Answer:
Step-by-step explanation:
if the store marks up the price by 40%
119*(40/100)=47.60
119+47.60=166.60
Answer:
Step-by-step explanation:
The correct question should be:
We have been told that all we have to do to get profit is to subtract the cost from revenue.
Revenue = 
Cost <em>= </em>
<em />
<em />
Hence:
Profit = 
Putting it in ordered form:
This is an expression for the profit.
Answer:
G - number of arrangements
F. the amount of her allowance
B. y = x
H. the number of hours
How many jumps is it from the decimal point to the position to the right of the 6
The answer to that is 3 jumps. Since the move of the decimal made the number larger than it really is, it will take a negative value to put it back where it belongs.
So A) is the correct answer. It is in scientific notation and if you move the decimal back 3 place you get the number you stared with.
Answer:
Algorithm
Start
Int n // To represent the number of array
Input n
Int countsearch = 0
float search
Float [] numbers // To represent an array of non decreasing number
// Input array elements but first Initialise a counter element
Int count = 0, digit
Do
// Check if element to be inserted is the first element
If(count == 0) Then
Input numbers[count]
Else
lbl: Input digit
If(digit > numbers[count-1]) then
numbers[count] = digit
Else
Output "Number must be greater than the previous number"
Goto lbl
Endif
Endif
count = count + 1
While(count<n)
count = 0
// Input element to count
input search
// Begin searching and counting
Do
if(numbers [count] == search)
countsearch = countsearch+1;
End if
While (count < n)
Output count
Program to illustrate the above
// Written in C++
// Comments are used for explanatory purpose
#include<iostream>
using namespace std;
int main()
{
// Variable declaration
float [] numbers;
int n, count;
float num, searchdigit;
cout<<"Number of array elements: ";
cin>> n;
// Enter array element
for(int I = 0; I<n;I++)
{
if(I == 0)
{
cin>>numbers [0]
}
else
{
lbl: cin>>num;
if(num >= numbers [I])
{
numbers [I] = num;
}
else
{
goto lbl;
}
}
// Search for a particular number
int search;
cin>>searchdigit;
for(int I = 0; I<n; I++)
{
if(numbers[I] == searchdigit
search++
}
}
// Print result
cout<<search;
return 0;
}
