Answer:
int main() {
int _2dArray[32][32];
for (int i = 0; i < 32; i++) {
for (int j = 0; j < 32; j++) {
_2dArray[i][j] = j + i * 32;
}
}
return 0;
}
Explanation:
Here is a generic C/C++ 2d array traversal and main function example. The rest you'll have to figure out based on what kind of app you're making.
Good luck!
The acronym RFID (Radio Frequency Identification) describes
networked devices that contain microcomputers but are not thought of as
computing devices, such as refrigerators, automobile components, light
bulbs, and industrial control devices. RFIDs are battery-powered sensors that gather and transmit data to a reading device. Some sensor based technologies are scanning electron microscopes, LiDAR,radar, GPS, x-ray, sonar, infrared and seismic.
Try {
AutoFactory.shutdown();
} catch (ProductionInProgressException e) {
AutoFactory.reset();
}
Answer:
Number of packets ≈ 5339
Explanation:
let
X = no of packets that is not erased.
P ( each packet getting erased ) = 0.8
P ( each packet not getting erased ) = 0.2
P ( X ≥ 1000 ) = 0.99
E(x) = n * 0.2
var ( x ) = n * 0.2 * 0.8
∴ Z = X - ( n * 0.2 ) / 
~ N ( 0.1 )
attached below is the remaining part of the solution
note : For the value of <em>n</em> take the positive number
