<span>#include <iostream>
using namespace std;
class InventoryTag {
public:
InventoryTag();
int getQuantityRemaining() const;
void addInventory(int numItems);
private:
int quantityRemaining;
};
InventoryTag::InventoryTag() {
quantityRemaining = 0;
}
int InventoryTag::getQuantityRemaining() const {
return quantityRemaining;
}
void InventoryTag::addInventory(int numItems) {
if (numItems > 10) {
quantityRemaining = quantityRemaining + numItems;
}
}
int main() {
InventoryTag redSweater;
int sweaterShipment = 0;
int sweaterInventoryBefore = 0;
sweaterInventoryBefore = redSweater.getQuantityRemaining();
sweaterShipment = 25;
cout << "Beginning tests." << endl;
// FIXME add unit test for addInventory
/* Your solution goes here */
cout << "Tests complete." << endl;
return 0;
}</span>
To solve, isolate the x. Cross multiply
3/13(13)(5) = x/5(5)(13)
3(5) = (13)x
Simplify
13x = 15
Isolate the x. Divide by 13.
13x/13 = 15/13
x = 15/13
x = 1.2
1.2 is your answer
hope this helps
Answer:
D.A(n) = 250 + (n – 1)(0.03 • 250); $347.50
Step-by-step explanation:
we know that
The simple interest formula is equal to
where
A is the Final Investment Value
P is the Principal amount of money to be invested
r is the rate of interest
t is Number of Time Periods
in this problem we have
substitute in the formula above
Answer:
11
Step-by-step explanation:
88 ÷ (11 * 2) ÷ 4
88 ÷ 22 ÷ 4 = 11
