Answer:
its okay. i mean like you have to do a bunch off stuff. andsomtimes it gets a little boring.
Explanation:
i am doing online school.
Hi,
JVM - Java Virtual Machine
Hope this helps.
r3t40
Answer:
names = ['Peter', 'Bruce', 'Steve', 'Tony', 'Natasha', 'Clint', 'Wanda', 'Hope', 'Danny', 'Carol']
numbers = [100, 50, 10, 1, 2, 7, 11, 17, 53, -8, -4, -9, -72, -64, -80]
for index, element in enumerate(names):
if index % 2 == 0:
print(element)
for num in numbers:
if num >= 0:
print(num, end = " ")
count = 0
for i in numbers:
count += i
avg = count/len(numbers)
print("sum = ", count)
print("average = ", avg)
for num in numbers:
if num % 2 != 0:
print(num, end = " ")
Explanation:
I'm stuck on the last two.. I have to do those too for an assignment.
Complete Question:
Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For first collision, if K=100, how long does the adapter wait until sensing the channel again for a 1 Mbps broadcast channel? For a 10 Mbps broadcast channel?
Answer:
a) 51.2 msec. b) 5.12 msec
Explanation:
If K=100, the time that the adapter must wait until sensing a channel after detecting a first collision, is given by the following expression:
The bit time, is just the inverse of the channel bandwidh, expressed in bits per second, so for the two instances posed by the question, we have:
a) BW = 1 Mbps = 10⁶ bps
⇒ Tw = 100*512*(1/10⁶) bps = 51.2*10⁻³ sec. = 51.2 msec
b) BW = 10 Mbps = 10⁷ bps
⇒ Tw = 100*512*(1/10⁷) bps = 5.12*10⁻³ sec. = 5.12 msec
Answer:
ummm what does this have to do with school work?
Explanation:
