Mark, a programmer, is testing his program in order to locate syntax errors. Which skill will help Mark most in detecting and re

Answer:

input price

set change to one hundred minus price

set quarter to change divided by 25

set change to change mod 25

set dime to change divided by 10

set change to change mod 10

set nickel to change divided by 5

set change to change mod 5

set penny to change

print "You acquired an item for price cents and paid me a dollar. Your change is: "

if quarter is equal to 1

print quarter

else if quarter is greater than 1

print quarter with "quarters"

if dime is equal to 1

print dime

else if dime is greater than 1

print dime with "dimes"

if nickel is equal to 1

print nickel

else if nickel is greater than 1

print nickel with "nickels"

if penny is equal to 1

print penny

else if penny is greater than 1

print penny with "pennies"

Explanation:

A clearer way to express the pseudocode above could involve using alternative names for change as:

input price

set change to one hundred minus price

set quarter to change divided by 25

set quarter_remainder to change mod 25

set dime to quarter_remainder divided by 10

set dimes_remainder to quarter_remainder mod 10

set nickel to dimes_remainder divided by 5

set nickel_remainder to dimes_remainder mod 5

set penny to nickel_remainder

print "You purchased an item for" price"cents and provided a dollar. Your change is: "

if quarter is equal to 1

print quarter "quarter"

else if quarter is greater than 1

print quarter "quarters"

if dime is equal to 1

print dime "dime"

else if dime is greater than 1

print dime "dimes"

if nickel is equal to 1

print nickel "nickel"

else if nickel is greater than 1

print nickel "nickels"

if penny is equal to 1

print penny "penny"

else if penny is greater than 1

print penny "pennies"

Next, I will illustrate the pseudocode through an example:

Let's assume price = 75

Thus, change = 100 - price = 100 - 75 = 25

change = 25

quarter = change/25 = 25/25 = 1

quarter = 1

quarter_remainder = change % 25 = 25%25 = 0

dime = 0/10 = 0

At this point, all additional values are 0.

The following message will be displayed on the screen:

You acquired an item for 75 cents and paid me a dollar. Your change is:

Now the program proceeds to the conditional statement

if quarter == 1

This condition holds true because the value of quarter is 1

print quarter "quarter"

Subsequently, this line will appear on the screen:

1 quarter

Thus, the full output of the pseudocode is:

You acquired an item for 75 cents and paid me a dollar. Your change is:

1 quarter

Answer: Sole Proprietorship. This question seeks to identify the business type mentioned in the scenario. In a Sole Proprietorship, the owner effectively is the business, managing all elements from operations to finance. Typically, this is the initial business form in entrepreneurship. The description fits Doug well; he operates the farming business independently, taking care of all its functions. Thus, it qualifies as a sole proprietorship.

Answer:

He ought to report to a trusted adult and seek a resolution to the issue.

Explanation:

The situation could escalate and potentially involve others as well.

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;

}