Answer:
16-bit wide
Explanation:
In order to find the width of the address bus, we need first to know how many memory cells it is needed to address.
If the size memory is 64 KB, this means that the memory size, in bytes, is equal to the following quantity:
64 KB = 2⁶ * 2¹⁰ bytes = 2¹⁶ bytes.
In order to address this quantity of cell positions, the address bus must be able to address 2¹⁶ bytes, so it must have 16-bit wide.
Answer:
C = 292 Mbps
Explanation:
Given:
- Signal Transmitted Power P = 250mW
- The noise in channel N = 10 uW
- The signal bandwidth W = 20 MHz
Find:
what is the maximum capacity of the channel?
Solution:
-The capacity of the channel is given by Shannon's Formula:
C = W*log_2 ( 1 + P/N)
- Plug the values in:
C = (20*10^6)*log_2 ( 1 + 250*10^-3/10)
C = (20*10^6)*log_2 (25001)
C = (20*10^6)*14.6096
C = 292 Mbps
Answer:
Explanation:
class Pet:
def __init__(self):
self.name = ''
self.age = 0
def print_info(self):
print('Pet Information:')
print(' Name:', self.name)
print(' Age:', self.age)
class Dog(Pet):
def __init__(self):
Pet.__init__(self)
self.breed = ''
def main():
my_pet = Pet()
my_dog = Dog()
pet_name = input()
pet_age = int(input())
dog_name = input()
dog_age = int(input())
dog_breed = input()
my_pet.name = pet_name
my_pet.age = pet_age
my_pet.print_info()
my_dog.name = dog_name
my_dog.age = dog_age
my_dog.breed = dog_breed
my_dog.print_info()
print(' Breed:', my_dog.breed)
main()
Answer:
b i think i dont see any dial caliper
Explanation: