Answer:
Plain text, Rich Text, and HTML format
Explanation:
In Outlook 2016, three formats are allowed. You can send the plain text only or you can send using the Rich text format. However, there is another sending format as well that is allowed, and it is the HTML format. And by default, if you will let the Outlook choose the most appropriate sending format then the email message will be sent using the HTML format.
Answer:
Hub
Explanation:
A Hub is a networking tool linking of the different devices or other networking devices.In comparison to router or other a Hub has not routing tables or the guidance where it will sends the information and transmits the data in the network in the respective link.
The hub is used in the Microsoft Edge that provides a panel where we can access favorite web pages. The hub is also used in the history of the previously visited web pages .
Answer:
Time required is 0.007 s
Explanation:
As per the question:
Analog signal to digital bit stream conversion by Host A =64 kbps
Byte packets obtained by Host A = 56 bytes
Rate of transmission = 2 Mbps
Propagation delay = 10 ms = 0.01 s
Now,
Considering the packets' first bit, as its transmission is only after the generation of all the bits in the packet.
Time taken to generate and convert all the bits into digital signal is given by;
t = 
t =
(Since, 1 byte = 8 bits)
t = 7 ms = 0.007 s
Time Required for transmission of the packet, t':


In python:
lst = ([])
largest = 0
while len(lst) != 6:
user_number = int(input("Enter a number: "))
lst.append(user_number)
for i in lst:
if i > largest:
largest = i
print(largest)
I hope this helps