Answer:
=vlookup(h14, a5:h11,8,false)
Explanation:
Here, h15 is the cell in which we need the output and the value to be matched in rows to find the exact row is h14, and the range is expressed as a5:h11. Now to find the column number, we need to figure out the first and the concerned column. So, the concerned column is in which the total attendance is being listed and it is h(h14), and the first column is a. So, the column number is a=1 h=8, =8. And since we need the exact match, the value of the fourth argument is false. And thus, we have the above formula. Remember. vlookup formula is:
=vlookup(cell where the result is to be placed, range, column number in the same row as h14, exact match or approximate match). For exact match it's false, and for the approximate match, it's true.
V
JJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJ
Pros:
-They can share information between different medical offices quicker than they could before hand.
-It is cheaper to send information via the internet than it is to hand it by hand
Cons:
-Hackers can get access to the medical patient's information.
-The upkeep costsof the internet connection may lower their income.
Answer:
- def Lambda(strList):
- return list(filter(lambda s: (s.startswith("e")), strList))
-
- print(Lambda(["meaning", "cart", "engine", "egg"]))
Explanation:
The solution code is written in Python 3.
Lambda function is an anonymous function. It is a function without any name and it is started with keyword lambda. To write a lambda function to filter the string started with an 'e', we can write a lambda expression, s.startswith("e") that work on one input strList. If any word from strList started with letter "e", the word will be added into a list generated by filter function. At the end, the Lambda function will return the list of strings as output.
When we test the Lambda function using the sample string list (Line 4), we shall get ['engine', 'egg'] printed to terminal.
Yes , it’s true. In a known-plaintext attack (kpa), the cryptanalyst can only view a small portion of encrypted data, and he or she has no control over what that data might be.
The attacker also has access to one or more pairs of plaintext/ciphertext in a Known Plaintext Attack (KPA). Specifically, consider the scenario where key and plaintext were used to derive the ciphertext (either of which the attacker is trying to find). The attacker is also aware of what are the locations of the output from key encrypting. That is, the assailant is aware of a pair. They might be familiar with further pairings (obtained with the same key).
A straightforward illustration would be if the unencrypted messages had a set expiration date after which they would become publicly available. such as the location of a planned public event. The coordinates are encrypted and kept secret prior to the event. But when the incident occurs, the attacker has discovered the value of the coordinates /plaintext while the coordinates were decrypted (without knowing the key).
In general, a cipher is easier to break the more plaintext/ciphertext pairs that are known.
To learn more about Plaintext Attack click here:
brainly.com/question/28445346
#SPJ4