Answer:
Step-by-step explanation:
yesssss
<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>
Answer:
Step-by-step explanation:
So 11 times 2=22 so you times 11 times 2 on all sides so the answers is 22
Answer:
0.5372
Step-by-step explanation:
Given that the number of births that occur in a hospital can be assumed to have a Poisson distribution with parameter = the average birth rate of 1.8 births per hour.
Let X be the no of births in the hospital per hour
X is Poisson
with mean = 1.8
the probability of observing at least two births in a given hour at the hospital
= 
the probability of observing at least two births in a given hour at the hospital = 0.5372
Answer:
Do you mean y? if you did it should be 1/6 or 1.666666 repeating.
Step-by-step explanation:
I needed the 6y to = 1 so I did the inverse of it by and got 1 so y should = 1/6
