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yanalaym [24]
3 years ago
9

Most colleges require blank years of foriegn language for admission

Computers and Technology
1 answer:
Sladkaya [172]3 years ago
3 0

They require about 4 years of foreign language if you want to learn more afterwards then it is another 2 years or 3.

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What is the result of the following code?<br><br> x=7//2+10%2**4<br><br> print(x)
spin [16.1K]

Answer:

3

Explanation:

We can split the expression into two sections

First Section:

7 // 2 = 3. Using the floor division we only utilize the integer part.

Second Section:

10 % 2**4 = 0**4 = 0

Bringing back the full version expression we have

3 + 0 = 3

7 0
2 years ago
Josephine is in the process of creating ads within her Standard Display campaign. She finds that there are two main ad formats t
Dennis_Churaev [7]

Answer:

d.Responsive Display ads

e.Uploaded ads (Image & AMPHTML)

Explanation:

The two main ad formats used in a Standard Display campaign are;

1. Responsive Display ads: they are Google's new default ad format and are automatically created by Google using assets provided by users in a square and landscape format.

2. Uploaded ads (Image & AMPHTML): they are created using a tool like Google Web Designer to create adverts outside of Google Ads.

They can be uploaded as a jpeg, gif, zip file and png extension into Google Ads.

3 0
3 years ago
PYTHON:Given the dictionary, d, find the largest key in the dictionary and associate the corresponding value with the variable v
lapo4ka [179]

di = {5: 3, 4: 1, 12: 2}

val_of_max = di[max(di)]

print(val_of_max)

I hope this helps!

4 0
3 years ago
QUESTION: Which is not an example of a video port or cable?
Vikki [24]

Answer:

Radio Corporation of America (RCA)

Explanation:

A corporation is not a video connector.

8 0
3 years ago
A direct-mapped cache holds 64KB of useful data (not including tag or control bits). Assuming that the block size is 32-byte and
Cerrena [4.2K]

Answer:

A) Number of bits for byte = 6 bits

B) number of bits for index = 17 bits

C) number of bits for tag = 15 bits

Explanation:

Given data :

cache size = 64 kB

block size = 32 -byte

block address = 32 -bit

number of blocks in cache memory  

cache size / block size = 64 kb / 32 b = 2^11 hence the number of blocks in cache memory = 11 bits = block offset

A) Number of bits for byte

log _{2} (6)^2 = 6  bits

B) number of bits for index

block offset + byte number

= 11 + 6 = 17 bits

c ) number of bits for tag

= 32 - number of bits for index

= 32 - 17 = 15 bits

5 0
3 years ago
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