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Gwar [14]
4 years ago
11

The main trade-off that all investors must consider is

Computers and Technology
2 answers:
Talja [164]4 years ago
3 0
<span>The main trade-off that all investors must consider is Risk vs Return
In the end, all the techniques that implemented in investing process is aimed for nothing other than profit.
Current market trend dictates that potential return tend to be higher the riskier the investment is and vice versa.
</span>
Inessa05 [86]4 years ago
3 0
<h2>Answer:</h2>

The main trade-off that all investors must consider is the <u>risk involved and the return gained</u>.

<h2>Explanation:</h2>

For the income to multiply at a faster rate, investors put their money in finance securities like stocks, bonds, fixed deposits, banks, mutual funds, etc. Whatever be the finance element, every investor must look at potential risks involved.

Despite the interest rates, diversification, uniformity, performance of the securities over a long period of time, it is always safe to consider the risk aspect as it can wash out the whole money and put the individual under crisis. So, every investor should understand the policies of the securities better where potential risks can be taken care along with the returns.

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Write an abstract data type for a queue in Java whose elements are generic elements of type T that have an associated integer pr
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Answer:

See explaination

Explanation:

#include<iostream>

const int MAX=20;

class dque

{

private:

int arr[MAX];

int front,rear;

public:

dque();

void addqatbeg(char item);

void addqatend(char item);

int delqatbeg();

int delqatend();

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};

dque::dque()

{

front=rear=-1;

for(int i=0;i<MAX;i++)

arr[i]=0;

}

void dque::addqatbeg(int item)

{

if(front==0&&rear==MAX-1){cout<<"\nDeque is full"<<endl;return;}

if(front==-1){front=rear=0;arr[front]=item;return;}

if(rear!=MAX-1)

{int c=count();int k=rear+1;

for(int i=1;i<=c;i++){arr[k]==arr[k-1];k--;}

arr[k]=item;front=k;rear++;

}

else{front--;arr[front]=item;}

}

void dque::addqatend(int item)

{

if(front==0&&rear==MAX-1){cout<<"\nDeque is full"<<endl;return;}

if(front==-1){front=rear=0;arr[rear]=item;return;}

if(rear==MAX-1){

int k=front-1;

for(int i=front-1;i<rear;i++){k=i;

if(k==MAX-1)arr[k]=0;

else arr[k]=arr[i+1];

}rear++;

arr[rear]=item;

}

int dque::delqatbeg()

{

if(front==-1){cout<<"\nDeque is empty"<<endl;return0;}

int item=arr[front];

arr[front]=0;

if(front==rear)front=rear=-1;

else

front++;

return item;

}

int dque::delqatend()

{

if(front==-1){cout<<"\nDeque is empty"<<endl;return 0;}

int item=arr[rear];

arr[rear]=0;

rear--;

if(rear==-1)front=-1;return item;

}

void dque::display()

{

cout<<endl<<"front->";

for(int i=0;i<MAX;i++)cout<<""<<arr[i];cout<<"<-rear";

}

int dque::count()

{

int c=0;

for (int i=0;i<MAX;i++){if(arr[i]!=0)c++;}

return c;}

}

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