true if it’s a leap year, or false otherwise. Stubs (i.e. method signatures without any
code) for both of these methods have been provided for you.
The calendar should be printed in the form below. In this example, January starts on a Saturday
(day 6). Note that February starts on a Tuesday in this example because January ended on a
Monday.
January
1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 2829
30 31
February
1 2 3 4 5
6 7 8 9 10 11 12 ...
The format he gave us is as follows:
public class PA6a {
/**
* Error to output if year is not positive
*/
static final String E_YEAR = \"The year must be positive!\";
/**
* Error to output if the day is not between 0 and 6
*/
static final String E_DAY = \"The day of January 1st must be between 0 and 6!\";
/**
* Determines if an input is a leap year
*
* @param year year in question
* @ereturn true if a leap year
*/
public static boolean isLeapYear(int year) {
return false; // TODO: replace with your code
}
/**
* Outputs a month to the console
*
* @param month title
* @param startDay 0=Sunday ... 6=Saturday
* @param numDays number of days in the month
* @return day of the week of the last day of the month
*/
public static int printMonth(String month, int startDay, int numDays) {
return 0; // TODO: replace with your code
}
/**
* Program execution point:
* input year, day of the week (0-6) of january 1
* output calendar for that year
*
* @param args command-line arguments (ignored)
*/
public static void main(String[] args) {
// TODO: write your code here
}
I am confused especially with the print month method, help is greatly appreciated.
Solution
package org.students;
import java.util.Scanner;
import com.sun.org.apache.bcel.internal.generic.IUSHR;
public class Calander {
//declaring static variables
static final String E_YEAR = \"The year must be positive!\";
static final String E_DAY = \"The day of January 1st must be between 0 and 6!\";
public static void main(String[] args) {
//Declaring variables
int getFirstDay;
int year;
Scanner scanner = new Scanner(System.in);
//This while loop continue to execute until user enters valid year
while (true) {
//Getting the year entered by the user
System.out.print(\"Enter the year:.
Write the interval notation for the set of numbers graphed. Solu.pdf
Write the interval notation for the set of numbers graphed.
Solution
The open circle represents that the value is not included in the interval while a solid (filled) circle
represents that the value is included in the interval.
So, we can see that there is open circle for (-2) and closed for (5).
Hence the required interval is (-2,5]..
Hyenas are diploid organisms, and their gametes contain 20 chromosome.pdf
Hyenas are diploid organisms, and their gametes contain 20 chromosomes. Fill in the blank in
each sentences using the terms below. Each term may be used more than once or not at all.
When the somatic cells of a hyena are in interphase, the ploidy level is denoted as The ploidy
level of the hyena\'s gametes is denoted as The 2n number of a hyena is 40 20 1n 80 2n
Solution
When the somatic cells of hyena in interphase = diploid [2n] because they are not divided and
tend to divide by preparing enzymes and multiplying DNA.
The ploidy level of hyena is haploid or 1n because they are developed through meiosis process.
The 2n number of chromosomes of hyena is 40 because gametes are haploid so n=20 [as given]
and the 2n = 2x 20 = 40.
Why doesnt hemoglobin have any intermediate conformations between .pdf
Why doesn\'t hemoglobin have any intermediate conformations between R-state and T-state?
Solution
Answer:
X-ray crystallogarphy has determined two major conformations of hemoglobin- the T state
which is stabilized in the absence of oxygen and the R state which is relatively more stable in the
presence of oxygen.
The pair of identical dimers of Hemoglobin are linked by several ion pairs at the interface that
stabilize it in its deoxy-state. Many of these interactions are disrupted upon binding to oxygen
and new ones are formed.
There are intermediate conformations between T and R states, but the intermediate
conformations have intermediate binding affinities. Also, these transition states are
thermodynamically unfavourable and short lived. These states have been utilized in studying the
allosteric pathways of hemoglobin. The binding of the ligand destabilizes the T state and makes
it more likely to switch to the high-affinity oxygenated (oxy)-R state..
4. What is gyanandromorphy How does this happenSolutionAnswe.pdf
4. What is gyanandromorphy? How does this happen?
Solution
Answer : Gynandromorphy is a phenamenan in which there is an occurence of both male and
female charectorstics in a single individual (part of the body expresses as male charectors, where
as the other part express female charectors. Individuals with this phenotype are called
gynandromorphs.
It results due to the loss of an X chromosome in a particular cell during development through
mitosis.For instance, a cell with XY through mitosis produces XXYY upon duplication which in
turn divides to form XY cells. But in gynandromorphs, this cell divides into an X cell and an
XYY cells. Due to this the organism develops with both male and female charectors..
Which of the following requires the mitochondria to create contact po.pdf
Which of the following requires the mitochondria to create contact points with the ER? mt
fission mt fusion mt biogenesis of more mitochondria autophagy of mitochondria (Mitophagy)
apoptosis
Solution
Answer= The mitochondria requires the following to create a contact with ER is
mt fission - mitochondrial fission
Explaination= The mitochondria & RE are tubular organelles with a characteristic network
structure that facilitates the formation of interorganeller connections
The mito.& ER joins together at multiple contact sites to form specific domains called as
mitochondria ER associated membranes.
why nitrogen fixation is so importantSolutionNitrogen fixation.pdf
why nitrogen fixation is so important
Solution
Nitrogen fixation
It is characteristic and manufactured, is basic for all types of life since nitrogen is required to
biosynthesize fundamental building pieces of plants, creatures and other living things, e.g.,
nucleotides for DNA and RNA, the coenzyme nicotinamide adenine dinucleotide for its part in
digestion system (exchanging electrons amongst atoms), and amino acids for proteins. Along
these lines, as a major aspect of the nitrogen cycle, it is fundamental for horticulture and the
fabricate of compost. It is additionally, in a roundabout way, important to the make of every
concoction intensify that contain nitrogen, which incorporates explosives, most pharmaceuticals,
colors, and so on.
Nitrogen fixation : it happens normally in the dirt by nitrogen settling microscopic organisms
(for instance, Azotobacter partnered with a few plants, particularly vegetables). Some nitrogen-
settling microbes have cozy associations with plants, alluded to as harmonious nitrogen
obsession. Looser connections between nitrogen-settling microorganisms and plants are regularly
alluded to as cooperative or non-advantageous, as found in nitrogen obsession happening on rice
roots. It additionally happens actually noticeable all around by method for lightning.
All natural nitrogen fixation is finished by method for metalloenzymes called nitrogenases.
These catalysts contain press, frequently with a moment metal, generally molybdenum yet here
and there vanadium. Microorganisms that can settle nitrogen are prokaryotes (both microscopic
organisms and archaea, conveyed all through their particular Domains) called diazotrophs. Some
higher plants, and a few creatures (termites), have shaped affiliations (advantageous interaction)
with diazotrophs..
what will happen to the photon if two photons enter the same atom si.pdf
what will happen to the photon if two photons enter the same atom simultaneously? when their
energies are combined they have exactly the right amount of energy to cause an electron to jump
to an excited state.
emission
absorption
nothing
Solution
absorption
There are certain energy levels in atom.An electron is a part of atom moving in its particular
level.So when an electron absorbs energy,it jumps to higher energy level from its
groundstate.And when it emits energy it comes back to its original state.
Which of the following is not a function of the skeletonA. Storag.pdf
Which of the following is not a function of the skeleton?
A. Storage of calcium and phosphate ions
B. Storage of red blood cells
C. Providing support for most muscles
D. Protecting the brain
E. Protecting the spinal cord
Solution
B. Storage of red blood cells ... FACT BASED QUESTION ALL OTHERS ARE FUNCTION
OF SKELTON....bones store calcium....and provide support skelton... cranium protects the brain
and vertebral column protects the spinal cord ....
Hope it helps ... Regards.
3. Which of the following will completely dissolve in water (More t.pdf
3. Which of the following will completely dissolve in water? (More than one answer may be
correct)
1. A large polar molecule
2. An ionically bonded salt
3. A small polar molecule
4. A small non-polar molecule
5. A large non-polar molecule
Solution
3. Which of the following will completely dissolve in water?
1. A large polar molecule - will completely dissolve in water (some under Heat conditions ) So
True
Lets take starch as Example for large polar molecule.Starch becomes soluble in water when
heated through a process is called starch gelatinization.
2. An ionically bonded salt - will completely dissolve in water So True
Lets take Nacl as Example
3. A small polar molecule - will completely dissolve in water So True
Ex: Glucose or Urea
4. A small non-polar molecule - No - Due to the presence of hydrophobic groups these are not
soluble in water
5. A large non-polar molecule- No - Due to the presence of hydrophobic groups these are not
soluble in water. But at the same time they have hydrophilic groups too. So they dissolve
partially..
True or False-A correlational study is used to examine the relatio.pdf
True or False-
A correlational study is used to examine the relationship between two variables but cannot
determine wheather it is a cause-and-effect relationship.
Solution
True
Correlation only tells whether the two variables are associated, i.e, if there is a linear relationship
between the two variables. But correlation does not imply causation.
what are some good current research proposal topic in psychology..pdf
what are some good current research proposal topic in psychology.
Solution
Current research topics of psychology are as follows
Is Child Obesity a Reflection of Parental Negligence?
Narcissistic Mother\'s Influence on Child\'s Mental Health
Schizophrenia in Young Women and Men
Social Group Dynamics in Infancy: Understanding the factors that contribute and/or influence
social interaction which shape infants\' development.
Cortical auditory evoked potentials and behavioural measures of auditory processing in children
with unilateral hearing loss due to congenital atresia
The Involvement of Psychologists in Military Interrogations
Effects of Postpartum Depression on the Mother and Child
The Psychological Profile of a Terrorist
The Impact of Hate Crimes on Victims and Communities
Efficacy of memory improvement in healthy older adults and adults with Mild Cognitive
Impairment with Memory Tune.
these are some topics that are good and interestong to work on.
Hope you find it useful.
Save Write a program to implement Binary search using recursive algo.pdf
Save Write a program to implement Binary search using recursive algorithm. Submit a
screenshot of your Java source codes and IDE/Comment line results in PDF file.
Solution
BinarySearchTest.java
import java.util.Scanner;
public class BinarySearchTest {
public static void main(String[] args) {
int[] arr1 = { 1,2,3,4,5,6,7,8,9,10};
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter the key: \");
int key = scan.nextInt();
int index = binarySearch(arr1,0,arr1.length,key);
System.out.println(\"Index is \"+index);
}
public static int binarySearch(int[] array, int a, int e, int key) {
if (a < e) {
int mid = a + (e - a) / 2;
if (key < array[mid]) {
return binarySearch(array, a, mid, key);
} else if (key > array[mid]) {
return binarySearch(array, mid+1, e , key);
} else {
return mid;
}
}
return -1;
}
}
Output:
Enter the key:
4
Index is 3.
1. Which of the following is responsible for problem-solving skills .pdf
1. Which of the following is responsible for problem-solving skills? A. central nervous system
B. peripheral nervous system C, somatic nervous system D. autonomic nervous system E. None
of these choices is correct 2. The sensory (afferent) division of the peripheral nervous system A.
transmits action potentials to sensory organs. B. conveys action potentials to the CNS. C.
stimulates glands to release hormones D. stimulates muscle contractions. E. does not involve
sensory receptors. 3. The autonomic nervous system A. stimulates skeletal muscle contractions,
B. has two sets ofneurons in a series. C. is involved in problem solving. D. is under voluntary
control. E. does not include the central nervous system 4. Neurons that have a single axon and a
single dendrite are A. unipolar, B. bipolar. C. multipolar. D. pseudopolar. E. monopolar 5. Which
of the following statements regarding neuroglia is false? A. Neuroglia serve as the major
supporting tissue in the CNS B. Neuroglia form part of the blood-brain barrier. C. Neuroglia
produce action potentials for skeletal muscles. D. Neuroglia form myelin sheaths around some
axons. E. Neuroglia produce cerebrospinal fluid.
Solution
1. The answer is: A. Central nervous system.
The central nervous system is responsible for problem solving skills of the brain.
2. The answer is: B. Conveys action potential to the CNS.
The afferent nerves conveys action potential to the CNS.
3. The answer is: B. has two sets of neurons in a series.
The autonomic nervous system has two sets of neurons in a series.
4. The answer is: B. Bipolar.
Neurons that have a single axon and a single dendrite are bipolar..
Some scientist and science fiction writers have envisioned life on ot.pdf
Some scientist and science fiction writers have envisioned life on other Wanes based on silicon
not carbon. Based on your understanding of the chemical nature of organic molecules why is
silicon a logical substitute for carbon as basis of biomolecules?
Solution
although silicon based life is not well accepted theory of life, I will try to explain why silicon has
been chosen as an alternative to carbon.
initially, the search for carbon initiated to find life or aliens on any other planets, since carbon is
the major component of biomolecules in all forms of life on Earth.
However, it may not be true that life forms in other planets too would contain carbon as its major
component. there might be some other component as well.
then research began to find an alternative for carbon search. some scientists believe that there is a
possibility of siliicon based life. on an extreme world like Titan, the atmosphere contains no
oxygen, all the water had been frozen to solid. But silicon is not oxidized to form inert rock.Titan
has liquid methane and ethane on its surface. methane acts a good solvent for silicon. silicon
molecules mimics organisc chemistry on Earth. They are stable and may be a start of alien
biochemistry..
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