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777dan777 [17]
3 years ago
9

Ask what kind of pet the user has. If they enter cat, print "Too bad...", if they enter dog, print "Lucky you!" (You can change

the messages if you like). Once this works, add other pets. (Iguana, Pig, Rabbit...)
Computers and Technology
1 answer:
nirvana33 [79]3 years ago
8 0

Answer:

Explanation:

a = input("what kind of pet the user has")

if a == 'cat':

  print("Too bad")

elif a == 'dog':

   print("Lucky you!")

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When on a LAN switch DHCP snooping is configured the networks that can be accessed by which clients?
kompoz [17]

Answer:

The network is accesible just for the whitelisted clients configured in the switch connected to the DHCP server

Explanation:

DHCP snooping main task is to prevent an unauthorized DHCP server from entering the our network.

Basically, in the switch we define the ports on which the traffic of the reliable DHCP server can travel. That is, we define as “trust” the ports where we have DHCP servers, DHCP relays and trunks between the switches.

8 0
4 years ago
The illustration shows different types of text language.
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nawww dude im good look it up

7 0
3 years ago
Read 2 more answers
Suppose we want an error-correcting code that will allow all single-bit errors to be corrected for memory words of length 10.
Greeley [361]

Answer:

See attached picture.

Explanation:

See attached picture.

5 0
4 years ago
(a) How many locations of memory can you address with 12-bit memory address? (b) How many bits are required to address a 2-Mega-
I am Lyosha [343]

Answer:

Follows are the solution to this question:

Explanation:

In point a:

Let,

The address of 1-bit  memory  to add in 2 location:

\to \frac{0}{1}  =2^1  \ (\frac{m}{m}  \ location)

The address of 2-bit  memory to add in 4 location:

\to \frac{\frac{00}{01}}{\frac{10}{11}}  =2^2  \ (\frac{m}{m}  \ location)

similarly,

Complete 'n'-bit memory address' location number is = 2^n.Here, 12-bit memory address, i.e. n = 12, hence the numeral. of the addressable locations of the memory:

= 2^n \\\\ = 2^{12} \\\\ = 4096

In point b:

\to Let \  Mega= 10^6

              =10^3\times 10^3\\\\= 2^{10} \times 2^{10}

So,

\to 2 \ Mega =2 \times 2^{20}

                 = 2^1 \times 2^{20}\\\\= 2^{21}

The memory position for '2^n' could be 'n' m bits'  

It can use 2^{21} bits to address the memory location of 21.  

That is to say, the 2-mega-location memory needs '21' bits.  

Memory Length = 21 bit Address

In point c:

i^{th} element array addresses are given by:

\to address [i] = B+w \times (i-LB)

_{where}, \\\\B = \text {Base  address}\\w= \text{size of the element}\\L B = \text{lower array bound}

\to B=\$ 52\\\to w= 4 byte\\ \to L B= 0\\\to address  = 10

\to address  [10] = \$ 52 + 4 \times (10-0)\\

                       =   \$ 52   + 40 \ bytes\\

1 term is 4 bytes in 'MIPS,' that is:

= \$ 52  + 10 \ words\\\\ = \$ 512

In point d:

\to  base \ address = \$ t 5

When MIPS is 1 word which equals to 32 bit :

In Unicode, its value is = 2 byte

In ASCII code its value is = 1 byte

both sizes are  < 4 byte

Calculating address:

\to address  [5] = \$ t5 + 4 \times (5-0)\\

                     = \$ t5 + 4 \times 5\\ \\ = \$ t5 + 20 \\\\= \$ t5 + 20  \ bytes  \\\\= \$ t5 + 5 \ words  \\\\= \$ t 10  \ words  \\\\

3 0
4 years ago
What do you find when you first open a word processor and how do you adjust the document Indentitions
Ilia_Sergeevich [38]

When you first open a word processor document you will find a document with the default indents and margins.

In order to adjust the document indentations the easiest way to do this is to place the cursor where you want the indent to be and then adjust the indent arrow on the ruler at the top of the document to the appropriate value.

7 0
3 years ago
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