Answer:
1/21.
Step-by-step explanation:
There are 9 digits and the total number of permutations of 3 from 9 is 9P3
= 9!/6! = 504.
There are 4 even digits so the number of permutations of 3 from these 4 is 4! (4-3)! = 4*3*2 = 24.
So the required probability = 24/504
= 1/21.
I would say the answer is D. 7837. Since the common ratio between the x’s and y’s were are 39.5 so multiplying that times 200 gives something around 7900 so, (rounding down) the answer would be D. I really hope that helped! Also please correct me if I am wrong... thanks!
9
Same thing applies to this answer as to the last answer I posted
<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>
