Answer:
Step 1: Create a table reflecting the three potential scenarios and two possible outcomes. There should be a total of 8 distinct rules for the three varying conditions, structured similarly to the discount table shown here in the example: attached is the discount table
Step 2: The rules can now be made simpler by taking the following aspects into account:
a) When a consumer completes the survey form AND opts into the newsletter, according to Rules 1 and 2, they qualify for a discount if their order exceeds $100. This leads to two distinct rules being formulated while the third condition (order quantity) holds significance.
b) If the buyer fills out the survey form OR subscribes to the newsletter, as indicated by Laws 3, 4, 5, and 6, they will benefit from free shipping, regardless of the order amount. As a result, this situation can be divided into two individual rules, where at least one requirement is satisfied, but not both.
c) When a customer fails to meet any requirements, corresponding to Rules 7 and 8, the order value will not qualify for either free shipping or discount. This can be seen as a single law. The linked simplification table illustrates this.
Answer: Random fluctuations.
Explanation: A time series consists of data points ordered chronologically, arranged in equal intervals. Typically, this data sequence is systematic and has defined intervals. However, there’s no allowance for randomness, making unpredictable variations, or random fluctuations, absent in such series. Thus, option (D) is the correct choice.
<span>To enter global configuration mode, you need to type the enable command.</span><span>
The prompt indicating this mode concludes with the ">" symbol.</span>
Response:
C++ code provided below with suitable annotations
Clarification:
pattern.cpp
#include<iostream>
using namespace std;
void printCross(int n)
{
int i,j,k;
if(n%2) //odd number of lines
{
for(int i=n;i>=1;i--)
{
for(int j=n;j>=1;j--)
{
if(j==i || j==(n-i+1))
cout<<j;
else
cout<<" ";
}
cout<<"\n";
}
}
else //even number of lines
{
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if(j==i || j==(n-i+1))
{
cout<<" "<<j<<" ";
}
else
cout<<" ";
}
cout<<"\n";
}
}
void printForwardSlash(int n)
{
if(n%2)
{
for(int i=n;i>=1;i--)
{
for(int j=n;j>=1;j--)
{
if(j==n-i+1)
{
cout<<j;
}
else
cout<<" ";
}
cout<<"\n";
}
}
else
{
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if(j==(n-i+1))
{
cout<<j;
}
else
cout<<" ";
}
cout<<"\n";
}
}
}
void printBackwardSlash(int n)
{
if(n%2) // odd number of lines
{
for(int i=n;i>=1;i--)
{
for(int j=n;j>=1;j--)
{
if(j==i)
{
cout<<j;
}
else
cout<<" ";
}
cout<<"\n";
}
}
else //even number of lines
{
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if(j==i)
{
cout<<j;
}
else
cout<<" ";
}
cout<<"\n";
}
}
}
int main()
{
int num;
char ch;
cout<<"Create a numberes shape that can be sized."<<endl;
cout<<"Input an integer [1,50] and a character [x,b,f]."<<endl;
cin>>num>>ch;
if(ch=='x' || ch=='X')
printCross(num);
else if(ch=='f' || ch=='F')
printForwardSlash(num);
else if(ch=='b' || ch=='B')
printBackwardSlash(num);
else
cout<<"\nWrong input"<<endl;
return 0;
}
The question offers several choices;
<span>A) </span>Incremental budgeting.
B) Performance budgeting.
C) Program budgeting.
D) Target based budgeting.
I'd argue that D is the correct option; Target based budgeting.
Target based budgeting centers on achieving objectives and competing alternatives. This budgeting approach uses pricing as a strategic tool to influence sales outcomes. It involves market research to determine the precise selling price of a product.
