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PriorityQueue.cs Jim Mischel using System; using Sy.pdf
// PriorityQueue.cs
//
// Jim Mischel
using System;
using System.Collections;
using System.Collections.Generic;
using System.Runtime.InteropServices;
namespace Mischel.Collections
{
[Serializable]
[ComVisible(false)]
public struct PriorityQueueItem
{
private TValue _value;
public TValue Value
{
get { return _value; }
}
private TPriority _priority;
public TPriority Priority
{
get { return _priority; }
}
internal PriorityQueueItem(TValue val, TPriority pri)
{
this._value = val;
this._priority = pri;
}
}
[Serializable]
[ComVisible(false)]
public class PriorityQueue : ICollection,
IEnumerable>
{
private PriorityQueueItem[] items;
private const Int32 DefaultCapacity = 16;
private Int32 capacity;
private Int32 numItems;
private Comparison compareFunc;
///
/// Initializes a new instance of the PriorityQueue class that is empty,
/// has the default initial capacity, and uses the default IComparer.
///
public PriorityQueue()
: this(DefaultCapacity, Comparer.Default)
{
}
public PriorityQueue(Int32 initialCapacity)
: this(initialCapacity, Comparer.Default)
{
}
public PriorityQueue(IComparer comparer)
: this(DefaultCapacity, comparer)
{
}
public PriorityQueue(int initialCapacity, IComparer comparer)
{
Init(initialCapacity, new Comparison(comparer.Compare));
}
public PriorityQueue(Comparison comparison)
: this(DefaultCapacity, comparison)
{
}
public PriorityQueue(int initialCapacity, Comparison comparison)
{
Init(initialCapacity, comparison);
}
private void Init(int initialCapacity, Comparison comparison)
{
numItems = 0;
compareFunc = comparison;
SetCapacity(initialCapacity);
}
public int Count
{
get { return numItems; }
}
public int Capacity
{
get { return items.Length; }
set { SetCapacity(value); }
}
private void SetCapacity(int newCapacity)
{
int newCap = newCapacity;
if (newCap < DefaultCapacity)
newCap = DefaultCapacity;
// throw exception if newCapacity < NumItems
if (newCap < numItems)
throw new ArgumentOutOfRangeException(\"newCapacity\", \"New capacity is less
than Count\");
this.capacity = newCap;
if (items == null)
{
items = new PriorityQueueItem[newCap];
return;
}
// Resize the array.
Array.Resize>(ref items, newCap);
}
public void Enqueue(PriorityQueueItem newItem)
{
if (numItems == capacity)
{
// need to increase capacity
// grow by 50 percent
SetCapacity((3 * Capacity) / 2);
}
int i = numItems;
++numItems;
while ((i > 0) && (compareFunc(items[(i - 1) / 2].Priority, newItem.Priority) < 0))
{
items[i] = items[(i - 1) / 2];
i = (i - 1) / 2;
}
items[i] = newItem;
//if (!VerifyQueue())
//{
// Console.WriteLine(\"ERROR: Queue out of order!\");
//}
}
public void Enqueue(TValue value, TPriority priority)
{
Enqueue(new PriorityQueueItem(value, priority));
}
private PriorityQueueItem RemoveAt(Int32 index)
{
PriorityQueueItem o = items[index];
--numItems;
// move the last item to fill the hole
PriorityQueueItem tmp = items[numItems];
// If you forget to clear this, you have a potential memory leak.
items[numItems] = default(PriorityQueueItem);
if (num.
the balanced equations are H3PO4(aq) + 3[NaOH](.pdf
the balanced equations are: H3PO4(aq) + 3[NaOH](aq) = Na3PO4(aq) + 3[H2O](l)
2[HI](aq) + LiO(aq) = LiI2(aq) + H2O(l) 2[HNO3](aq) + Ca(OH)2(s) = Ca(NO3)2(aq) +
2[H2O](l)
Solution
the balanced equations are: H3PO4(aq) + 3[NaOH](aq) = Na3PO4(aq) + 3[H2O](l)
2[HI](aq) + LiO(aq) = LiI2(aq) + H2O(l) 2[HNO3](aq) + Ca(OH)2(s) = Ca(NO3)2(aq) +
2[H2O](l).
Molecular wt. of NaOH = 40 No. of moles of NaOH =.pdf
Molecular wt. of NaOH = 40 No. of moles of NaOH = 0.3/40 = 0.0075 moles
Molarity = (No. of moles of NaOH)/(Volume of
Solution
) = 0.0075/0.17 = 0.0441 moles/lt.
images are not visible. post the question again .pdf
images are not visible. post the question again
Solution
images are not visible. post the question again.
VSEPR is easy if you follow the right steps. 1. .pdf
VSEPR is easy if you follow the right steps. 1. Find the number of valence (outer)
electrons around the central atom (remember the charge) 2. Work out how many of these are
involved in bonding (remember you may have single or double bonds) 3. Remember the lone
pairs. 4. Work out the shape from the number of atoms bonded to the central atom and the
number of lone pairs. So for SeO2, Se is in group 6, and has 6 valence electrons. Se=O is a
double bond, so there are 2 of Se\'s electrons in a double bond There are two Se=O bonds, so 4
electrons are involved in Se=O bonds. This leaves another 2 electrons which are a lone pair. So
overall you have three groups around the central atom (2 Se=O bonds and a lone pair) so it is
trigonal planar if you drew in the lone pair. Removing the lone pair gives you a bent molecule.
For PF4+: P has 5 valence electrons, but it is a positive ion, so you knock off an electron to get 4
P-F bonds are single bonds, and there are 4 of them, which accounts for all the electrons.
Therefore PF4+ is a tetrahedral shape. Finally, ICl3: I has 7 valence electrons. I-Cl bonds are
single bonds, so the I uses one electron in each bond. Therefore 3 of the 7 electrons are involved
in I-Cl bonds, the other 4 must be in 2 lone pairs. This gives 5 electron groups around the central
ion, so the shape is trigonal bipyramid if you draw in the lone pairs (trigonal plane with 2 groups
above and below the plane. You can determine that these two groups must be the lone pairs since
they repel more than the I-Cl bonds). Removing the lone pairs gives you a trigonal planar
molecule. The number of groups around the central ion determines the shape. 2 is linear 3 is
trigonal planar
Solution
VSEPR is easy if you follow the right steps. 1. Find the number of valence (outer)
electrons around the central atom (remember the charge) 2. Work out how many of these are
involved in bonding (remember you may have single or double bonds) 3. Remember the lone
pairs. 4. Work out the shape from the number of atoms bonded to the central atom and the
number of lone pairs. So for SeO2, Se is in group 6, and has 6 valence electrons. Se=O is a
double bond, so there are 2 of Se\'s electrons in a double bond There are two Se=O bonds, so 4
electrons are involved in Se=O bonds. This leaves another 2 electrons which are a lone pair. So
overall you have three groups around the central atom (2 Se=O bonds and a lone pair) so it is
trigonal planar if you drew in the lone pair. Removing the lone pair gives you a bent molecule.
For PF4+: P has 5 valence electrons, but it is a positive ion, so you knock off an electron to get 4
P-F bonds are single bonds, and there are 4 of them, which accounts for all the electrons.
Therefore PF4+ is a tetrahedral shape. Finally, ICl3: I has 7 valence electrons. I-Cl bonds are
single bonds, so the I uses one electron in each bond. Therefore 3 of the 7 electrons are involved
in I-Cl bonds, the other 4 must be in 2 lone pairs. This gives 5 electro.
This is quite a broad question. I will try to acc.pdf
This is quite a broad question. I will try to account all the important observations. 1.
Metal ions which do not have d-orbital electrons do not have color. (alkali, alkali-earth, Sc3+)
This is because of energy separation of s and p orbitals are not in the range of visible colors. 2.
Metal with fully-filled d orbitals do not have color. (Zn2+) This is because electrons in d-orbitals
cannot be excited to higher energy d-orbitals (all d-orbitals have the same energies, but when
influenced by other atom in complex, d-orbitals can split into different energy levels) 3. For ions
of the same metal, the ion with larger charge has a color of lower energy than the ion with
smaller charge. Example: Cr2+ (blue) and Cr3+ (green). This is because larger charge
contributes to larger split of the d-orbital. The more the orbitals split, the higher energy of light it
traps, allowing low-energy photon to enter our eyes. Note that you cannot be sure that the color
you see is a single color, not a combination of many colors. 4. For metals in the same column, a
metal ion which is below has a color of lower energy than an ion above. This can be explained
like no.3, larger ion causes more orbitals splitting. 5. Color of metal ions also depends on
\"ligands\" (complexing molecule/ion). e.g. Co2+ in water is pink [Co(H2O)6]2+, while Co2+ in
NH3 is blue [Co(NH3)4]2+. See \"spectrochemical series\" for more detail about this. 6. To find
\"exact\" color of metal ions, you need a supercomputer to do so. (I\'m not sure whether today
computer can calculate it). So, it\'s better to memorize the colors. Personally, I think you\'ll
understand about metal\'s color when you study transition metals, like Crystal Field Theory, and
Ligand Field Theory.
Solution
This is quite a broad question. I will try to account all the important observations. 1.
Metal ions which do not have d-orbital electrons do not have color. (alkali, alkali-earth, Sc3+)
This is because of energy separation of s and p orbitals are not in the range of visible colors. 2.
Metal with fully-filled d orbitals do not have color. (Zn2+) This is because electrons in d-orbitals
cannot be excited to higher energy d-orbitals (all d-orbitals have the same energies, but when
influenced by other atom in complex, d-orbitals can split into different energy levels) 3. For ions
of the same metal, the ion with larger charge has a color of lower energy than the ion with
smaller charge. Example: Cr2+ (blue) and Cr3+ (green). This is because larger charge
contributes to larger split of the d-orbital. The more the orbitals split, the higher energy of light it
traps, allowing low-energy photon to enter our eyes. Note that you cannot be sure that the color
you see is a single color, not a combination of many colors. 4. For metals in the same column, a
metal ion which is below has a color of lower energy than an ion above. This can be explained
like no.3, larger ion causes more orbitals splitting. 5. Color of metal ions also depends .
}public staatic boolean isValidElement(String token){public vo.pdf
}
public staatic boolean isValidElement(String token)
{
public void setParts(String IP)
{
IP=\"192.168.20.40\";
}
public void setParts(int[], data)
{
data[0]=toString(IP);
data[1]=toString(IP);
data[2]=toString(IP);
data[3]=toString(IP);
}
Solution
}
public staatic boolean isValidElement(String token)
{
public void setParts(String IP)
{
IP=\"192.168.20.40\";
}
public void setParts(int[], data)
{
data[0]=toString(IP);
data[1]=toString(IP);
data[2]=toString(IP);
data[3]=toString(IP);
}.
What is an example of big data either from your personal experience .pdf
What is an example of big data either from your personal experience or outside research?
Answer:
Big Data: Big data means really a big data, it is a collection of large datasets that cannot be
processed using traditional computing techniques. Big data is not merely a data, rather it has
become a complete subject, which involves various tools, technqiues and frameworks.
Big data involves the data produced by different devices and applications. Given below are some
of the fields that come under the Big Data.
Black Box Data : It is a component of helicopter, airplanes, and jets, etc. It captures voices of the
flight crew, recordings of microphones and earphones, and the performance information of the
aircraft.
Social Media Data : Social media such as Facebook and Twitter hold information and the views
posted by millions of people across the globe.
Stock Exchange Data : The stock exchange data holds information about the ‘buy’ and ‘sell’
decisions made on a share of different companies made by the customers.
Power Grid Data : The power grid data holds information consumed by a particular node with
respect to a base station.
Transport Data : Transport data includes model, capacity, distance and availability of a vehicle.
Search Engine Data : Search engines retrieve lots of data from different databases.
How is it similar or different to the example in the case? In response to your peers’ posts, discuss
how big data in the examples provided impact an organization in terms of the advantages and
disadvantages?
Big data is really critical to our life and its emerging as one of the most important technologies in
modern world. Follow are just few benefits which are very much known to all of us:
Using the information kept in the social network like Facebook, the marketing agencies are
learning about the response for their campaigns, promotions, and other advertising mediums.
Using the information in the social media like preferences and product perception of their
consumers, product companies and retail organizations are planning their production.
Disadvantages of Big Data:
- Unknown population representation
- Issues of data quality
- Typically not very multivariate (at the person level)
- Privacy and confidentiality issues
- Difficult to assess accuracy and uncertainty
Solution
What is an example of big data either from your personal experience or outside research?
Answer:
Big Data: Big data means really a big data, it is a collection of large datasets that cannot be
processed using traditional computing techniques. Big data is not merely a data, rather it has
become a complete subject, which involves various tools, technqiues and frameworks.
Big data involves the data produced by different devices and applications. Given below are some
of the fields that come under the Big Data.
Black Box Data : It is a component of helicopter, airplanes, and jets, etc. It captures voices of the
flight crew, recordings of microphones and earphones, and the performance.
Recommended
PriorityQueue.cs Jim Mischel using System; using Sy.pdf
// PriorityQueue.cs
//
// Jim Mischel
using System;
using System.Collections;
using System.Collections.Generic;
using System.Runtime.InteropServices;
namespace Mischel.Collections
{
[Serializable]
[ComVisible(false)]
public struct PriorityQueueItem
{
private TValue _value;
public TValue Value
{
get { return _value; }
}
private TPriority _priority;
public TPriority Priority
{
get { return _priority; }
}
internal PriorityQueueItem(TValue val, TPriority pri)
{
this._value = val;
this._priority = pri;
}
}
[Serializable]
[ComVisible(false)]
public class PriorityQueue : ICollection,
IEnumerable>
{
private PriorityQueueItem[] items;
private const Int32 DefaultCapacity = 16;
private Int32 capacity;
private Int32 numItems;
private Comparison compareFunc;
///
/// Initializes a new instance of the PriorityQueue class that is empty,
/// has the default initial capacity, and uses the default IComparer.
///
public PriorityQueue()
: this(DefaultCapacity, Comparer.Default)
{
}
public PriorityQueue(Int32 initialCapacity)
: this(initialCapacity, Comparer.Default)
{
}
public PriorityQueue(IComparer comparer)
: this(DefaultCapacity, comparer)
{
}
public PriorityQueue(int initialCapacity, IComparer comparer)
{
Init(initialCapacity, new Comparison(comparer.Compare));
}
public PriorityQueue(Comparison comparison)
: this(DefaultCapacity, comparison)
{
}
public PriorityQueue(int initialCapacity, Comparison comparison)
{
Init(initialCapacity, comparison);
}
private void Init(int initialCapacity, Comparison comparison)
{
numItems = 0;
compareFunc = comparison;
SetCapacity(initialCapacity);
}
public int Count
{
get { return numItems; }
}
public int Capacity
{
get { return items.Length; }
set { SetCapacity(value); }
}
private void SetCapacity(int newCapacity)
{
int newCap = newCapacity;
if (newCap < DefaultCapacity)
newCap = DefaultCapacity;
// throw exception if newCapacity < NumItems
if (newCap < numItems)
throw new ArgumentOutOfRangeException(\"newCapacity\", \"New capacity is less
than Count\");
this.capacity = newCap;
if (items == null)
{
items = new PriorityQueueItem[newCap];
return;
}
// Resize the array.
Array.Resize>(ref items, newCap);
}
public void Enqueue(PriorityQueueItem newItem)
{
if (numItems == capacity)
{
// need to increase capacity
// grow by 50 percent
SetCapacity((3 * Capacity) / 2);
}
int i = numItems;
++numItems;
while ((i > 0) && (compareFunc(items[(i - 1) / 2].Priority, newItem.Priority) < 0))
{
items[i] = items[(i - 1) / 2];
i = (i - 1) / 2;
}
items[i] = newItem;
//if (!VerifyQueue())
//{
// Console.WriteLine(\"ERROR: Queue out of order!\");
//}
}
public void Enqueue(TValue value, TPriority priority)
{
Enqueue(new PriorityQueueItem(value, priority));
}
private PriorityQueueItem RemoveAt(Int32 index)
{
PriorityQueueItem o = items[index];
--numItems;
// move the last item to fill the hole
PriorityQueueItem tmp = items[numItems];
// If you forget to clear this, you have a potential memory leak.
items[numItems] = default(PriorityQueueItem);
if (num.
the balanced equations are H3PO4(aq) + 3[NaOH](.pdf
the balanced equations are: H3PO4(aq) + 3[NaOH](aq) = Na3PO4(aq) + 3[H2O](l)
2[HI](aq) + LiO(aq) = LiI2(aq) + H2O(l) 2[HNO3](aq) + Ca(OH)2(s) = Ca(NO3)2(aq) +
2[H2O](l)
Solution
the balanced equations are: H3PO4(aq) + 3[NaOH](aq) = Na3PO4(aq) + 3[H2O](l)
2[HI](aq) + LiO(aq) = LiI2(aq) + H2O(l) 2[HNO3](aq) + Ca(OH)2(s) = Ca(NO3)2(aq) +
2[H2O](l).
Molecular wt. of NaOH = 40 No. of moles of NaOH =.pdf
Molecular wt. of NaOH = 40 No. of moles of NaOH = 0.3/40 = 0.0075 moles
Molarity = (No. of moles of NaOH)/(Volume of
Solution
) = 0.0075/0.17 = 0.0441 moles/lt.
images are not visible. post the question again .pdf
images are not visible. post the question again
Solution
images are not visible. post the question again.
VSEPR is easy if you follow the right steps. 1. .pdf
VSEPR is easy if you follow the right steps. 1. Find the number of valence (outer)
electrons around the central atom (remember the charge) 2. Work out how many of these are
involved in bonding (remember you may have single or double bonds) 3. Remember the lone
pairs. 4. Work out the shape from the number of atoms bonded to the central atom and the
number of lone pairs. So for SeO2, Se is in group 6, and has 6 valence electrons. Se=O is a
double bond, so there are 2 of Se\'s electrons in a double bond There are two Se=O bonds, so 4
electrons are involved in Se=O bonds. This leaves another 2 electrons which are a lone pair. So
overall you have three groups around the central atom (2 Se=O bonds and a lone pair) so it is
trigonal planar if you drew in the lone pair. Removing the lone pair gives you a bent molecule.
For PF4+: P has 5 valence electrons, but it is a positive ion, so you knock off an electron to get 4
P-F bonds are single bonds, and there are 4 of them, which accounts for all the electrons.
Therefore PF4+ is a tetrahedral shape. Finally, ICl3: I has 7 valence electrons. I-Cl bonds are
single bonds, so the I uses one electron in each bond. Therefore 3 of the 7 electrons are involved
in I-Cl bonds, the other 4 must be in 2 lone pairs. This gives 5 electron groups around the central
ion, so the shape is trigonal bipyramid if you draw in the lone pairs (trigonal plane with 2 groups
above and below the plane. You can determine that these two groups must be the lone pairs since
they repel more than the I-Cl bonds). Removing the lone pairs gives you a trigonal planar
molecule. The number of groups around the central ion determines the shape. 2 is linear 3 is
trigonal planar
Solution
VSEPR is easy if you follow the right steps. 1. Find the number of valence (outer)
electrons around the central atom (remember the charge) 2. Work out how many of these are
involved in bonding (remember you may have single or double bonds) 3. Remember the lone
pairs. 4. Work out the shape from the number of atoms bonded to the central atom and the
number of lone pairs. So for SeO2, Se is in group 6, and has 6 valence electrons. Se=O is a
double bond, so there are 2 of Se\'s electrons in a double bond There are two Se=O bonds, so 4
electrons are involved in Se=O bonds. This leaves another 2 electrons which are a lone pair. So
overall you have three groups around the central atom (2 Se=O bonds and a lone pair) so it is
trigonal planar if you drew in the lone pair. Removing the lone pair gives you a bent molecule.
For PF4+: P has 5 valence electrons, but it is a positive ion, so you knock off an electron to get 4
P-F bonds are single bonds, and there are 4 of them, which accounts for all the electrons.
Therefore PF4+ is a tetrahedral shape. Finally, ICl3: I has 7 valence electrons. I-Cl bonds are
single bonds, so the I uses one electron in each bond. Therefore 3 of the 7 electrons are involved
in I-Cl bonds, the other 4 must be in 2 lone pairs. This gives 5 electro.
This is quite a broad question. I will try to acc.pdf
This is quite a broad question. I will try to account all the important observations. 1.
Metal ions which do not have d-orbital electrons do not have color. (alkali, alkali-earth, Sc3+)
This is because of energy separation of s and p orbitals are not in the range of visible colors. 2.
Metal with fully-filled d orbitals do not have color. (Zn2+) This is because electrons in d-orbitals
cannot be excited to higher energy d-orbitals (all d-orbitals have the same energies, but when
influenced by other atom in complex, d-orbitals can split into different energy levels) 3. For ions
of the same metal, the ion with larger charge has a color of lower energy than the ion with
smaller charge. Example: Cr2+ (blue) and Cr3+ (green). This is because larger charge
contributes to larger split of the d-orbital. The more the orbitals split, the higher energy of light it
traps, allowing low-energy photon to enter our eyes. Note that you cannot be sure that the color
you see is a single color, not a combination of many colors. 4. For metals in the same column, a
metal ion which is below has a color of lower energy than an ion above. This can be explained
like no.3, larger ion causes more orbitals splitting. 5. Color of metal ions also depends on
\"ligands\" (complexing molecule/ion). e.g. Co2+ in water is pink [Co(H2O)6]2+, while Co2+ in
NH3 is blue [Co(NH3)4]2+. See \"spectrochemical series\" for more detail about this. 6. To find
\"exact\" color of metal ions, you need a supercomputer to do so. (I\'m not sure whether today
computer can calculate it). So, it\'s better to memorize the colors. Personally, I think you\'ll
understand about metal\'s color when you study transition metals, like Crystal Field Theory, and
Ligand Field Theory.
Solution
This is quite a broad question. I will try to account all the important observations. 1.
Metal ions which do not have d-orbital electrons do not have color. (alkali, alkali-earth, Sc3+)
This is because of energy separation of s and p orbitals are not in the range of visible colors. 2.
Metal with fully-filled d orbitals do not have color. (Zn2+) This is because electrons in d-orbitals
cannot be excited to higher energy d-orbitals (all d-orbitals have the same energies, but when
influenced by other atom in complex, d-orbitals can split into different energy levels) 3. For ions
of the same metal, the ion with larger charge has a color of lower energy than the ion with
smaller charge. Example: Cr2+ (blue) and Cr3+ (green). This is because larger charge
contributes to larger split of the d-orbital. The more the orbitals split, the higher energy of light it
traps, allowing low-energy photon to enter our eyes. Note that you cannot be sure that the color
you see is a single color, not a combination of many colors. 4. For metals in the same column, a
metal ion which is below has a color of lower energy than an ion above. This can be explained
like no.3, larger ion causes more orbitals splitting. 5. Color of metal ions also depends .
}public staatic boolean isValidElement(String token){public vo.pdf
}
public staatic boolean isValidElement(String token)
{
public void setParts(String IP)
{
IP=\"192.168.20.40\";
}
public void setParts(int[], data)
{
data[0]=toString(IP);
data[1]=toString(IP);
data[2]=toString(IP);
data[3]=toString(IP);
}
Solution
}
public staatic boolean isValidElement(String token)
{
public void setParts(String IP)
{
IP=\"192.168.20.40\";
}
public void setParts(int[], data)
{
data[0]=toString(IP);
data[1]=toString(IP);
data[2]=toString(IP);
data[3]=toString(IP);
}.
What is an example of big data either from your personal experience .pdf
What is an example of big data either from your personal experience or outside research?
Answer:
Big Data: Big data means really a big data, it is a collection of large datasets that cannot be
processed using traditional computing techniques. Big data is not merely a data, rather it has
become a complete subject, which involves various tools, technqiues and frameworks.
Big data involves the data produced by different devices and applications. Given below are some
of the fields that come under the Big Data.
Black Box Data : It is a component of helicopter, airplanes, and jets, etc. It captures voices of the
flight crew, recordings of microphones and earphones, and the performance information of the
aircraft.
Social Media Data : Social media such as Facebook and Twitter hold information and the views
posted by millions of people across the globe.
Stock Exchange Data : The stock exchange data holds information about the ‘buy’ and ‘sell’
decisions made on a share of different companies made by the customers.
Power Grid Data : The power grid data holds information consumed by a particular node with
respect to a base station.
Transport Data : Transport data includes model, capacity, distance and availability of a vehicle.
Search Engine Data : Search engines retrieve lots of data from different databases.
How is it similar or different to the example in the case? In response to your peers’ posts, discuss
how big data in the examples provided impact an organization in terms of the advantages and
disadvantages?
Big data is really critical to our life and its emerging as one of the most important technologies in
modern world. Follow are just few benefits which are very much known to all of us:
Using the information kept in the social network like Facebook, the marketing agencies are
learning about the response for their campaigns, promotions, and other advertising mediums.
Using the information in the social media like preferences and product perception of their
consumers, product companies and retail organizations are planning their production.
Disadvantages of Big Data:
- Unknown population representation
- Issues of data quality
- Typically not very multivariate (at the person level)
- Privacy and confidentiality issues
- Difficult to assess accuracy and uncertainty
Solution
What is an example of big data either from your personal experience or outside research?
Answer:
Big Data: Big data means really a big data, it is a collection of large datasets that cannot be
processed using traditional computing techniques. Big data is not merely a data, rather it has
become a complete subject, which involves various tools, technqiues and frameworks.
Big data involves the data produced by different devices and applications. Given below are some
of the fields that come under the Big Data.
Black Box Data : It is a component of helicopter, airplanes, and jets, etc. It captures voices of the
flight crew, recordings of microphones and earphones, and the performance.
S-Se=S with 3 lone pairs on the first S, 1 lone.pdf
S-Se=S with 3 lone pairs on the first S, 1 lone pair on the Se, and 2 lone pairs on
the second S
Solution
S-Se=S with 3 lone pairs on the first S, 1 lone pair on the Se, and 2 lone pairs on
the second S.
The sample program for above series in JAVA will be like belowimpo.pdf
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}
Solution
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}.
The formula for calculating Alveolar ventilation is as followsAlv.pdf
The formula for calculating Alveolar ventilation is as follows:
Alveolar ventilation = (tidal volume - dead space) * respiratory rate
Alveolar ventilation = ( 500 – 100) * 12
Alveolar ventilation = 4800
Solution
The formula for calculating Alveolar ventilation is as follows:
Alveolar ventilation = (tidal volume - dead space) * respiratory rate
Alveolar ventilation = ( 500 – 100) * 12
Alveolar ventilation = 4800.
D. This element can only have a -2 oxidation stat.pdf
D. This element can only have a -2 oxidation state in covalent compounds. is false.
note: such as S can be +4 in SO2
Solution
D. This element can only have a -2 oxidation state in covalent compounds. is false.
note: such as S can be +4 in SO2.
solAn object will allocated statically when that object is needed.pdf
sol:
An object will allocated statically when that object is needed for entire life time of that program
1)if object needs recusrion and storage of object conservly then it is stored in stack
2)f object needs dynamic storage then it should be allocated in heap
Solution
sol:
An object will allocated statically when that object is needed for entire life time of that program
1)if object needs recusrion and storage of object conservly then it is stored in stack
2)f object needs dynamic storage then it should be allocated in heap.
carbon moves by A. interstitial diffusion So.pdf
carbon moves by? A. interstitial diffusion
Solution
carbon moves by? A. interstitial diffusion.
NH4Cl = acid NaOH = base They react 11 Leftover = 0.26-0..pdf
NH4Cl = acid
NaOH = base
They react 1:1
Leftover = 0.26-0.192 = 0.068M NH4Cl
pH = -log(0.068) = 1.1675
pH + pOH = 14
pOH = 14 - 1.1675 = 12.8325
[OH-] = 10^-12.8325 = 1.471 * 10^-13 M
Solution
NH4Cl = acid
NaOH = base
They react 1:1
Leftover = 0.26-0.192 = 0.068M NH4Cl
pH = -log(0.068) = 1.1675
pH + pOH = 14
pOH = 14 - 1.1675 = 12.8325
[OH-] = 10^-12.8325 = 1.471 * 10^-13 M.
D is the transition represent!! Solution .pdf
D is the transition represent!!
Solution
D is the transition represent!!.
CO2 is non polar because C has no lone pairs, so .pdf
CO2 is non polar because C has no lone pairs, so the structure is linear and hence
dipole cancell each other SO2 has lone pair, so the structure is bent due to the repulsion and
hence the structure has a net dipole moment, so polar.
Solution
CO2 is non polar because C has no lone pairs, so the structure is linear and hence
dipole cancell each other SO2 has lone pair, so the structure is bent due to the repulsion and
hence the structure has a net dipole moment, so polar..
As more halide ions are added, it loses its color.pdf
As more halide ions are added, it loses its color and becomes more white due to the
ion concentration.
Solution
As more halide ions are added, it loses its color and becomes more white due to the
ion concentration..
AppointmentDemo.javaimport java.util.Scanner; Demonstrat.pdf
AppointmentDemo.java
import java.util.Scanner;
/**
Demonstration of the appointment classes
*/
public class AppointmentDemo
{
public static void main(String[] args)
{
Appointment[] appointments = new Appointment[4];
appointments[0] = new Daily(\"Brush your teeth.\");
appointments[1] = new Monthly(1, \"Visit grandma.\");
appointments[2] = new Onetime(2015, 11, 1, \"Dentist appointment.\");
appointments[3] = new Onetime(2015, 10, 31, \"Trick or Treat.\");
Scanner in = new Scanner(System.in);
int year = in.nextInt();
int month = in.nextInt();
int day = in.nextInt();
for (Appointment a : appointments)
{
if (a.occursOn(year, month, day))
{
System.out.println(a);
}
}
}
}
Appointment.java
/**
A class to keep track of an appointment.
*/
public abstract class Appointment
{
private String description;
/**
Constructs an appointment without a description.
*/
public Appointment()
{
description = \"\";
}
/**
Sets the description of this appointment.
*/
public void setDescription(String description)
{
this.description = description;
}
/**
Determines if this appointment occurs on the given date.
*/
public abstract boolean occursOn(int year, int month, int day);
/**
Converts appointment to string description.
*/
public String toString()
{
return description;
}
}
Daily.java
public class Daily extends Appointment
{
public Daily (String description)
{
setDescription(description);
}
public boolean occursOn(int year, int month, int day)
{
return true;
}
}
Onetime.java
public class Onetime extends Appointment
{
private int monthApp;
private int yearApp;
private int dayApp;
public Onetime(int yearAppInput, int monthAppInput, int dayAppInput, String description)
{
yearApp = yearAppInput;
monthApp = monthAppInput;
dayApp = dayAppInput;
setDescription(description);
}
public boolean occursOn(int year, int month, int day)
{
if ((year == yearApp) && (month == monthApp) && (day == dayApp))
{
return true;
}
else
{
return false;
}
}
}
Monthly.java
public class Monthly extends Appointment
{
private int dayApp;
public Monthly (int dayAppInput, String description)
{
dayApp = dayAppInput;
setDescription(description);
}
public boolean occursOn(int year, int month, int day)
{
if (day == dayApp)
{
return true;
}
else
{
return false;
}
}
}
Solution
AppointmentDemo.java
import java.util.Scanner;
/**
Demonstration of the appointment classes
*/
public class AppointmentDemo
{
public static void main(String[] args)
{
Appointment[] appointments = new Appointment[4];
appointments[0] = new Daily(\"Brush your teeth.\");
appointments[1] = new Monthly(1, \"Visit grandma.\");
appointments[2] = new Onetime(2015, 11, 1, \"Dentist appointment.\");
appointments[3] = new Onetime(2015, 10, 31, \"Trick or Treat.\");
Scanner in = new Scanner(System.in);
int year = in.nextInt();
int month = in.nextInt();
int day = in.nextInt();
for (Appointment a : appointments)
{
if (a.occursOn(year, month, day))
{
System.out.println(a);
}
}
}
}
Appointment.java
/**
A class to keep track of an appointment.
*/
public ab.
Answer.Ltext0 .local _ZStL8__ioinit .comm .pdf
Answer:
.Ltext0:
.local _ZStL8__ioinit
.comm _ZStL8__ioinit,1,1
.section .rodata
.LC1:
0000 66696C6C .string \"fill in the \"
.LC2:
000d 206E756D .string \" number. :\"
.text
.globl main
main:
.LFB1021:
.cfi_startproc
0000 55 pushq %rbp
.cfi_def_cfa_offset 16
.cfi_offset 6, -16
0001 4889E5 movq %rsp, %rbp
.cfi_def_cfa_register 6
0004 53 pushq %rbx
0005 4883EC78 subq $120, %rsp
.cfi_offset 3, -24
0009 64488B04 movq %fs:40, %rax
0012 488945E8 movq %rax, -24(%rbp)
0016 31C0 xorl %eax, %eax
0018 660FEFC0 pxor %xmm0, %xmm0
001c F30F1145 movss %xmm0, -120(%rbp)
0021 C7458C14 movl $20, -116(%rbp)
.LBB2:
0028 C7458400 movl $0, -124(%rbp)
.L3:
002f 8B4584 movl -124(%rbp), %eax
0032 3B458C cmpl -116(%rbp), %eax
0035 7D50 jge .L2
0081 83458401 addl $1, -124(%rbp)
0085 EBA8 jmp .L3
0037 8B4584 movl -124(%rbp), %eax
003a 8D5801 leal 1(%rax), %ebx
003d BE000000 movl $.LC1, %esi
0042 BF000000 movl $_ZSt4cout, %edi
0047 E8000000 call _ZStlsISt11char_traitsIcEERSt13basic_ostreamIcT_ES5_PKc
004c 89DE movl %ebx, %esi
004e 4889C7 movq %rax, %rdi
0051 E8000000 call _ZNSolsEi
0056 BE000000 movl $.LC2, %esi
005b 4889C7 movq %rax, %rdi
005e E8000000 call _ZStlsISt11char_traitsIcEERSt13basic_ostreamIcT_ES5_PKc
0063 488D4590 leaq -112(%rbp), %rax
0067 8B5584 movl -124(%rbp), %edx
006a 4863D2 movslq %edx, %rdx
006d 48C1E202 salq $2, %rdx
0071 4801D0 addq %rdx, %rax
0074 4889C6 movq %rax, %rsi
0077 BF000000 movl $_ZSt3cin, %edi
007c E8000000 call _ZNSirsERi
.L2:
.LBE2:
0087 8B558C movl -116(%rbp), %edx
008a 488D4590 leaq -112(%rbp), %rax
008e 89D6 movl %edx, %esi
0090 4889C7 movq %rax, %rdi
0093 E8000000 call _Z4sortPii
0098 B8000000 movl $0, %eax
009d 488B4DE8 movq -24(%rbp), %rcx
00a1 6448330C xorq %fs:40, %rcx
00aa 7405 je .L5
00ac E8000000 call __stack_chk_fail
.L5:
00b1 4883C478 addq $120, %rsp
00b5 5B popq %rbx
00b6 5D popq %rbp
.cfi_def_cfa 7, 8
00b7 C3 ret
.cfi_endproc
.LFE1021:
.section .rodata
.LC3:
0018 54686520 .string \"The mean is \"
.text
.globl _Z4meanPii
_Z4meanPii:
.LFB1022:
.cfi_startproc
00b8 55 pushq %rbp
.cfi_def_cfa_offset 16
.cfi_offset 6, -16
00b9 4889E5 movq %rsp, %rbp
.cfi_def_cfa_register 6
00bc 4883EC20 subq $32, %rsp
00c0 48897DE8 movq %rdi, -24(%rbp)
00c4 8975E4 movl %esi, -28(%rbp)
00c7 660FEFC0 pxor %xmm0, %xmm0
00cb F30F1145 movss %xmm0, -8(%rbp)
.LBB3:
00d0 C745FC00 movl $0, -4(%rbp)
.L8:
00d7 8B45FC movl -4(%rbp), %eax
00da 3B45E4 cmpl -28(%rbp), %eax
00dd 7D32 jge .L7
010b 8345FC01 addl $1, -4(%rbp)
010f EBC6 jmp .L8
00df 8B45FC movl -4(%rbp), %eax
00e2 4898 cltq
00e4 488D1485 leaq 0(,%rax,4), %rdx
00ec 488B45E8 movq -24(%rbp), %rax
00f0 4801D0 addq %rdx, %rax
00f3 8B00 movl (%rax), %eax
00f5 660FEFC0 pxor %xmm0, %xmm0
00f9 F30F2AC0 cvtsi2ss %eax, %xmm0
00fd F30F104D movss -8(%rbp), %xmm1
0102 F30F58C1 addss %xmm1, %xmm0
0106 F30F1145 movss %xmm0, -8(%rbp)
.L7:
.LBE3:
0111 660FEFC0 pxor %xmm0, %xmm0
0115 F30F2A45 cvtsi2ss -28(%rbp), %xmm0
011a F30F104D movss -8(%rbp), %xmm1
011f F30F5EC8 divss %xmm0, %xmm1
0123 F30F114D movss %xmm1, .
Debt-equity ratio=DebtEQuityHence debt=1.1EquityTotal assets=De.pdf
Debt-equity ratio=Debt/EQuity
Hence debt=1.1Equity
Total assets=Debt+Equity
2975 =2.1Equity
Hence Equity=(2975/2.1)=$1416.67(Approx)
ROE=Net income/Equity
Hence net income=(1416.67*11%)
=$155.83(Approx).
Solution
Debt-equity ratio=Debt/EQuity
Hence debt=1.1Equity
Total assets=Debt+Equity
2975 =2.1Equity
Hence Equity=(2975/2.1)=$1416.67(Approx)
ROE=Net income/Equity
Hence net income=(1416.67*11%)
=$155.83(Approx)..
Gaseous chemical element, chemical symbol O, atom.pdf
Gaseous chemical element, chemical symbol O, atomic number 8. It constitutes
21% (by volume) of air and more than 46% (by weight) of Earth\'s crust, where it is the most
plentiful element. It is a colourless, odourless, tasteless gas, occurring as the diatomic molecule
O2. In respiration, it is taken up by animals and some bacteria (and by plants in the dark), which
give off carbon dioxide (CO2). In photosynthesis, green plants assimilate carbon dioxide in the
presence of sunlight and give off oxygen. The small amount of oxygen that dissolves in water is
essential for the respiration of fish and other aquatic life. Oxygen takes part in combustion and in
corrosion but does not itself burn. It has valence 2 in compounds; the most important is water. It
forms oxides and is part of many other molecules and functional groups, including nitrate,
sulfate, phosphate, and carbonate; alcohols, aldehydes, carboxylic acids, and ketones; and
peroxides. Obtained for industrial use by distillation of liquefied air, oxygen is used in
steelmaking and other metallurgical processes and in the chemical industry. Medical uses include
respiratory therapy, incubators, and inhaled anesthetics. Oxygen is part of all gas mixtures for
manned spacecraft, scuba divers, workers in closed environments, and hyperbaric chambers. It is
also used in rocket engines as an oxidizer (in liquefied form) and in water and waste treatment
processes.
Solution
Gaseous chemical element, chemical symbol O, atomic number 8. It constitutes
21% (by volume) of air and more than 46% (by weight) of Earth\'s crust, where it is the most
plentiful element. It is a colourless, odourless, tasteless gas, occurring as the diatomic molecule
O2. In respiration, it is taken up by animals and some bacteria (and by plants in the dark), which
give off carbon dioxide (CO2). In photosynthesis, green plants assimilate carbon dioxide in the
presence of sunlight and give off oxygen. The small amount of oxygen that dissolves in water is
essential for the respiration of fish and other aquatic life. Oxygen takes part in combustion and in
corrosion but does not itself burn. It has valence 2 in compounds; the most important is water. It
forms oxides and is part of many other molecules and functional groups, including nitrate,
sulfate, phosphate, and carbonate; alcohols, aldehydes, carboxylic acids, and ketones; and
peroxides. Obtained for industrial use by distillation of liquefied air, oxygen is used in
steelmaking and other metallurgical processes and in the chemical industry. Medical uses include
respiratory therapy, incubators, and inhaled anesthetics. Oxygen is part of all gas mixtures for
manned spacecraft, scuba divers, workers in closed environments, and hyperbaric chambers. It is
also used in rocket engines as an oxidizer (in liquefied form) and in water and waste treatment
processes..
B. Remember this - hydroboration attacks alkenes .pdf
B. Remember this - hydroboration attacks alkenes but not rings and it follows the
anti Markonikov rules where the H will attach to the carbon with the fewest alkyl substitiutions.
The oh with will by default (after 2)H2O2, OH-) attach to the other carbon.
Solution
B. Remember this - hydroboration attacks alkenes but not rings and it follows the
anti Markonikov rules where the H will attach to the carbon with the fewest alkyl substitiutions.
The oh with will by default (after 2)H2O2, OH-) attach to the other carbon..
a). EconomiserAn economiser is a mechanical device which is used a.pdf
a). Economiser
An economiser is a mechanical device which is used as a heat exchanger by preheating a fluid to
reduce energy consumption. In a steam boiler, it is a heat ex-changer device that heats up fluids
or recovers residual heat from the combustion product i.e. flue gases in thermal power plant
before being released through the chimney. Flue gases are the combustion exhaust gases
produced at power plants consist of mostly nitrogen, carbon dioxide, water vapor, soot carbon
monoxide etc. Hence, the economiser in thermal power plants, is used to economise the process
of electrical power generation, as the name of the device is suggestive of. The recovered heat is
in turn used to preheat the boiler feed water, that will eventually be converted to super-heated
steam. Thus, saving on fuel consumption and economising the process to a large extent, as we
are essentially gathering the waste heat and applying it to, where it is required. Nowadays
however, in addition to that, the heat available in the exhaust flue gases can be economically
recovered using air pre-heater which are essential in all pulverized coal fired boiler.
b). Reheater
ins some heat either from the combustion gases leaving the boiler (if there is still much
temperature difference) or by adding more heat by burning small amount of fuel. The energy of
the exhaust steam with the gained heat increases the steam enthalpy of the steam and give a good
opportunity to generate much work with the low pressure turbine. The two works (High pressure
W_H and Low pressure W_L) are much greater than the heat added by fuel burned (Q1 and Q2),
so the efficiency increases. Another issue, the reheat helps in saving the turbine blades from
corrosion due to low dryness fraction x<0.88 which it may occur if one single stage turbine is
used.
c). Cogeneration
Cogeneration (Combined Heat and Power or CHP) is the simultaneous production of electricity
and heat, both of which are used. The central and most fundamental principle of cogeneration is
that, in order to maximise the many benefits that arise from it, systems should be based on the
heat demand of the application. This can be an individual building, an industrial factory or a
town/city served by district heat/cooling. Through the utilisation of the heat, the efficiency of a
cogeneration plant can reach 90% or more.
Cogeneration therefore offers energy savings ranging between 15-40% when compared against
the supply of electricity and heat from conventional power stations and boilers.
Cogeneration optimises the energy supply to all types of consumers, with the followingbenefits
for both users and society at large:
d). Turbocharging
A turbocharger, or turbo (colloquialism), from Greek \"\" (\"wake\"),[1] (also from Latin
\"turbo\" (\"spinning top\"),[2]) is aturbine-driven forced induction device that increases an
internal combustion engine\'s efficiency and power output by forcing extra air into the
combustion chamber.[3][4] This improvement over a.
A) As it is shown this disease should be partly autosomal recessive .pdf
A) As it is shown this disease should be partly autosomal recessive one. Because it is not sex
linked disease and both the first row parents should be carriers. Because neither the square nor
circle is colourd.
B) Both the parents in generation (ii) can be carriers. But as far as no diseased person in
generation (iii), it is more likely to be only one of them is a carrier. If it was a sex linked trait the
female should have been the carrier.
C) Marriage between cousins led the occurrence of disease.
Solution
A) As it is shown this disease should be partly autosomal recessive one. Because it is not sex
linked disease and both the first row parents should be carriers. Because neither the square nor
circle is colourd.
B) Both the parents in generation (ii) can be carriers. But as far as no diseased person in
generation (iii), it is more likely to be only one of them is a carrier. If it was a sex linked trait the
female should have been the carrier.
C) Marriage between cousins led the occurrence of disease..
Overview on Edible Vaccine: Pros & Cons with Mechanism
This ppt include the description of the edible vaccine i.e. a new concept over the traditional vaccine administered by injection.
Digital Tools and AI for Teaching Learning and Research
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S-Se=S with 3 lone pairs on the first S, 1 lone.pdf
S-Se=S with 3 lone pairs on the first S, 1 lone pair on the Se, and 2 lone pairs on
the second S
Solution
S-Se=S with 3 lone pairs on the first S, 1 lone pair on the Se, and 2 lone pairs on
the second S.
The sample program for above series in JAVA will be like belowimpo.pdf
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}
Solution
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}.
The formula for calculating Alveolar ventilation is as followsAlv.pdf
The formula for calculating Alveolar ventilation is as follows:
Alveolar ventilation = (tidal volume - dead space) * respiratory rate
Alveolar ventilation = ( 500 – 100) * 12
Alveolar ventilation = 4800
Solution
The formula for calculating Alveolar ventilation is as follows:
Alveolar ventilation = (tidal volume - dead space) * respiratory rate
Alveolar ventilation = ( 500 – 100) * 12
Alveolar ventilation = 4800.
D. This element can only have a -2 oxidation stat.pdf
D. This element can only have a -2 oxidation state in covalent compounds. is false.
note: such as S can be +4 in SO2
Solution
D. This element can only have a -2 oxidation state in covalent compounds. is false.
note: such as S can be +4 in SO2.
solAn object will allocated statically when that object is needed.pdf
sol:
An object will allocated statically when that object is needed for entire life time of that program
1)if object needs recusrion and storage of object conservly then it is stored in stack
2)f object needs dynamic storage then it should be allocated in heap
Solution
sol:
An object will allocated statically when that object is needed for entire life time of that program
1)if object needs recusrion and storage of object conservly then it is stored in stack
2)f object needs dynamic storage then it should be allocated in heap.
carbon moves by A. interstitial diffusion So.pdf
carbon moves by? A. interstitial diffusion
Solution
carbon moves by? A. interstitial diffusion.
NH4Cl = acid NaOH = base They react 11 Leftover = 0.26-0..pdf
NH4Cl = acid
NaOH = base
They react 1:1
Leftover = 0.26-0.192 = 0.068M NH4Cl
pH = -log(0.068) = 1.1675
pH + pOH = 14
pOH = 14 - 1.1675 = 12.8325
[OH-] = 10^-12.8325 = 1.471 * 10^-13 M
Solution
NH4Cl = acid
NaOH = base
They react 1:1
Leftover = 0.26-0.192 = 0.068M NH4Cl
pH = -log(0.068) = 1.1675
pH + pOH = 14
pOH = 14 - 1.1675 = 12.8325
[OH-] = 10^-12.8325 = 1.471 * 10^-13 M.
D is the transition represent!! Solution .pdf
D is the transition represent!!
Solution
D is the transition represent!!.
CO2 is non polar because C has no lone pairs, so .pdf
CO2 is non polar because C has no lone pairs, so the structure is linear and hence
dipole cancell each other SO2 has lone pair, so the structure is bent due to the repulsion and
hence the structure has a net dipole moment, so polar.
Solution
CO2 is non polar because C has no lone pairs, so the structure is linear and hence
dipole cancell each other SO2 has lone pair, so the structure is bent due to the repulsion and
hence the structure has a net dipole moment, so polar..
As more halide ions are added, it loses its color.pdf
As more halide ions are added, it loses its color and becomes more white due to the
ion concentration.
Solution
As more halide ions are added, it loses its color and becomes more white due to the
ion concentration..
AppointmentDemo.javaimport java.util.Scanner; Demonstrat.pdf
AppointmentDemo.java
import java.util.Scanner;
/**
Demonstration of the appointment classes
*/
public class AppointmentDemo
{
public static void main(String[] args)
{
Appointment[] appointments = new Appointment[4];
appointments[0] = new Daily(\"Brush your teeth.\");
appointments[1] = new Monthly(1, \"Visit grandma.\");
appointments[2] = new Onetime(2015, 11, 1, \"Dentist appointment.\");
appointments[3] = new Onetime(2015, 10, 31, \"Trick or Treat.\");
Scanner in = new Scanner(System.in);
int year = in.nextInt();
int month = in.nextInt();
int day = in.nextInt();
for (Appointment a : appointments)
{
if (a.occursOn(year, month, day))
{
System.out.println(a);
}
}
}
}
Appointment.java
/**
A class to keep track of an appointment.
*/
public abstract class Appointment
{
private String description;
/**
Constructs an appointment without a description.
*/
public Appointment()
{
description = \"\";
}
/**
Sets the description of this appointment.
*/
public void setDescription(String description)
{
this.description = description;
}
/**
Determines if this appointment occurs on the given date.
*/
public abstract boolean occursOn(int year, int month, int day);
/**
Converts appointment to string description.
*/
public String toString()
{
return description;
}
}
Daily.java
public class Daily extends Appointment
{
public Daily (String description)
{
setDescription(description);
}
public boolean occursOn(int year, int month, int day)
{
return true;
}
}
Onetime.java
public class Onetime extends Appointment
{
private int monthApp;
private int yearApp;
private int dayApp;
public Onetime(int yearAppInput, int monthAppInput, int dayAppInput, String description)
{
yearApp = yearAppInput;
monthApp = monthAppInput;
dayApp = dayAppInput;
setDescription(description);
}
public boolean occursOn(int year, int month, int day)
{
if ((year == yearApp) && (month == monthApp) && (day == dayApp))
{
return true;
}
else
{
return false;
}
}
}
Monthly.java
public class Monthly extends Appointment
{
private int dayApp;
public Monthly (int dayAppInput, String description)
{
dayApp = dayAppInput;
setDescription(description);
}
public boolean occursOn(int year, int month, int day)
{
if (day == dayApp)
{
return true;
}
else
{
return false;
}
}
}
Solution
AppointmentDemo.java
import java.util.Scanner;
/**
Demonstration of the appointment classes
*/
public class AppointmentDemo
{
public static void main(String[] args)
{
Appointment[] appointments = new Appointment[4];
appointments[0] = new Daily(\"Brush your teeth.\");
appointments[1] = new Monthly(1, \"Visit grandma.\");
appointments[2] = new Onetime(2015, 11, 1, \"Dentist appointment.\");
appointments[3] = new Onetime(2015, 10, 31, \"Trick or Treat.\");
Scanner in = new Scanner(System.in);
int year = in.nextInt();
int month = in.nextInt();
int day = in.nextInt();
for (Appointment a : appointments)
{
if (a.occursOn(year, month, day))
{
System.out.println(a);
}
}
}
}
Appointment.java
/**
A class to keep track of an appointment.
*/
public ab.
Answer.Ltext0 .local _ZStL8__ioinit .comm .pdf
Answer:
.Ltext0:
.local _ZStL8__ioinit
.comm _ZStL8__ioinit,1,1
.section .rodata
.LC1:
0000 66696C6C .string \"fill in the \"
.LC2:
000d 206E756D .string \" number. :\"
.text
.globl main
main:
.LFB1021:
.cfi_startproc
0000 55 pushq %rbp
.cfi_def_cfa_offset 16
.cfi_offset 6, -16
0001 4889E5 movq %rsp, %rbp
.cfi_def_cfa_register 6
0004 53 pushq %rbx
0005 4883EC78 subq $120, %rsp
.cfi_offset 3, -24
0009 64488B04 movq %fs:40, %rax
0012 488945E8 movq %rax, -24(%rbp)
0016 31C0 xorl %eax, %eax
0018 660FEFC0 pxor %xmm0, %xmm0
001c F30F1145 movss %xmm0, -120(%rbp)
0021 C7458C14 movl $20, -116(%rbp)
.LBB2:
0028 C7458400 movl $0, -124(%rbp)
.L3:
002f 8B4584 movl -124(%rbp), %eax
0032 3B458C cmpl -116(%rbp), %eax
0035 7D50 jge .L2
0081 83458401 addl $1, -124(%rbp)
0085 EBA8 jmp .L3
0037 8B4584 movl -124(%rbp), %eax
003a 8D5801 leal 1(%rax), %ebx
003d BE000000 movl $.LC1, %esi
0042 BF000000 movl $_ZSt4cout, %edi
0047 E8000000 call _ZStlsISt11char_traitsIcEERSt13basic_ostreamIcT_ES5_PKc
004c 89DE movl %ebx, %esi
004e 4889C7 movq %rax, %rdi
0051 E8000000 call _ZNSolsEi
0056 BE000000 movl $.LC2, %esi
005b 4889C7 movq %rax, %rdi
005e E8000000 call _ZStlsISt11char_traitsIcEERSt13basic_ostreamIcT_ES5_PKc
0063 488D4590 leaq -112(%rbp), %rax
0067 8B5584 movl -124(%rbp), %edx
006a 4863D2 movslq %edx, %rdx
006d 48C1E202 salq $2, %rdx
0071 4801D0 addq %rdx, %rax
0074 4889C6 movq %rax, %rsi
0077 BF000000 movl $_ZSt3cin, %edi
007c E8000000 call _ZNSirsERi
.L2:
.LBE2:
0087 8B558C movl -116(%rbp), %edx
008a 488D4590 leaq -112(%rbp), %rax
008e 89D6 movl %edx, %esi
0090 4889C7 movq %rax, %rdi
0093 E8000000 call _Z4sortPii
0098 B8000000 movl $0, %eax
009d 488B4DE8 movq -24(%rbp), %rcx
00a1 6448330C xorq %fs:40, %rcx
00aa 7405 je .L5
00ac E8000000 call __stack_chk_fail
.L5:
00b1 4883C478 addq $120, %rsp
00b5 5B popq %rbx
00b6 5D popq %rbp
.cfi_def_cfa 7, 8
00b7 C3 ret
.cfi_endproc
.LFE1021:
.section .rodata
.LC3:
0018 54686520 .string \"The mean is \"
.text
.globl _Z4meanPii
_Z4meanPii:
.LFB1022:
.cfi_startproc
00b8 55 pushq %rbp
.cfi_def_cfa_offset 16
.cfi_offset 6, -16
00b9 4889E5 movq %rsp, %rbp
.cfi_def_cfa_register 6
00bc 4883EC20 subq $32, %rsp
00c0 48897DE8 movq %rdi, -24(%rbp)
00c4 8975E4 movl %esi, -28(%rbp)
00c7 660FEFC0 pxor %xmm0, %xmm0
00cb F30F1145 movss %xmm0, -8(%rbp)
.LBB3:
00d0 C745FC00 movl $0, -4(%rbp)
.L8:
00d7 8B45FC movl -4(%rbp), %eax
00da 3B45E4 cmpl -28(%rbp), %eax
00dd 7D32 jge .L7
010b 8345FC01 addl $1, -4(%rbp)
010f EBC6 jmp .L8
00df 8B45FC movl -4(%rbp), %eax
00e2 4898 cltq
00e4 488D1485 leaq 0(,%rax,4), %rdx
00ec 488B45E8 movq -24(%rbp), %rax
00f0 4801D0 addq %rdx, %rax
00f3 8B00 movl (%rax), %eax
00f5 660FEFC0 pxor %xmm0, %xmm0
00f9 F30F2AC0 cvtsi2ss %eax, %xmm0
00fd F30F104D movss -8(%rbp), %xmm1
0102 F30F58C1 addss %xmm1, %xmm0
0106 F30F1145 movss %xmm0, -8(%rbp)
.L7:
.LBE3:
0111 660FEFC0 pxor %xmm0, %xmm0
0115 F30F2A45 cvtsi2ss -28(%rbp), %xmm0
011a F30F104D movss -8(%rbp), %xmm1
011f F30F5EC8 divss %xmm0, %xmm1
0123 F30F114D movss %xmm1, .
Debt-equity ratio=DebtEQuityHence debt=1.1EquityTotal assets=De.pdf
Debt-equity ratio=Debt/EQuity
Hence debt=1.1Equity
Total assets=Debt+Equity
2975 =2.1Equity
Hence Equity=(2975/2.1)=$1416.67(Approx)
ROE=Net income/Equity
Hence net income=(1416.67*11%)
=$155.83(Approx).
Solution
Debt-equity ratio=Debt/EQuity
Hence debt=1.1Equity
Total assets=Debt+Equity
2975 =2.1Equity
Hence Equity=(2975/2.1)=$1416.67(Approx)
ROE=Net income/Equity
Hence net income=(1416.67*11%)
=$155.83(Approx)..
Gaseous chemical element, chemical symbol O, atom.pdf
Gaseous chemical element, chemical symbol O, atomic number 8. It constitutes
21% (by volume) of air and more than 46% (by weight) of Earth\'s crust, where it is the most
plentiful element. It is a colourless, odourless, tasteless gas, occurring as the diatomic molecule
O2. In respiration, it is taken up by animals and some bacteria (and by plants in the dark), which
give off carbon dioxide (CO2). In photosynthesis, green plants assimilate carbon dioxide in the
presence of sunlight and give off oxygen. The small amount of oxygen that dissolves in water is
essential for the respiration of fish and other aquatic life. Oxygen takes part in combustion and in
corrosion but does not itself burn. It has valence 2 in compounds; the most important is water. It
forms oxides and is part of many other molecules and functional groups, including nitrate,
sulfate, phosphate, and carbonate; alcohols, aldehydes, carboxylic acids, and ketones; and
peroxides. Obtained for industrial use by distillation of liquefied air, oxygen is used in
steelmaking and other metallurgical processes and in the chemical industry. Medical uses include
respiratory therapy, incubators, and inhaled anesthetics. Oxygen is part of all gas mixtures for
manned spacecraft, scuba divers, workers in closed environments, and hyperbaric chambers. It is
also used in rocket engines as an oxidizer (in liquefied form) and in water and waste treatment
processes.
Solution
Gaseous chemical element, chemical symbol O, atomic number 8. It constitutes
21% (by volume) of air and more than 46% (by weight) of Earth\'s crust, where it is the most
plentiful element. It is a colourless, odourless, tasteless gas, occurring as the diatomic molecule
O2. In respiration, it is taken up by animals and some bacteria (and by plants in the dark), which
give off carbon dioxide (CO2). In photosynthesis, green plants assimilate carbon dioxide in the
presence of sunlight and give off oxygen. The small amount of oxygen that dissolves in water is
essential for the respiration of fish and other aquatic life. Oxygen takes part in combustion and in
corrosion but does not itself burn. It has valence 2 in compounds; the most important is water. It
forms oxides and is part of many other molecules and functional groups, including nitrate,
sulfate, phosphate, and carbonate; alcohols, aldehydes, carboxylic acids, and ketones; and
peroxides. Obtained for industrial use by distillation of liquefied air, oxygen is used in
steelmaking and other metallurgical processes and in the chemical industry. Medical uses include
respiratory therapy, incubators, and inhaled anesthetics. Oxygen is part of all gas mixtures for
manned spacecraft, scuba divers, workers in closed environments, and hyperbaric chambers. It is
also used in rocket engines as an oxidizer (in liquefied form) and in water and waste treatment
processes..
B. Remember this - hydroboration attacks alkenes .pdf
B. Remember this - hydroboration attacks alkenes but not rings and it follows the
anti Markonikov rules where the H will attach to the carbon with the fewest alkyl substitiutions.
The oh with will by default (after 2)H2O2, OH-) attach to the other carbon.
Solution
B. Remember this - hydroboration attacks alkenes but not rings and it follows the
anti Markonikov rules where the H will attach to the carbon with the fewest alkyl substitiutions.
The oh with will by default (after 2)H2O2, OH-) attach to the other carbon..
a). EconomiserAn economiser is a mechanical device which is used a.pdf
a). Economiser
An economiser is a mechanical device which is used as a heat exchanger by preheating a fluid to
reduce energy consumption. In a steam boiler, it is a heat ex-changer device that heats up fluids
or recovers residual heat from the combustion product i.e. flue gases in thermal power plant
before being released through the chimney. Flue gases are the combustion exhaust gases
produced at power plants consist of mostly nitrogen, carbon dioxide, water vapor, soot carbon
monoxide etc. Hence, the economiser in thermal power plants, is used to economise the process
of electrical power generation, as the name of the device is suggestive of. The recovered heat is
in turn used to preheat the boiler feed water, that will eventually be converted to super-heated
steam. Thus, saving on fuel consumption and economising the process to a large extent, as we
are essentially gathering the waste heat and applying it to, where it is required. Nowadays
however, in addition to that, the heat available in the exhaust flue gases can be economically
recovered using air pre-heater which are essential in all pulverized coal fired boiler.
b). Reheater
ins some heat either from the combustion gases leaving the boiler (if there is still much
temperature difference) or by adding more heat by burning small amount of fuel. The energy of
the exhaust steam with the gained heat increases the steam enthalpy of the steam and give a good
opportunity to generate much work with the low pressure turbine. The two works (High pressure
W_H and Low pressure W_L) are much greater than the heat added by fuel burned (Q1 and Q2),
so the efficiency increases. Another issue, the reheat helps in saving the turbine blades from
corrosion due to low dryness fraction x<0.88 which it may occur if one single stage turbine is
used.
c). Cogeneration
Cogeneration (Combined Heat and Power or CHP) is the simultaneous production of electricity
and heat, both of which are used. The central and most fundamental principle of cogeneration is
that, in order to maximise the many benefits that arise from it, systems should be based on the
heat demand of the application. This can be an individual building, an industrial factory or a
town/city served by district heat/cooling. Through the utilisation of the heat, the efficiency of a
cogeneration plant can reach 90% or more.
Cogeneration therefore offers energy savings ranging between 15-40% when compared against
the supply of electricity and heat from conventional power stations and boilers.
Cogeneration optimises the energy supply to all types of consumers, with the followingbenefits
for both users and society at large:
d). Turbocharging
A turbocharger, or turbo (colloquialism), from Greek \"\" (\"wake\"),[1] (also from Latin
\"turbo\" (\"spinning top\"),[2]) is aturbine-driven forced induction device that increases an
internal combustion engine\'s efficiency and power output by forcing extra air into the
combustion chamber.[3][4] This improvement over a.
A) As it is shown this disease should be partly autosomal recessive .pdf
A) As it is shown this disease should be partly autosomal recessive one. Because it is not sex
linked disease and both the first row parents should be carriers. Because neither the square nor
circle is colourd.
B) Both the parents in generation (ii) can be carriers. But as far as no diseased person in
generation (iii), it is more likely to be only one of them is a carrier. If it was a sex linked trait the
female should have been the carrier.
C) Marriage between cousins led the occurrence of disease.
Solution
A) As it is shown this disease should be partly autosomal recessive one. Because it is not sex
linked disease and both the first row parents should be carriers. Because neither the square nor
circle is colourd.
B) Both the parents in generation (ii) can be carriers. But as far as no diseased person in
generation (iii), it is more likely to be only one of them is a carrier. If it was a sex linked trait the
female should have been the carrier.
C) Marriage between cousins led the occurrence of disease..
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