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Ghella [55]
2 years ago
14

You are between flights at the airport and you want to check your email. Your wireless settings show two options:

Computers and Technology
1 answer:
Likurg_2 [28]2 years ago
8 0
The correct answers Is B
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Q10: Putting it all together (0.75 points) Here, we'll update the values in dictionary, storing the output in a dictionary calle
Molodets [167]

Answer:

# Code in Python

dictionary={'A':1,'B':2,'C':3,'D':4}

other_dictionary={}

for keys in dictionary:

if dictionary[keys]&1==1:

temp=dictionary[keys]*dictionary[keys]-10*10

other_dictionary[keys]=temp

else:

other_dictionary[keys]=dictionary[keys]

print(other_dictionary)  

assert other_dictionary

Explanation:

  • Initialize a sample example dictionary  and other_dictionary.
  • Do a binary comparision for checking odd number .
  • Update the the value stored in the dictionary to store the squared difference of the original value and '10'.
  • For even: store the original value (from dictionary).
4 0
3 years ago
For those that play pc games how do you go outside in The Sims 2?
yanalaym [24]

Answer:

My sister plays a lot of the sims 4 but I do recall she also played the Sims 2 so when she would go outside she would command them to do that.

Explanation:

If this does not work I would suggest you get a video that can explain it better than I possibly would? Considering the Sims 2 was made a while back you would maybe want somebody with a little more experience.

5 0
2 years ago
Read 2 more answers
The 7-bit ASCII code for the character ‘&’ is: 0100110 An odd parity check bit is now added to this code so 8 bits are trans
Alex787 [66]

Answer

First part:

The transmitted 8-bit sequence for ASCII character '&' with odd parity will be 00100110. Here leftmost bit is odd parity bit.

Second part:

The invalid bit sequence are option a. 01001000 and d. 11100111

Explanation:

Explanation for first part:

In odd parity, check bit of either 0 or 1 is added to the binary number as leftmost bit for making the number of 1s in binary number odd.

If there are even number of 1s present in the original number then 1 is added as leftmost bit to make total number of 1s odd.

If there are odd number of 1s present in the original number then 0 is added as leftmost bit to keep the total number of 1s odd.  

Explanation for second part:

A valid odd parity bit sequence will always have odd number of 1s.

Since in option a and d,  total number of 1s are 2 and 6 i.e. even number. Therefore they are invalid odd parity check bit sequences.

And since in option b and c, total number of 1s are 5 and 7 i.e. odd numbers respectively. Therefore they are valid odd parity check bit sequences.

7 0
3 years ago
Describe how keyboard software works and which problems does keyboard software<br> need to handle?
Anastaziya [24]

Answer:

The target computer's operating system and gain unauthorized access to the hardware, hook into the keyboard with functions provided by the OS, or use remote access software to transmit recorded data out of the target computer to a remote location.

No physical sensation of pressing a button results in misfires. Lack of physical divisions between keys makes touch typing impossible. Smaller key spacing results in typos.

3 0
2 years ago
Output values below an amount
valina [46]

Answer:

Explanation:

The following code is written in Python and it is a function called filter, which takes in the requested inputs. It then places the threshold in its own variable and pops it out of the array of inputs. Then it loops through the array comparing each element with the threshold and printing out all of the ones that are less than or equal to the threshold.

def filter():

   num_of_elements = input("Enter a number: ")

   count = 0

   arr_of_nums = []

   while count <= int(num_of_elements):

       arr_of_nums.append(input("Enter a number: "))

       count += 1

   threshold = arr_of_nums[-1]

   arr_of_nums.pop(-1)

   print(threshold)

   for element in arr_of_nums:

       if int(element) < int(threshold):

           print(element)

       else:

           continue

7 0
2 years ago
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