Which one is the answer How do you know thatOccupancy type is a .pdf
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Which one is the answer? How do you know that? Occupancy type is a surrogate measure for Fire Load Ventilation Combustibility ToxicityA. Fire LoadB. VentilationC. CombustibilityD. Toxicity Solution Occupancy type is a surrogate measure for ventilation. Surrogate measures of human occupancy and activity in the 10 patient rooms using both indoor CO2 concentrations and infrared (IR) beam-break counters installed at the patient room doorways. Outdoor air fractions in the HVAC systems serving the two floors..
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Questions1) How does multimedia transmission differ than messag.pdf
Questions:
1) How does multimedia transmission differ than \'message\' transmission?
2) Describe compression and include the different types used in multimedia.
Solution
1)
Message transmission includes the standard text communication using standard communication
protocols. Here the messages are generally short and fixed in length. Message Service Centres
(MSCs) are responsible for sending and receiving of messages. These messages are stored and
forwarded and in case of any failure, they are queued for resending. The message if successfully
sent, is then discarded from storage.
Multimedia transmission is a way to send messages that includes multimedia content like
images, music or video files. Message Service Centres along with carriers are responsible for
successful sending and receiving of messages. Carriers are mainly used to change the format of
the Multimedia message because the formats supported on different targeted devices are
different. Internet is required to transfer multimedia messages between carrier and carrier. If the
target device does not support multimedia, these messages can be viewed on the browser through
an URL send as text message.
2)
Compression is reduction in size of a file by reducing the number of bits that are used to
represent without any loss of data. Generally, compression is employed to save storage space and
increase the pace of file sending over a network which in turn saves bandwidth. An algorithm is
used to compress a file.
Multimedia compression techniques are classified as
Lossy Compression: These recover the somewhat a similar representation to the original file.
They have high compression ratios and used more practically.
Examples of Lossy compression techniques are
Lossless Compression: These techniques recover the original file.
Examples of lossless techniques are.
Question 1 of 10 Map sapling learning Based on the chart to the right.pdf
Question 1 of 10 Map sapling learning Based on the chart to the right Ratio of Exports to GDP
for Major Countries and the World (Percent) in which year did world globalization hit its lowest
point 35 30 O 1890 O 1913 25 France O 1960 Germany 20 O 1970 Japan 15 LUK. O 1990 US
10 In 1913, what country attained the World highest amount of globalization? 5 O France 1890
1970 1913 1960 1990 O Japan O Germany United O Kingdom (UK) United States (US)
Solution
A yardstick for measuring globalization is Exports/GDP ratio. The higher this ratio, the higher
the magnitude of globalization.
(a) World globalization was lowest in 1960, when Export/GDP ratio was lowest at 2.5%
(b) In 1913, UK had highest globalization because its Export/GDP ratio was highest at 29%..
please send edited code of RCBug.javaRCBUG.javaimport java.util..pdf
please send edited code of RCBug.java
RCBUG.java
import java.util.Random;
public class RCBug {
private EZImage bugPicture; // Member variable to store bug picture
private int x, y, startx, starty; // Member variables to store x, y, startx, starty
private int destx, desty; // Member variables to store destination values
private long starttime; // Member variable to store the start time
private long duration; // Member variable to store the duration
private boolean interpolation; // Member variable that is set to true if it is in the
interpolation state
//1) add the private static final variables here
// Constructor for RCBug takes 3 parameters
public RCBug(String filename,int posx, int posy){
// Set the current position of the bug
x=posx;y=posy;
// Create the image of the bug
bugPicture = EZ.addImage(filename, posx, posy);
// Move it to the starting position. This is actually redundant.
bugPicture.translateTo(x,y);
// Set interpolation mode to false initially
interpolation = false;
}
// Set the destination by giving it 3 variables
// Dur means duration and is measured in seconds
public void setDestination(int posx, int posy, long dur){
// Set destination position and duration
// Convert seconds to miliseconds
destx = posx; desty = posy; duration = dur*1000;
// Get the startting time (i.e. time according to your computer)
starttime = System.currentTimeMillis();
// Set the starting position of your bug
startx=x; starty=y;
// Set interolation mode to true
interpolation = true;
}
// If you?re in interpolation state then return true, else false.
public boolean moving() {
return interpolation;
}
// This moves the bug based on the current time and elapsed time according to the
interpolation equation
public void go(){
// If interpolation mode is true then do interpolation
if (interpolation == true) {
// Normalize the time (i.e. make it a number from 0 to 1)
float normTime = (float) (System.currentTimeMillis() - starttime)/ (float) duration;
// Calculate the interpolated position of the bug
x = (int) (startx + ((float) (destx - startx) * normTime));
y = (int) (starty + ((float) (desty - starty) * normTime));
// If the difference between current time and start time has exceeded the duration
// then the animation/interpolation is over.
if ((System.currentTimeMillis() - starttime) >= duration) {
// Set interpolation to false
interpolation = false;
// Move the bug all the way to the destination
x = destx; y = desty;
}
//2)add a random amount to the x and y coordinates
// Don?t forget to move the image itself.
bugPicture.translateTo(x,y);
}
}
}
RCBugMain.java
import java.awt.Color;
import java.io.InputStreamReader;
import java.util.Scanner;
public class RCBugMain {
public static void main(String[] args) {
// Setup EZ graphics system.
EZ.initialize(512, 512);
// Set background color to dark green
EZ.setBackgroundColor(new Color(0, 100, 0));
// Make an InputStreamReader object which reads from the keyboard.
// Make a Scanner object that uses the InputStreamRe.
On January 1, 2002, Germany officially adopted the euro as its curre.pdf
On January 1, 2002, Germany officially adopted the euro as its currency, and deutsche mark
stopped being legal tender. According to the Article in the Wall Street Journal, even 10 years
later many stores in Germany continued to accept it. Briefly explain how possible for people to
continue to use a currency when government replaced it with another currency.
Source: Who needs the euro When You Can Pay with Deutsche Marks? Wall Street Journal July,
2012
Solution
The most important function of money is that it is used as medium of exchange. In other words,
it can be used for payment of goods and services.
Money is able to act as medium of exchange because it possesses the quality of general
acceptability.
Anything that is generally acceptable can act as money and thus can be used as medium of
exchange.
Quality of general acceptability is the key and not the government undertaking.
Many people across Germany are still accepting deutsche mark as medium of exchange and due
to this general acceptability it is still being used as medium of exchange even though German
government has discontinued deutsche mark and adopted the Euro.
Adoption of a currency by government makes it legal tender in the country concerned but it does
not guarantee that it will be used as medium of exchange unless and until it is generally
acceptable.
Thus, it is possible for people to continue using a currency when government replaced it with
another currency if the currency being used by the people is still generally acceptable as medium
of exchange. This quality of currency keeps it in circulation and prolongs its usage even though
government has dumped it..
Networking problem help Consider the following TCP session between h.pdf
Networking problem help Consider the following TCP session between hosts A and host B on a
perfect channel: A has a file of size 2.5MSS to send to B. A desires to send segments of
alternating sizes MSS and MSS/2 (sequence number = 0 for the first segment of size MSS). B
also has a file of size 2.5MSS to send to A. B desires to send segments of alternating sizes
MSS/2 and MSS (sequence number = 0 for the first segment of size MSS/2).
A sends first.Note, A and B don’t have to fully fill up their “desired” segment size if there is no
sufficient bytes to send, especially at the end of their conversation, where it is even possible for
them to send a segment whose data size is equal to 0. Draw all the segment exchanges until the
Both A and B think they don’t have any more bytes to send and there is no byte to acknowledge.
For each segment, clearly mark the sequence number, the ACK number, and the size of data it
carries. If the data size is equal to 0 in a segment, the sequence number field is irrelevant. You
may just mark N/A.
Solution
Source port: 16 Bit number which identifies the Source Port number (Sending Computer\'s TCP
Port).
Destination port: 16 Bit number which identifies the Destination Port number (Receiving Port).
Sequence number: 32 Bit number used for byte level numbering of TCP segments. If you are
using TCP, each byte of data is assigned a sequence number. If SYN flag is set (during the initial
three way handshake connection initiation), then this is the initial sequence number. The
sequence number of the actual first data byte will then be this sequence number plus 1. For
example, let the first byte of data by a device in a particular TCP header will have its sequence
number in this field 50000. If this packet has 500 bytes of data in it, then the next packet sent by
this device will have the sequence number of 50000 + 500 + 1 = 50501.
Acknowledgment Number: 32 Bit number field which indicates the next sequence number that
the sending device is expecting from the other device.
Header Length: 4 Bit field which shows the number of 32 Bit words in the header. Also known
as the Data Offset field. The minimum size header is 5 words (binary pattern is 0101).
Reserved: Always set to 0 (Size 6 bits).
Control Bit Flags: We have seen before that TCP is a Connection Oriented Protocol. The
meaning of Connection Oriented Protocol is that, before any data can be transmitted, a reliable
connection must be obtained and acknowledged. Control Bits govern the entire process of
connection establishment, data transmissions and connection termination. The control bits are
listed as follows: They are:
URG: Urgent Pointer.
ACK: Acknowledgement.
PSH: This flag means Push function. Using this flag, TCP allows a sending application to
specify that the data must be pushed immediately. When an application requests the TCP to push
data, the TCP should send the data that has accumulated without waiting to fill the segment.
RST: Reset the connection. The RST bit.
Lipid bilayers can only be synthesized by biological organisms. This .pdf
Lipid bilayers, which form cell membranes, can only be synthesized by biological organisms. It is believed that the internal compartments of eukaryotic cells, like organelles, developed from invaginations of the external prokaryotic cell membrane. Abiogenesis hypothesizes that life exists throughout the universe and was distributed to Earth via meteorites and asteroids, bringing extremophile bacteria trapped in debris. However, abiogenesis proposes that life arose from non-living organic compounds on Earth, not bacteria from space.
Is a neutral trait (meaning it is neither deleterious nor advantageo.pdf
Is a neutral trait (meaning it is neither deleterious nor advantageous to the survival and
reproduction of the organism) likely to be eliminated from a population by adaptive evolution?
(such as the presence of pelvic bones in snakes, which are legless) Why or why not?
Solution
There are four major methods by which evolution happens:
mutation, migration, drift, and selection
Mutation brings about changes or variations in the populaton.
Migration increases or decreases the number of the population.
Drift: when the individuals do not reproduce or they die, there could be disappearance of genes
which changes the frequency of occurrence of some genotypes in a population.
Selection occurs when the superior, better traits are selected to move forward in the generations.
When we consider evolution of neutral traits, we cover all the first three : Mutation, migration
and drift. These neutral traits generally stay in the population for a long time. They may
disappear eventually but its longevity is uncertain. It also could be that the genes responsible for
the neutral trait do not come under direct selection or it could also be the result of plieotropy
where one gene is responsible for many traits..
Let T R^n rightarrow R^m be a linear transformation. Given a subspa.pdf
Let T : R^n rightarrow R^m be a linear transformation. Given a subspace Z of R^m, let 5 {v
elementof R^n | T(v) elementof Z). Show that S is a subspace of R^n.
Solution
Let v,w in S
Then T(v) and T(w) belongs to Z
Consider av+bw , a,b are scalars
We need to see if av+bw are in S
For this consider T(av+bw) = aT(v)+bT(w) (because T is linear transformtaion)
Now T(v) and T(w) are in Z
And so aT(v) and bT(w) are in Z
This => aT(v)+bT(w) are in Z
Hence we get that T(av+bw) is in Z
So we get av+bw is in S
Hence S is subspace of Rn.
Recommended
Questions1) How does multimedia transmission differ than messag.pdf
Questions:
1) How does multimedia transmission differ than \'message\' transmission?
2) Describe compression and include the different types used in multimedia.
Solution
1)
Message transmission includes the standard text communication using standard communication
protocols. Here the messages are generally short and fixed in length. Message Service Centres
(MSCs) are responsible for sending and receiving of messages. These messages are stored and
forwarded and in case of any failure, they are queued for resending. The message if successfully
sent, is then discarded from storage.
Multimedia transmission is a way to send messages that includes multimedia content like
images, music or video files. Message Service Centres along with carriers are responsible for
successful sending and receiving of messages. Carriers are mainly used to change the format of
the Multimedia message because the formats supported on different targeted devices are
different. Internet is required to transfer multimedia messages between carrier and carrier. If the
target device does not support multimedia, these messages can be viewed on the browser through
an URL send as text message.
2)
Compression is reduction in size of a file by reducing the number of bits that are used to
represent without any loss of data. Generally, compression is employed to save storage space and
increase the pace of file sending over a network which in turn saves bandwidth. An algorithm is
used to compress a file.
Multimedia compression techniques are classified as
Lossy Compression: These recover the somewhat a similar representation to the original file.
They have high compression ratios and used more practically.
Examples of Lossy compression techniques are
Lossless Compression: These techniques recover the original file.
Examples of lossless techniques are.
Question 1 of 10 Map sapling learning Based on the chart to the right.pdf
Question 1 of 10 Map sapling learning Based on the chart to the right Ratio of Exports to GDP
for Major Countries and the World (Percent) in which year did world globalization hit its lowest
point 35 30 O 1890 O 1913 25 France O 1960 Germany 20 O 1970 Japan 15 LUK. O 1990 US
10 In 1913, what country attained the World highest amount of globalization? 5 O France 1890
1970 1913 1960 1990 O Japan O Germany United O Kingdom (UK) United States (US)
Solution
A yardstick for measuring globalization is Exports/GDP ratio. The higher this ratio, the higher
the magnitude of globalization.
(a) World globalization was lowest in 1960, when Export/GDP ratio was lowest at 2.5%
(b) In 1913, UK had highest globalization because its Export/GDP ratio was highest at 29%..
please send edited code of RCBug.javaRCBUG.javaimport java.util..pdf
please send edited code of RCBug.java
RCBUG.java
import java.util.Random;
public class RCBug {
private EZImage bugPicture; // Member variable to store bug picture
private int x, y, startx, starty; // Member variables to store x, y, startx, starty
private int destx, desty; // Member variables to store destination values
private long starttime; // Member variable to store the start time
private long duration; // Member variable to store the duration
private boolean interpolation; // Member variable that is set to true if it is in the
interpolation state
//1) add the private static final variables here
// Constructor for RCBug takes 3 parameters
public RCBug(String filename,int posx, int posy){
// Set the current position of the bug
x=posx;y=posy;
// Create the image of the bug
bugPicture = EZ.addImage(filename, posx, posy);
// Move it to the starting position. This is actually redundant.
bugPicture.translateTo(x,y);
// Set interpolation mode to false initially
interpolation = false;
}
// Set the destination by giving it 3 variables
// Dur means duration and is measured in seconds
public void setDestination(int posx, int posy, long dur){
// Set destination position and duration
// Convert seconds to miliseconds
destx = posx; desty = posy; duration = dur*1000;
// Get the startting time (i.e. time according to your computer)
starttime = System.currentTimeMillis();
// Set the starting position of your bug
startx=x; starty=y;
// Set interolation mode to true
interpolation = true;
}
// If you?re in interpolation state then return true, else false.
public boolean moving() {
return interpolation;
}
// This moves the bug based on the current time and elapsed time according to the
interpolation equation
public void go(){
// If interpolation mode is true then do interpolation
if (interpolation == true) {
// Normalize the time (i.e. make it a number from 0 to 1)
float normTime = (float) (System.currentTimeMillis() - starttime)/ (float) duration;
// Calculate the interpolated position of the bug
x = (int) (startx + ((float) (destx - startx) * normTime));
y = (int) (starty + ((float) (desty - starty) * normTime));
// If the difference between current time and start time has exceeded the duration
// then the animation/interpolation is over.
if ((System.currentTimeMillis() - starttime) >= duration) {
// Set interpolation to false
interpolation = false;
// Move the bug all the way to the destination
x = destx; y = desty;
}
//2)add a random amount to the x and y coordinates
// Don?t forget to move the image itself.
bugPicture.translateTo(x,y);
}
}
}
RCBugMain.java
import java.awt.Color;
import java.io.InputStreamReader;
import java.util.Scanner;
public class RCBugMain {
public static void main(String[] args) {
// Setup EZ graphics system.
EZ.initialize(512, 512);
// Set background color to dark green
EZ.setBackgroundColor(new Color(0, 100, 0));
// Make an InputStreamReader object which reads from the keyboard.
// Make a Scanner object that uses the InputStreamRe.
On January 1, 2002, Germany officially adopted the euro as its curre.pdf
On January 1, 2002, Germany officially adopted the euro as its currency, and deutsche mark
stopped being legal tender. According to the Article in the Wall Street Journal, even 10 years
later many stores in Germany continued to accept it. Briefly explain how possible for people to
continue to use a currency when government replaced it with another currency.
Source: Who needs the euro When You Can Pay with Deutsche Marks? Wall Street Journal July,
2012
Solution
The most important function of money is that it is used as medium of exchange. In other words,
it can be used for payment of goods and services.
Money is able to act as medium of exchange because it possesses the quality of general
acceptability.
Anything that is generally acceptable can act as money and thus can be used as medium of
exchange.
Quality of general acceptability is the key and not the government undertaking.
Many people across Germany are still accepting deutsche mark as medium of exchange and due
to this general acceptability it is still being used as medium of exchange even though German
government has discontinued deutsche mark and adopted the Euro.
Adoption of a currency by government makes it legal tender in the country concerned but it does
not guarantee that it will be used as medium of exchange unless and until it is generally
acceptable.
Thus, it is possible for people to continue using a currency when government replaced it with
another currency if the currency being used by the people is still generally acceptable as medium
of exchange. This quality of currency keeps it in circulation and prolongs its usage even though
government has dumped it..
Networking problem help Consider the following TCP session between h.pdf
Networking problem help Consider the following TCP session between hosts A and host B on a
perfect channel: A has a file of size 2.5MSS to send to B. A desires to send segments of
alternating sizes MSS and MSS/2 (sequence number = 0 for the first segment of size MSS). B
also has a file of size 2.5MSS to send to A. B desires to send segments of alternating sizes
MSS/2 and MSS (sequence number = 0 for the first segment of size MSS/2).
A sends first.Note, A and B don’t have to fully fill up their “desired” segment size if there is no
sufficient bytes to send, especially at the end of their conversation, where it is even possible for
them to send a segment whose data size is equal to 0. Draw all the segment exchanges until the
Both A and B think they don’t have any more bytes to send and there is no byte to acknowledge.
For each segment, clearly mark the sequence number, the ACK number, and the size of data it
carries. If the data size is equal to 0 in a segment, the sequence number field is irrelevant. You
may just mark N/A.
Solution
Source port: 16 Bit number which identifies the Source Port number (Sending Computer\'s TCP
Port).
Destination port: 16 Bit number which identifies the Destination Port number (Receiving Port).
Sequence number: 32 Bit number used for byte level numbering of TCP segments. If you are
using TCP, each byte of data is assigned a sequence number. If SYN flag is set (during the initial
three way handshake connection initiation), then this is the initial sequence number. The
sequence number of the actual first data byte will then be this sequence number plus 1. For
example, let the first byte of data by a device in a particular TCP header will have its sequence
number in this field 50000. If this packet has 500 bytes of data in it, then the next packet sent by
this device will have the sequence number of 50000 + 500 + 1 = 50501.
Acknowledgment Number: 32 Bit number field which indicates the next sequence number that
the sending device is expecting from the other device.
Header Length: 4 Bit field which shows the number of 32 Bit words in the header. Also known
as the Data Offset field. The minimum size header is 5 words (binary pattern is 0101).
Reserved: Always set to 0 (Size 6 bits).
Control Bit Flags: We have seen before that TCP is a Connection Oriented Protocol. The
meaning of Connection Oriented Protocol is that, before any data can be transmitted, a reliable
connection must be obtained and acknowledged. Control Bits govern the entire process of
connection establishment, data transmissions and connection termination. The control bits are
listed as follows: They are:
URG: Urgent Pointer.
ACK: Acknowledgement.
PSH: This flag means Push function. Using this flag, TCP allows a sending application to
specify that the data must be pushed immediately. When an application requests the TCP to push
data, the TCP should send the data that has accumulated without waiting to fill the segment.
RST: Reset the connection. The RST bit.
Lipid bilayers can only be synthesized by biological organisms. This .pdf
Lipid bilayers, which form cell membranes, can only be synthesized by biological organisms. It is believed that the internal compartments of eukaryotic cells, like organelles, developed from invaginations of the external prokaryotic cell membrane. Abiogenesis hypothesizes that life exists throughout the universe and was distributed to Earth via meteorites and asteroids, bringing extremophile bacteria trapped in debris. However, abiogenesis proposes that life arose from non-living organic compounds on Earth, not bacteria from space.
Is a neutral trait (meaning it is neither deleterious nor advantageo.pdf
Is a neutral trait (meaning it is neither deleterious nor advantageous to the survival and
reproduction of the organism) likely to be eliminated from a population by adaptive evolution?
(such as the presence of pelvic bones in snakes, which are legless) Why or why not?
Solution
There are four major methods by which evolution happens:
mutation, migration, drift, and selection
Mutation brings about changes or variations in the populaton.
Migration increases or decreases the number of the population.
Drift: when the individuals do not reproduce or they die, there could be disappearance of genes
which changes the frequency of occurrence of some genotypes in a population.
Selection occurs when the superior, better traits are selected to move forward in the generations.
When we consider evolution of neutral traits, we cover all the first three : Mutation, migration
and drift. These neutral traits generally stay in the population for a long time. They may
disappear eventually but its longevity is uncertain. It also could be that the genes responsible for
the neutral trait do not come under direct selection or it could also be the result of plieotropy
where one gene is responsible for many traits..
Let T R^n rightarrow R^m be a linear transformation. Given a subspa.pdf
Let T : R^n rightarrow R^m be a linear transformation. Given a subspace Z of R^m, let 5 {v
elementof R^n | T(v) elementof Z). Show that S is a subspace of R^n.
Solution
Let v,w in S
Then T(v) and T(w) belongs to Z
Consider av+bw , a,b are scalars
We need to see if av+bw are in S
For this consider T(av+bw) = aT(v)+bT(w) (because T is linear transformtaion)
Now T(v) and T(w) are in Z
And so aT(v) and bT(w) are in Z
This => aT(v)+bT(w) are in Z
Hence we get that T(av+bw) is in Z
So we get av+bw is in S
Hence S is subspace of Rn.
Just a general question on some vocab words I cannot piece together .pdf
Just a general question on some vocab words I cannot piece together myself. Can you provide
me with some relevance to some terms used in my biology class literature that I do not
understand.
Oxidized/oxidation. I understand what these are, but I do not understand their relevance... When
you oxidize a compound like ammonia NH3 to NO3...what is the point? does it make energy... is
it like an enzyme.. like what does that mean?
electron donor/acceptor....why does this matter which is which....? what is the purpose of this
besides donating and excepting electrons....?
NAD, NADPH... okay I know what these are composed of but what are they exactly and what is
their purpose...?
some of my definitions say things like \"use CO2\" as main carbon source\" ???? okay.. but what
does carbon have to do with anything... how is that energy or food or whatever?
Solution
1. ANS:
Oxidation: Oxidation is defined as the loss of electrons, hydrogen’s or gain of oxygen atom. If
the oxidation taken placed in a cell, then it is called as oxidized. Both the Oxidized/oxidation are
same, but the using situation is different.
Reduction: Reduction is defined as gain of electrons, hydrogen’s or loss of oxygen atom.
Depends upon reaction the redox (oxidation or reduction) will change, sometimes it explains
about electrons or in another time it states about oxygen but the sole of the meaning will never
change.
Ex: Oxidation and Reduction in terms of oxygen transfer
Oxidation and Reduction in terms of electrons transfer
Oxidation and Reduction in terms of hydrogen transfer
2. ANS: Nitrogen oxidation:
It is a process of biological oxidation, in which the anaerobic ammonia (NH4+) and nitrite
(NO2-) converts into nitrogen gas (N2).
Ammonia is converted into nitrite (NO2-) due to the effect of nitrosomonas bacteria. Nitrite
(NO2-) then is converted into nitrate (NO3-) by the influence of nitrobacter bacteria.
In this reaction ATP is not released, so the energy is not produced.
3. ANS: Electron donor/acceptor:
Electron acceptor: It accept ions that act as oxidizing agents in chemical reactions.
Electron donor: It donates electrons, these are reducing agents.
The main purpose of this donating and accepting electrons to stabilizes chemical reactions. In
biological point of view a series of electron transfer can produce ATP molecules. ATP act like a
universal currency. ATP is produced by photon gradient, which is created by ETC in
mitochondria.
4. ANS: NAD and NADPH:
NAD is a primary oxidizing agents of cell and the NADPH is a primary reducing agents of cell.
NAD in oxidation reactions, it accepts two electrons and one photon from the substrate.
In reduced form NADPH acts as the primary reducing agent of cell, it donates two electrons and
one photon to the medium.
5. ANS:
In the photosynthesis reactions plants make glucose by incorporating CO2 and water. The
glucose molecule later oxidized in organisms and produced energy in the form of ATP. Foods
are mostly composed with carbohydrates.
In order to be useful to managers, management accounting reports shou.pdf
In order to be useful to managers, management accounting reports should possess all of the
following cs EXCEPT a, be prepared to report information for any unit of the business to support
decision making. b. be ed at any time management needs information d. provide objective
measures of past operations and subjective estimates about future decisions
Solution
The correct choice is b
Explanation : - Management accounting is not required to adhere to generally accepted
accounting principle because it provides managerial information for internal purposes i.e. it is
used by the members within the organization and not used by members outside the organizations
such as regulators, creditors etc..
Im also confused on question 9 and 10 On which point(s) would Charle.pdf
Im also confused on question 9 and 10 On which point(s) would Charles Darwin and Georges
Button have likely agreed Select one: the idea that all of life shared a common ancestor the
idea that species change over time both are correct neither is correct On which point(s) would
Charles Darwin and Georges Button have likely disagreed: Select one: the idea that the earth
was formed according to the laws of physics the idea that all life shares a common ancestor the
idea that species change over time they would have agreed on all of these points Which of the
following ideas was NOT accepted by Jean-Batiste Lamarck? Select one: bacteria represent
some of the youngest species on earth life arises spontaneously organisms go extinct none of
the above
Solution
Answer c : Both are correct
Answer d: Earth is quite old both agree, common ancestor in species tree
Answer c:Lamarck did not believe that species could become extinct.
In about six billion years, our suns luminosity (a.k.a. power outpu.pdf
In about six billion years, our sun\'s luminosity (a.k.a. power output) will be 2000 times more
than it is now. At this stage, our sun will be considered a \"red giant\" star. The surface
temperature of a red giant is only about two thirds of our sun\'s current temperature. Between
what two famous celestials objects will be the location of our sun\'s surface in six billion years?
Solution
Luminosity,L = k*R^2 * T^4
so, L2/L1 = [R2/R1]^2 * [T2/T1]^4
2000 = [R2/R1]^2 * 0.1975
R2 = 100.62 * R1 = 100.62 * 6.96*10^8 = 7.003 * 10 ^10m
Radius of orbit of mercury = 4.6*10^10 m - 7 * 10 ^ 7m
Radius of orbit of venus = 1.08*10^11 m
Hence sun\'s surface will be between venus amd mercury.
I have to write a java program . So basically I have to take my in.pdf
I have to write a java program . So basically I have to take my input in only cms. Inches is also
accepted but the inches should be converted into the cms before the program take in the input .
So when the program takes the input , the input must be in cm .
I have to write a java program . So basically I have to take my input in only cms. Inches is also
accepted but the inches should be converted into the cms before the program take in the input .
So when the program takes the input , the input must be in cm .
Solution
import java.io.*;
import java.util.*;
class InchesToCms{
public static void main (String[] args)throws Exception{
//1 Inches = 2.54 Centimeters;
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));//to get
input creating bf is an object.
System.out.print(\"Enter the inches: \");//asking the input is inches
float inches = Float.parseFloat(bf.readLine());//geting input first in string,so convert into
float using parseFloat function.
float cms = inches * 2.54f;//multiplying given input inches and 2.54.
System.out.println(inches + \" inches is \" + cms + \" centimeter(s).\");//display the inches
and centimeters.
System.out.print(\"Enter the Centimeters: \");//asking the input is Centimeters.
float CMS = Float.parseFloat(bf.readLine());
System.out.println(\"You entered \" +CMS+ \" centimeter(s).\");//display out put in
Centimeters only.
}
}
/* Output:
1) Enter the inches: 11.0
11.0 inches is 27.939999 centimeter(s).
Enter the Centimeters: 2.0
You entered 2.0 centimeter(s).
Explanatin:
This program 1) It takes input in inches and convert into centimeters
2) and Next It takes input in centimters and display the out in the same centimeters.
*/.
Fill in the blanks.Intensity of orangebrownred color of reduced .pdf
Fill in the blanks.
Intensity of orange/brown/red color of reduced DNS is ____________ related to the amount of
_______________ in solution.
Solution
Intensity of orange/brown/red color of reduced DNS is proportionally related to the amount of
maltose in solution.
DNS is used to study the enzymatic activity of amylase, which breaks starch into maltose.
Higher amylase activity produces more amount of maltose, which reduces 3, 5-dinitrosalicylic
acid (DNS), which is yellow in color, to 3-amino, 5-nitrosalicylic acid (commonly termed
\"reduced DNS\"), which may be orange, brown or red depending on the concentration of
maltose..
Explain how cellular respiration (O2 consumption and CO2 production).pdf
Explain how cellular respiration (O2 consumption and CO2 production) is linked to ATP
production. Is the coupling of these two processes perfect? Why or why not?
Solution
The primary function of cellular respiration is to generate ATP, which traps some of the
chemical energy of food molecules in its high- energy bonds (adenosine triphosphate). The
process of generation of ATP is via glycolysis and Krebs’s cycle finally through oxidative
phosphorylation.
The overall balanced reaction of cellular respiration is:
CHO + 6O 6CO + 6HO + ATP
Glucose + oxygen carbon dioxide + water + energy
In this reaction, glucose oxidized and oxygen reduced.
Cellular respiration (O2 consumption and CO2 production) is linked to ATP production because
it is different from cellular respiration as it involves metabolic activities using molecular oxygen
such as Citric acid cycle and electron transport to synthesize energy in the form of ATP to
perform respiration. CO2 is the output of the cellular respiration from mitochondria and expelled
out of the body by carrying through hemoglobin in the form of carboxy- hemoglobin to the
alveoli of lungs.
The coupling of these two processes such as O2 consumption and CO2 production inked to ATP
production is perfect. Mitochondrial chemiosmosis –cellular respiration (through Fernadez
Moran inner mitochondrial oxysomes particles) during cellular respiration obtain energy from
chemical enzymatic breakdown of the organic food molecules (glucose, pyruvate, acetylcoA) to
produce ATP in the presence of O2 consumption. During electron transport, the final electron
acceptor is oxygen for oxidative phosphorylation via energy producing molecules such as NADH
(3 ATP), FADH2 (2ATP) and ATP. Therefore, it is going to \"oxygen\" that accepting electrons
at the end of the electron transport chain to generate ATP via mitochondrial respiratory complex-
I to V. Every 3 protons used to produce one ATP molecule. Inner membrane possess small
protein channels known as porins in mitochondria & these channels promote the movement of
any small molecules such as ATP through them
However, the coupling is perfect between the O2 consumption and CO2 production for ATP
production but whole-organism oxygen consumption is mainly related to rate of substrate
oxidation & it cannot be accurate measurement for ATP production due to variable P/O ratio
occurs..
Entrepreneurship isA. one of the basic rights of the private enter.pdf
Entrepreneurship is
A. one of the basic rights of the private enterprise system.
B. the willingness to take risks.
C. required for capital expansion.
D. not found in the private enterprise system.
Solution
B. the willingness to take risks.
entrepenuership is the willingness to take risk. even if u have money and idea of business, but u
don\'t have willpower to invest in that idea and take risk than u can\'t be entrepreneure. so risk
taker can only become entrepenure than only he can set up a business by using his own or
borrowed capital..
Discrete Random Variables we are working with commonly used discre.pdf
Discrete Random Variables
we are working with commonly used discrete and continuous distribution In a hospital there are
16 patients, 4 of whom have AIDS. A doctor is assigned to 6 of these patients at random. What is
the probability that he gets 2 of the AIDS patients?
Solution
There are 16C6 ways to choose any 6 patients.
There are 4C2 ways to get 2 with AIDS, and 12C4 to choose 4 without.
Thus,
P = [(12C4)(4C2)]/(16C6) = 0.370879121 [ANSWER].
Determine the open intervals over which the function is increasing an.pdf
Determine the open intervals over which the function is increasing and the intervals over which
the function is decreasing. Label your answers clearly and use interval notation. Increasing:
Decreasing:
Solution
function is increasing in (-4, -2) and then in (1,3)
function is decreasing in (-2, 1) and then in (3,5).
Discuss and define Isotropy. Anisotropy and Orthotropy with regards t.pdf
Discuss and define Isotropy. Anisotropy and Orthotropy with regards to engineering materials.
Regarding Finite Element Analysis: List the three main types of elements and explain how to
choose elements? Common sources of errors in FEA? After conducting a photoelastic
investigation of a sample, you remove the load and notice that a fringe pattern still exists. What
could this mean?
Solution
(i)
Isotropy is the property of the material in which the physical properties of a substance are
independent of the direction of measurement. Glass and metals are examples of isotropic
materials.
Anisotropy is the property of the material in which the physical properties of a substance varies
with the change in direction. Wood and Composites are examples of anisotropic materials.
Orthotropy is the property of the material in which the physical properties of a substance change
when measured from different directions. The properties differ along three mutually orthogonal
twofold axes of rotational symmetry. Wood is an example of orthotropic materials.
(ii)
(a )The three main elements in FEA are -.
Consider the circuits in figure 2.32. If we assume the transistors ar.pdf
Consider the circuits in figure 2.32. If we assume the transistors are well described by the
Schockley model and their respective S and B terminals are tied together, the two circuits are
equivalent (i.e. I_dsi=I_ds2). If the body effect is considered, will I_Ds2 be equal to, greater
than, or less than I_dsi? Consider only body effect: assume velocity saturation and mobility
degradation effects are negligible. Explain.
Solution
As source(S) and Substrate terminals are tied together, hence VSB= potential difference between
S and B=0.
Hence is no body effect.
Hence Id1= Id2..
Can you please debug this Thank you in advance! This program is sup.pdf
Can you please debug this? Thank you in advance! This program is supposed to use stacks and
find the paths of an airplane. It should tell you whether or not the plane goes from the requested
city to the requested destination. requestFile.txt contains the requests, flightFile.txt contains the
flights and cityFile.txt contains the city\'s that the airline serves.
My errors: my output is simply \"Hpair does not serve Los Angeles.\" Thats it and it stops.
It should look something like (these are not real values of the file just an example):
\"Request is to fly from cityA to CityB.
Hpair flight from cityA to CityB.
Request is to fly from cityC to cityD.
Sorry, HPAir does not serve City D
Request is to fly from cityE to CityF
Sorry, HPAir does not fly from cityE to CityF\"
**It said to use ispath.cpp but I got confused on how to implement it completely.\"
isPath.cpp
//***I did not include ispath.cpp in my program. Parts of it are in the map.h portion.
main.cpp
#include
#include
#include \"Map.h\"
using namespace std;
int main()
{
Map aMap;
aMap.readFlights();
aMap.verifyRequestedFile();
}
StackInterface.h
#ifndef StackInterface_h
#define StackInterface_h
template
class StackInterface
{
/** Checks if the stack is empty.
@return True if the stack is empty or false if the stack is not.*/
virtual bool isEmpty() const = 0;
/** Adds a new item to the stack
@post newEntry is at the top of the stack
@param newEntry is the object to be added to the stack
@return True if successfully added, otherwise false. */
virtual bool push(const ItemType& newEntry) = 0;
/** Removes item from stack
@post The top item in the stack is removed
@return True if the item was removed, false if it was not. */
virtual bool pop() = 0;
/** Returns the top item in the stack
@pre The stack is not empty
@post The top item in the stack is returned
@return The top of the stack */
virtual ItemType peek() const = 0;
};
#endif /* StackInterface_h */
Node.h
#ifndef node_h
#define node_h
template
class Node {
public:
Node();
Node (const ItemType &anItem);
Node(const ItemType &anItem, Node * nextNodePtr);
void setItem(const ItemType &anItem);
void setNext(Node * nextNodePtr);
ItemType getItem() const;
Node* getNext() const;
private:
ItemType item;
Node* next;
};
template
Node::Node():next(nullptr){}
template
Node::Node(const ItemType &anItem):item(anItem), next(nullptr){}
template
Node::Node(const ItemType &anItem, Node * nextNodePtr):item(anItem), next(nextNodePtr)
{
}
template
void Node::setItem(const ItemType &anItem){
item = anItem;
}
template
void Node::setNext(Node * nextNodePtr) {
next = nextNodePtr;
}
template
ItemType Node::getItem() const {
return item;
}
template
Node* Node::getNext()const{
return next;
}
#endif /* node_h */
LinkedStack.h
#include
#ifndef LinkedStack_h
#define LinkedStack_h
#include \"Node.h\"
#include \"StackInterface.h\"
template
class LinkedStack:public StackInterface{
public:
LinkedStack();
LinkedStack(const LinkedStack &aStack); //space matter?***********
virt.
Circulating blood cells that differentiate into macrophages in the ti.pdf
Circulating blood cells that differentiate into macrophages in the tissue M cells Monocytes
Common myeloid progenitors Eosinophils
Solution
Answer: b. Monocytes
Monocytes are the circulating blood cells which after production from the bone marrow,
circulate in the bloodstream for about 8 h, during which they enlarge; and then migrate into the
tissues and differentiate into specific tissue macrophages. The differentiation of a monocyte into
a tissue macrophage involves a number of changes such as :.
Between January 2005 and May 2007, the CDC had identified 138 cases .pdf
Between January 2005 and May 2007, the CDC had identified 138 cases of Acanthomelobe
Keratitis infections (a rare but potentially blinding infection of the cornea of the eye) in 35 states
and Puerto Rico. Although only 46 interviews of infected individuals have been acquired, the
results indicate a linkage to the use of Complete MoisturePlus Multipurpose contact lens
solution. How might public health officials have handled this outbreak?
Official letters to all registered healthcare practitioners announcing methods of treatment and
follow-up
Provide the option to health care providers to report the disease
Public health announcement
All of the abovea.
Official letters to all registered healthcare practitioners announcing methods of treatment and
follow-upb.
Provide the option to health care providers to report the diseasec.
Public health announcementd.
All of the above
Solution
D.All of above
Public health officials work in outbreak very precociously. Too often an initial suspicious
outbreak is just lots of different illnesses that are being confused for the same thing or an error
from the lab or other monitoring systems. So it’s important to study sign/symptoms and study the
number of cases. The case definition will create a standardized method for identifying past,
present and future cases. It should include clinical information, characteristics of those with the
disease, and geographic and timing information about the cluster of cases. A good case definition
starts out very inclusive so as not to miss any potential cases that do not have the typically
identified symptoms. As the investigation continues for an unknown disease, this definition may
become more restricted to ensure we focus on persons who truly have the illness of concern.
After confirmation, they take appropriate action.As in above cases in only 46 interviews have
been developed infection still part of their procedure they have to follow all three steps which
mention above..
A satellite is 112 miles above the earth, and the angle A to the hor.pdf
A satellite is 112 miles above the earth, and the angle A to the horizon is 77degree . The radius
of the earth is ______ miles
Solution
Here we will apply sine ratio
sin(77) = radius/(112+radius).
0.9743=r/(r+112)
0.9743(r+112) = r
0.9743r+109.129 =r
0.0257r =109.129
r= 109.129/0.0257
r= 4246.26 miles.
A car antenna is 1.5 m long. The car is moving at (40 mph) under a D.pdf
A car antenna is 1.5 m long. The car is moving at (40 mph) under a DC high-tension wire with
(100 amp) (the distance between the wire and the top of the antenna is 10m) . What emf is
induced in the antenna and what is its polarity ?
Solution
emf is induced in the antenna = VBL
40mph = 17.881 m/s , B = u0I/(2piR)
=> emf = 17.881*4*3.14*10^-7100*1.5)/(2*3.14*10) = 0.000053643 V = 53.64 uV
polarity will be in (VxB) direction.
832 AM ut.blackboard.com HW #1-The Federal Reserve and Monetary Po.pdf
8:32 AM ut.blackboard.com ? HW #1-The Federal Reserve and Monetary Policy This
homework assignment is worth up to 20 points toward your final grade. 1. What are the five (5)
risks common to all financial institutions? Briefly explain each. (5 points) 2. Who is the current
Chairman of the Federal Reserve? (2 points) 3. Explain the three uses of money. How do
cryptocurrencies, such as Bitcoin and Ripple, fit into this framework? (3 points) 4. What are the
three main goals of the Federal Reserve and how does the Fed use Monetary Policy to achieve
these goals? (3 points)
Solution
Credit risk According to the Bank for International Settlements (BIS), credit risk is defined as the
potential that a bank borrower or counterparty will fail to meet its obligations in accordance with
agreed terms. Credit risk is most likely caused by loans, acceptances, interbank transactions,
trade financing, foreign exchange transactions, financial futures, swaps, bonds, equities, options,
and in the extension of commitments and guarantees, and the settlement of transactions. In
simple words, if person A borrows loan from a bank and is not able to repay the loan because of
inadequate income, loss in business, death, unwillingness or any other reasons, the bank faces
credit risk. Similarly, if you do not pay your credit card bill, the bank faces a credit risk.
Hence, to minimize the credit risk on the bank’s end, the rate of interest will be higher for
borrowers if they are associated with high credit risk. Factors like unsteady income, low credit
score, employment type, collateral assets and others determine the credit risk associated with a
borrower. As stated earlier, credit risk can be associated with interbank transactions, foreign
transactions and other types of transactions happening outside the bank. If the transaction at one
end is successful but unsuccessful at the other end, loss occurs. If the transaction at one end is
settled but there are delays in settlement at the other end, there might be lost investment
opportunities.
Look at it like person A sending US dollars to his family in India at the rate of 60 INR (Indian
Rupee) per dollar. The person B, who is the recipient however receives the payment late and
doesn’t get the exchange rate of 60 INR. Instead he receives the money at the exchange rate of
58 INR. This means they incurred a loss in the transaction. Similar situations occur during big
transactions in banks. If the bank is not able to settle a transaction at an expected time or during
an expected time duration, they may incur a credit risk. However, this kind of risk is called
“Settlement Risk” and it is closely associated with credit risk. It depends on the timing of the
exchange of value, payment/settlement finality and the role of intermediaries and clearing
houses.
While some credit risk is a result of macro forces affecting the economy or specific markets or
even specific individuals, there is another important risk that can be classified un.
you will use cellular automata to create a 2D predator–prey simulati.pdf
you will use cellular automata to create a 2D predator–prey simulation in your program. The
preys are ants and the predators are doodlebugs. These critters live in a 20 * 20 grid of cells.
Only one critter may occupy a cell at a time. The grid is enclosed and no critter may move off
the grid. Time is simulated in steps. Each critter performs some action every time step. The ants
behave according to the following model: Move: For every time step, the ants randomly try to
move up, down, left, or right. If the neighboring cell in the selected direction is occupied or
would move the ant off the grid, then the ant stays in the current cell. Breed: If an ant survives
for three time steps (not been eaten by doodlebugs), at the end of the time step (i.e., after
moving) the ant will breed. This is simulated by creating a new ant in an adjacent (up, down, left,
or right) cell that is empty randomly. If there is no empty cell available, no breeding occurs.
Once an offspring is produced, an ant cannot produce an offspring again until it has survived
three more time steps.
The doodlebugs behave according to the following model:
Move: For every time step, the doodlebug will firstly try to move to an adjacent cell containing
an ant and eat the ant (you can decide if there are several ants in the adjacent cells, how the
doodlebug will choose to move). If there are no ants in adjacent cells, the doodlebug moves
according to the same rules as the ant. Note that a doodlebug cannot eat other doodlebugs.
Breed: If a doodlebug survives for eight time steps, at the end of the time step, it will spawn off a
new doodlebug in the same manner as the ant.
Starve: If a doodlebug has not eaten an ant within three time steps, at the end of the third time
step it will starve and die. The doodlebug should then be removed from the grid of cells.
Initialize the world with 5 doodlebugs and 100 ants. You will randomly place them. You should
see a cyclical pattern between the population of predators and prey, although random
perturbations may lead to the elimination of one or both species. You will prompt the user to
enter the number of time steps to run.
Create a class named Critter that contains data and functions common to ants and doodlebugs.
This class should have a virtual function named move that is defined in the derived classes of
Ant and Doodlebug. Each class will be in its own source file.
For each time step, do the following in your program: after moves, when breeding, eating, and
starving are resolved, display the resulting grid. Draw the world using ASCII characters of “O”
for an ant, “X” for a doodlebug and “E” for an empty space (the characters should be arranged to
look like a grid). The doodlebugs will move before the ants in each time step. When you reach
the time steps entered by the user, ask them to enter another number and start to run the
simulation again or to exit the program. You must maintain the state of the current grid while
creating the next displa.
Why should you use properties within the methods of a class in C# ev.pdf
Why should you use properties within the methods of a class in C# even if syntactically you
don’t have to ?
Solution
In general, methods represent actions and properties represent data. Properties are meant to be
used like fields, meaning that properties should not be computationally complex or produce side
effects. When it does not violate the following guidelines, consider using a property, rather than
a method, because less experienced developers find properties easier to use.That way use
properties..
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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Just a general question on some vocab words I cannot piece together .pdf
Just a general question on some vocab words I cannot piece together myself. Can you provide
me with some relevance to some terms used in my biology class literature that I do not
understand.
Oxidized/oxidation. I understand what these are, but I do not understand their relevance... When
you oxidize a compound like ammonia NH3 to NO3...what is the point? does it make energy... is
it like an enzyme.. like what does that mean?
electron donor/acceptor....why does this matter which is which....? what is the purpose of this
besides donating and excepting electrons....?
NAD, NADPH... okay I know what these are composed of but what are they exactly and what is
their purpose...?
some of my definitions say things like \"use CO2\" as main carbon source\" ???? okay.. but what
does carbon have to do with anything... how is that energy or food or whatever?
Solution
1. ANS:
Oxidation: Oxidation is defined as the loss of electrons, hydrogen’s or gain of oxygen atom. If
the oxidation taken placed in a cell, then it is called as oxidized. Both the Oxidized/oxidation are
same, but the using situation is different.
Reduction: Reduction is defined as gain of electrons, hydrogen’s or loss of oxygen atom.
Depends upon reaction the redox (oxidation or reduction) will change, sometimes it explains
about electrons or in another time it states about oxygen but the sole of the meaning will never
change.
Ex: Oxidation and Reduction in terms of oxygen transfer
Oxidation and Reduction in terms of electrons transfer
Oxidation and Reduction in terms of hydrogen transfer
2. ANS: Nitrogen oxidation:
It is a process of biological oxidation, in which the anaerobic ammonia (NH4+) and nitrite
(NO2-) converts into nitrogen gas (N2).
Ammonia is converted into nitrite (NO2-) due to the effect of nitrosomonas bacteria. Nitrite
(NO2-) then is converted into nitrate (NO3-) by the influence of nitrobacter bacteria.
In this reaction ATP is not released, so the energy is not produced.
3. ANS: Electron donor/acceptor:
Electron acceptor: It accept ions that act as oxidizing agents in chemical reactions.
Electron donor: It donates electrons, these are reducing agents.
The main purpose of this donating and accepting electrons to stabilizes chemical reactions. In
biological point of view a series of electron transfer can produce ATP molecules. ATP act like a
universal currency. ATP is produced by photon gradient, which is created by ETC in
mitochondria.
4. ANS: NAD and NADPH:
NAD is a primary oxidizing agents of cell and the NADPH is a primary reducing agents of cell.
NAD in oxidation reactions, it accepts two electrons and one photon from the substrate.
In reduced form NADPH acts as the primary reducing agent of cell, it donates two electrons and
one photon to the medium.
5. ANS:
In the photosynthesis reactions plants make glucose by incorporating CO2 and water. The
glucose molecule later oxidized in organisms and produced energy in the form of ATP. Foods
are mostly composed with carbohydrates.
In order to be useful to managers, management accounting reports shou.pdf
In order to be useful to managers, management accounting reports should possess all of the
following cs EXCEPT a, be prepared to report information for any unit of the business to support
decision making. b. be ed at any time management needs information d. provide objective
measures of past operations and subjective estimates about future decisions
Solution
The correct choice is b
Explanation : - Management accounting is not required to adhere to generally accepted
accounting principle because it provides managerial information for internal purposes i.e. it is
used by the members within the organization and not used by members outside the organizations
such as regulators, creditors etc..
Im also confused on question 9 and 10 On which point(s) would Charle.pdf
Im also confused on question 9 and 10 On which point(s) would Charles Darwin and Georges
Button have likely agreed Select one: the idea that all of life shared a common ancestor the
idea that species change over time both are correct neither is correct On which point(s) would
Charles Darwin and Georges Button have likely disagreed: Select one: the idea that the earth
was formed according to the laws of physics the idea that all life shares a common ancestor the
idea that species change over time they would have agreed on all of these points Which of the
following ideas was NOT accepted by Jean-Batiste Lamarck? Select one: bacteria represent
some of the youngest species on earth life arises spontaneously organisms go extinct none of
the above
Solution
Answer c : Both are correct
Answer d: Earth is quite old both agree, common ancestor in species tree
Answer c:Lamarck did not believe that species could become extinct.
In about six billion years, our suns luminosity (a.k.a. power outpu.pdf
In about six billion years, our sun\'s luminosity (a.k.a. power output) will be 2000 times more
than it is now. At this stage, our sun will be considered a \"red giant\" star. The surface
temperature of a red giant is only about two thirds of our sun\'s current temperature. Between
what two famous celestials objects will be the location of our sun\'s surface in six billion years?
Solution
Luminosity,L = k*R^2 * T^4
so, L2/L1 = [R2/R1]^2 * [T2/T1]^4
2000 = [R2/R1]^2 * 0.1975
R2 = 100.62 * R1 = 100.62 * 6.96*10^8 = 7.003 * 10 ^10m
Radius of orbit of mercury = 4.6*10^10 m - 7 * 10 ^ 7m
Radius of orbit of venus = 1.08*10^11 m
Hence sun\'s surface will be between venus amd mercury.
I have to write a java program . So basically I have to take my in.pdf
I have to write a java program . So basically I have to take my input in only cms. Inches is also
accepted but the inches should be converted into the cms before the program take in the input .
So when the program takes the input , the input must be in cm .
I have to write a java program . So basically I have to take my input in only cms. Inches is also
accepted but the inches should be converted into the cms before the program take in the input .
So when the program takes the input , the input must be in cm .
Solution
import java.io.*;
import java.util.*;
class InchesToCms{
public static void main (String[] args)throws Exception{
//1 Inches = 2.54 Centimeters;
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));//to get
input creating bf is an object.
System.out.print(\"Enter the inches: \");//asking the input is inches
float inches = Float.parseFloat(bf.readLine());//geting input first in string,so convert into
float using parseFloat function.
float cms = inches * 2.54f;//multiplying given input inches and 2.54.
System.out.println(inches + \" inches is \" + cms + \" centimeter(s).\");//display the inches
and centimeters.
System.out.print(\"Enter the Centimeters: \");//asking the input is Centimeters.
float CMS = Float.parseFloat(bf.readLine());
System.out.println(\"You entered \" +CMS+ \" centimeter(s).\");//display out put in
Centimeters only.
}
}
/* Output:
1) Enter the inches: 11.0
11.0 inches is 27.939999 centimeter(s).
Enter the Centimeters: 2.0
You entered 2.0 centimeter(s).
Explanatin:
This program 1) It takes input in inches and convert into centimeters
2) and Next It takes input in centimters and display the out in the same centimeters.
*/.
Fill in the blanks.Intensity of orangebrownred color of reduced .pdf
Fill in the blanks.
Intensity of orange/brown/red color of reduced DNS is ____________ related to the amount of
_______________ in solution.
Solution
Intensity of orange/brown/red color of reduced DNS is proportionally related to the amount of
maltose in solution.
DNS is used to study the enzymatic activity of amylase, which breaks starch into maltose.
Higher amylase activity produces more amount of maltose, which reduces 3, 5-dinitrosalicylic
acid (DNS), which is yellow in color, to 3-amino, 5-nitrosalicylic acid (commonly termed
\"reduced DNS\"), which may be orange, brown or red depending on the concentration of
maltose..
Explain how cellular respiration (O2 consumption and CO2 production).pdf
Explain how cellular respiration (O2 consumption and CO2 production) is linked to ATP
production. Is the coupling of these two processes perfect? Why or why not?
Solution
The primary function of cellular respiration is to generate ATP, which traps some of the
chemical energy of food molecules in its high- energy bonds (adenosine triphosphate). The
process of generation of ATP is via glycolysis and Krebs’s cycle finally through oxidative
phosphorylation.
The overall balanced reaction of cellular respiration is:
CHO + 6O 6CO + 6HO + ATP
Glucose + oxygen carbon dioxide + water + energy
In this reaction, glucose oxidized and oxygen reduced.
Cellular respiration (O2 consumption and CO2 production) is linked to ATP production because
it is different from cellular respiration as it involves metabolic activities using molecular oxygen
such as Citric acid cycle and electron transport to synthesize energy in the form of ATP to
perform respiration. CO2 is the output of the cellular respiration from mitochondria and expelled
out of the body by carrying through hemoglobin in the form of carboxy- hemoglobin to the
alveoli of lungs.
The coupling of these two processes such as O2 consumption and CO2 production inked to ATP
production is perfect. Mitochondrial chemiosmosis –cellular respiration (through Fernadez
Moran inner mitochondrial oxysomes particles) during cellular respiration obtain energy from
chemical enzymatic breakdown of the organic food molecules (glucose, pyruvate, acetylcoA) to
produce ATP in the presence of O2 consumption. During electron transport, the final electron
acceptor is oxygen for oxidative phosphorylation via energy producing molecules such as NADH
(3 ATP), FADH2 (2ATP) and ATP. Therefore, it is going to \"oxygen\" that accepting electrons
at the end of the electron transport chain to generate ATP via mitochondrial respiratory complex-
I to V. Every 3 protons used to produce one ATP molecule. Inner membrane possess small
protein channels known as porins in mitochondria & these channels promote the movement of
any small molecules such as ATP through them
However, the coupling is perfect between the O2 consumption and CO2 production for ATP
production but whole-organism oxygen consumption is mainly related to rate of substrate
oxidation & it cannot be accurate measurement for ATP production due to variable P/O ratio
occurs..
Entrepreneurship isA. one of the basic rights of the private enter.pdf
Entrepreneurship is
A. one of the basic rights of the private enterprise system.
B. the willingness to take risks.
C. required for capital expansion.
D. not found in the private enterprise system.
Solution
B. the willingness to take risks.
entrepenuership is the willingness to take risk. even if u have money and idea of business, but u
don\'t have willpower to invest in that idea and take risk than u can\'t be entrepreneure. so risk
taker can only become entrepenure than only he can set up a business by using his own or
borrowed capital..
Discrete Random Variables we are working with commonly used discre.pdf
Discrete Random Variables
we are working with commonly used discrete and continuous distribution In a hospital there are
16 patients, 4 of whom have AIDS. A doctor is assigned to 6 of these patients at random. What is
the probability that he gets 2 of the AIDS patients?
Solution
There are 16C6 ways to choose any 6 patients.
There are 4C2 ways to get 2 with AIDS, and 12C4 to choose 4 without.
Thus,
P = [(12C4)(4C2)]/(16C6) = 0.370879121 [ANSWER].
Determine the open intervals over which the function is increasing an.pdf
Determine the open intervals over which the function is increasing and the intervals over which
the function is decreasing. Label your answers clearly and use interval notation. Increasing:
Decreasing:
Solution
function is increasing in (-4, -2) and then in (1,3)
function is decreasing in (-2, 1) and then in (3,5).
Discuss and define Isotropy. Anisotropy and Orthotropy with regards t.pdf
Discuss and define Isotropy. Anisotropy and Orthotropy with regards to engineering materials.
Regarding Finite Element Analysis: List the three main types of elements and explain how to
choose elements? Common sources of errors in FEA? After conducting a photoelastic
investigation of a sample, you remove the load and notice that a fringe pattern still exists. What
could this mean?
Solution
(i)
Isotropy is the property of the material in which the physical properties of a substance are
independent of the direction of measurement. Glass and metals are examples of isotropic
materials.
Anisotropy is the property of the material in which the physical properties of a substance varies
with the change in direction. Wood and Composites are examples of anisotropic materials.
Orthotropy is the property of the material in which the physical properties of a substance change
when measured from different directions. The properties differ along three mutually orthogonal
twofold axes of rotational symmetry. Wood is an example of orthotropic materials.
(ii)
(a )The three main elements in FEA are -.
Consider the circuits in figure 2.32. If we assume the transistors ar.pdf
Consider the circuits in figure 2.32. If we assume the transistors are well described by the
Schockley model and their respective S and B terminals are tied together, the two circuits are
equivalent (i.e. I_dsi=I_ds2). If the body effect is considered, will I_Ds2 be equal to, greater
than, or less than I_dsi? Consider only body effect: assume velocity saturation and mobility
degradation effects are negligible. Explain.
Solution
As source(S) and Substrate terminals are tied together, hence VSB= potential difference between
S and B=0.
Hence is no body effect.
Hence Id1= Id2..
Can you please debug this Thank you in advance! This program is sup.pdf
Can you please debug this? Thank you in advance! This program is supposed to use stacks and
find the paths of an airplane. It should tell you whether or not the plane goes from the requested
city to the requested destination. requestFile.txt contains the requests, flightFile.txt contains the
flights and cityFile.txt contains the city\'s that the airline serves.
My errors: my output is simply \"Hpair does not serve Los Angeles.\" Thats it and it stops.
It should look something like (these are not real values of the file just an example):
\"Request is to fly from cityA to CityB.
Hpair flight from cityA to CityB.
Request is to fly from cityC to cityD.
Sorry, HPAir does not serve City D
Request is to fly from cityE to CityF
Sorry, HPAir does not fly from cityE to CityF\"
**It said to use ispath.cpp but I got confused on how to implement it completely.\"
isPath.cpp
//***I did not include ispath.cpp in my program. Parts of it are in the map.h portion.
main.cpp
#include
#include
#include \"Map.h\"
using namespace std;
int main()
{
Map aMap;
aMap.readFlights();
aMap.verifyRequestedFile();
}
StackInterface.h
#ifndef StackInterface_h
#define StackInterface_h
template
class StackInterface
{
/** Checks if the stack is empty.
@return True if the stack is empty or false if the stack is not.*/
virtual bool isEmpty() const = 0;
/** Adds a new item to the stack
@post newEntry is at the top of the stack
@param newEntry is the object to be added to the stack
@return True if successfully added, otherwise false. */
virtual bool push(const ItemType& newEntry) = 0;
/** Removes item from stack
@post The top item in the stack is removed
@return True if the item was removed, false if it was not. */
virtual bool pop() = 0;
/** Returns the top item in the stack
@pre The stack is not empty
@post The top item in the stack is returned
@return The top of the stack */
virtual ItemType peek() const = 0;
};
#endif /* StackInterface_h */
Node.h
#ifndef node_h
#define node_h
template
class Node {
public:
Node();
Node (const ItemType &anItem);
Node(const ItemType &anItem, Node * nextNodePtr);
void setItem(const ItemType &anItem);
void setNext(Node * nextNodePtr);
ItemType getItem() const;
Node* getNext() const;
private:
ItemType item;
Node* next;
};
template
Node::Node():next(nullptr){}
template
Node::Node(const ItemType &anItem):item(anItem), next(nullptr){}
template
Node::Node(const ItemType &anItem, Node * nextNodePtr):item(anItem), next(nextNodePtr)
{
}
template
void Node::setItem(const ItemType &anItem){
item = anItem;
}
template
void Node::setNext(Node * nextNodePtr) {
next = nextNodePtr;
}
template
ItemType Node::getItem() const {
return item;
}
template
Node* Node::getNext()const{
return next;
}
#endif /* node_h */
LinkedStack.h
#include
#ifndef LinkedStack_h
#define LinkedStack_h
#include \"Node.h\"
#include \"StackInterface.h\"
template
class LinkedStack:public StackInterface{
public:
LinkedStack();
LinkedStack(const LinkedStack &aStack); //space matter?***********
virt.
Circulating blood cells that differentiate into macrophages in the ti.pdf
Circulating blood cells that differentiate into macrophages in the tissue M cells Monocytes
Common myeloid progenitors Eosinophils
Solution
Answer: b. Monocytes
Monocytes are the circulating blood cells which after production from the bone marrow,
circulate in the bloodstream for about 8 h, during which they enlarge; and then migrate into the
tissues and differentiate into specific tissue macrophages. The differentiation of a monocyte into
a tissue macrophage involves a number of changes such as :.
Between January 2005 and May 2007, the CDC had identified 138 cases .pdf
Between January 2005 and May 2007, the CDC had identified 138 cases of Acanthomelobe
Keratitis infections (a rare but potentially blinding infection of the cornea of the eye) in 35 states
and Puerto Rico. Although only 46 interviews of infected individuals have been acquired, the
results indicate a linkage to the use of Complete MoisturePlus Multipurpose contact lens
solution. How might public health officials have handled this outbreak?
Official letters to all registered healthcare practitioners announcing methods of treatment and
follow-up
Provide the option to health care providers to report the disease
Public health announcement
All of the abovea.
Official letters to all registered healthcare practitioners announcing methods of treatment and
follow-upb.
Provide the option to health care providers to report the diseasec.
Public health announcementd.
All of the above
Solution
D.All of above
Public health officials work in outbreak very precociously. Too often an initial suspicious
outbreak is just lots of different illnesses that are being confused for the same thing or an error
from the lab or other monitoring systems. So it’s important to study sign/symptoms and study the
number of cases. The case definition will create a standardized method for identifying past,
present and future cases. It should include clinical information, characteristics of those with the
disease, and geographic and timing information about the cluster of cases. A good case definition
starts out very inclusive so as not to miss any potential cases that do not have the typically
identified symptoms. As the investigation continues for an unknown disease, this definition may
become more restricted to ensure we focus on persons who truly have the illness of concern.
After confirmation, they take appropriate action.As in above cases in only 46 interviews have
been developed infection still part of their procedure they have to follow all three steps which
mention above..
A satellite is 112 miles above the earth, and the angle A to the hor.pdf
A satellite is 112 miles above the earth, and the angle A to the horizon is 77degree . The radius
of the earth is ______ miles
Solution
Here we will apply sine ratio
sin(77) = radius/(112+radius).
0.9743=r/(r+112)
0.9743(r+112) = r
0.9743r+109.129 =r
0.0257r =109.129
r= 109.129/0.0257
r= 4246.26 miles.
A car antenna is 1.5 m long. The car is moving at (40 mph) under a D.pdf
A car antenna is 1.5 m long. The car is moving at (40 mph) under a DC high-tension wire with
(100 amp) (the distance between the wire and the top of the antenna is 10m) . What emf is
induced in the antenna and what is its polarity ?
Solution
emf is induced in the antenna = VBL
40mph = 17.881 m/s , B = u0I/(2piR)
=> emf = 17.881*4*3.14*10^-7100*1.5)/(2*3.14*10) = 0.000053643 V = 53.64 uV
polarity will be in (VxB) direction.
832 AM ut.blackboard.com HW #1-The Federal Reserve and Monetary Po.pdf
8:32 AM ut.blackboard.com ? HW #1-The Federal Reserve and Monetary Policy This
homework assignment is worth up to 20 points toward your final grade. 1. What are the five (5)
risks common to all financial institutions? Briefly explain each. (5 points) 2. Who is the current
Chairman of the Federal Reserve? (2 points) 3. Explain the three uses of money. How do
cryptocurrencies, such as Bitcoin and Ripple, fit into this framework? (3 points) 4. What are the
three main goals of the Federal Reserve and how does the Fed use Monetary Policy to achieve
these goals? (3 points)
Solution
Credit risk According to the Bank for International Settlements (BIS), credit risk is defined as the
potential that a bank borrower or counterparty will fail to meet its obligations in accordance with
agreed terms. Credit risk is most likely caused by loans, acceptances, interbank transactions,
trade financing, foreign exchange transactions, financial futures, swaps, bonds, equities, options,
and in the extension of commitments and guarantees, and the settlement of transactions. In
simple words, if person A borrows loan from a bank and is not able to repay the loan because of
inadequate income, loss in business, death, unwillingness or any other reasons, the bank faces
credit risk. Similarly, if you do not pay your credit card bill, the bank faces a credit risk.
Hence, to minimize the credit risk on the bank’s end, the rate of interest will be higher for
borrowers if they are associated with high credit risk. Factors like unsteady income, low credit
score, employment type, collateral assets and others determine the credit risk associated with a
borrower. As stated earlier, credit risk can be associated with interbank transactions, foreign
transactions and other types of transactions happening outside the bank. If the transaction at one
end is successful but unsuccessful at the other end, loss occurs. If the transaction at one end is
settled but there are delays in settlement at the other end, there might be lost investment
opportunities.
Look at it like person A sending US dollars to his family in India at the rate of 60 INR (Indian
Rupee) per dollar. The person B, who is the recipient however receives the payment late and
doesn’t get the exchange rate of 60 INR. Instead he receives the money at the exchange rate of
58 INR. This means they incurred a loss in the transaction. Similar situations occur during big
transactions in banks. If the bank is not able to settle a transaction at an expected time or during
an expected time duration, they may incur a credit risk. However, this kind of risk is called
“Settlement Risk” and it is closely associated with credit risk. It depends on the timing of the
exchange of value, payment/settlement finality and the role of intermediaries and clearing
houses.
While some credit risk is a result of macro forces affecting the economy or specific markets or
even specific individuals, there is another important risk that can be classified un.
you will use cellular automata to create a 2D predator–prey simulati.pdf
you will use cellular automata to create a 2D predator–prey simulation in your program. The
preys are ants and the predators are doodlebugs. These critters live in a 20 * 20 grid of cells.
Only one critter may occupy a cell at a time. The grid is enclosed and no critter may move off
the grid. Time is simulated in steps. Each critter performs some action every time step. The ants
behave according to the following model: Move: For every time step, the ants randomly try to
move up, down, left, or right. If the neighboring cell in the selected direction is occupied or
would move the ant off the grid, then the ant stays in the current cell. Breed: If an ant survives
for three time steps (not been eaten by doodlebugs), at the end of the time step (i.e., after
moving) the ant will breed. This is simulated by creating a new ant in an adjacent (up, down, left,
or right) cell that is empty randomly. If there is no empty cell available, no breeding occurs.
Once an offspring is produced, an ant cannot produce an offspring again until it has survived
three more time steps.
The doodlebugs behave according to the following model:
Move: For every time step, the doodlebug will firstly try to move to an adjacent cell containing
an ant and eat the ant (you can decide if there are several ants in the adjacent cells, how the
doodlebug will choose to move). If there are no ants in adjacent cells, the doodlebug moves
according to the same rules as the ant. Note that a doodlebug cannot eat other doodlebugs.
Breed: If a doodlebug survives for eight time steps, at the end of the time step, it will spawn off a
new doodlebug in the same manner as the ant.
Starve: If a doodlebug has not eaten an ant within three time steps, at the end of the third time
step it will starve and die. The doodlebug should then be removed from the grid of cells.
Initialize the world with 5 doodlebugs and 100 ants. You will randomly place them. You should
see a cyclical pattern between the population of predators and prey, although random
perturbations may lead to the elimination of one or both species. You will prompt the user to
enter the number of time steps to run.
Create a class named Critter that contains data and functions common to ants and doodlebugs.
This class should have a virtual function named move that is defined in the derived classes of
Ant and Doodlebug. Each class will be in its own source file.
For each time step, do the following in your program: after moves, when breeding, eating, and
starving are resolved, display the resulting grid. Draw the world using ASCII characters of “O”
for an ant, “X” for a doodlebug and “E” for an empty space (the characters should be arranged to
look like a grid). The doodlebugs will move before the ants in each time step. When you reach
the time steps entered by the user, ask them to enter another number and start to run the
simulation again or to exit the program. You must maintain the state of the current grid while
creating the next displa.
Why should you use properties within the methods of a class in C# ev.pdf
Why should you use properties within the methods of a class in C# even if syntactically you
don’t have to ?
Solution
In general, methods represent actions and properties represent data. Properties are meant to be
used like fields, meaning that properties should not be computationally complex or produce side
effects. When it does not violate the following guidelines, consider using a property, rather than
a method, because less experienced developers find properties easier to use.That way use
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