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Yakvenalex [24]
3 years ago
9

Let’s say you’re publishing a message with the Hootsuite Composer. The message contains a link to a landing page, and you want t

o add a UTM parameter to that link to track click-throughs. What feature in the Composer would you use to do this?
Computers and Technology
1 answer:
myrzilka [38]3 years ago
6 0

Answer:

Ow.ly shortener

Explanation:

Ow.ly is a good example of a link shortener that helps shorten the UTM link in Hootsuite composer.

Cheers

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Heya!!<br><br>•DEFINE DATA SCIENCE??<br>(∩_∩)<br><br>#kavya#<br>​
anastassius [24]

Answer:

Explanation:

Data science defined

Data science encompasses preparing data for analysis, including cleansing, aggregating, and manipulating the data to perform advanced data analysis. Analytic applications and data scientists can then review the results to uncover patterns and enable business leaders to draw informed insights.

5 0
2 years ago
On a printed circuit board, electronic components will be mounted from the
Likurg_2 [28]
On a printed circuit board, electronic parts will be mounted from the substrate side of the board. The leads jab through the substrate and the copper sheeting that has been carved. The leads are then soldered to the copper.

I hope the answer will help you. 
3 0
3 years ago
def getCharacterForward(char, key): """ Given a character char, and an integer key, the function shifts char forward `key` steps
zhenek [66]

Answer:

  1. def getCharacterForward(char, key):  
  2.    charList = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
  3.    if(len(char) > 1):
  4.        return None  
  5.    elif(not isinstance(key, int)):
  6.        return -1
  7.    else:
  8.        index = charList.find(char)
  9.        if(index + key <= 25):
  10.            return charList[index + key]
  11.        else:
  12.            return charList[(index + key)% 26]
  13. print(getCharacterForward("C", 4))
  14. print(getCharacterForward("X", 4))

Explanation:

Firstly, define a charList that includes all uppercase alphabets (Line 2). We presume this program will only handle uppercase characters.

Follow the question requirement and define necessary input validation such as checking if the char is a single character (Line 4). We can do the validation by checking if the length of the char is more than 1, if so, this is not a single character and should return None (Line 5). Next, validate the key by using isinstance function to see if this is an integer. If this is not an integer return -1 (Line 6 - 7).

Otherwise, the program will proceed to find the index of char in the charList using find method (Line 9). Next, we can add the key to index and use the result value to get forwarded character from the charList and return it as output (Line 11).

However, we need to deal a situation that the char is found at close end of the charList and the forward key steps will be out of range of alphabet list. For example the char is X and the key is 4, the four steps forward will result in out of range error. To handle this situation, we can move the last two forward steps from the starting point of the charList. So X move forward 4 will become B. We can implement this logic by having index + key modulus by 26 (Line 13).  

We can test the function will passing two sample set of arguments (Line 15 - 16) and we shall get the output as follows:

G

B

8 0
3 years ago
Answer if you play PS4 apex legends if you do answer username and first to answer is brainliest
Vinil7 [7]

Answer:

I do not, sorry

Explanation:

3 0
3 years ago
Read 2 more answers
A datagram network allows routers to drop packets whenever they need to. The probability of a router discarding a packetis p. Co
tresset_1 [31]

Answer:

a.) k² - 3k + 3

b.) 1/(1 - k)²

c.) k^{2}  - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }

Explanation:

a.) A packet can make 1,2 or 3 hops

probability of 1 hop = k  ...(1)

probability of 2 hops = k(1-k)  ...(2)

probability of 3 hops = (1-k)²...(3)

Average number of probabilities = (1 x prob. of 1 hop) + (2 x prob. of 2 hops) + (3 x prob. of 3 hops)

                                                       = (1 × k) + (2 × k × (1 - k)) + (3 × (1-k)²)

                                                       = k + 2k - 2k² + 3(1 + k² - 2k)

∴mean number of hops                = k² - 3k + 3

b.) from (a) above, the mean number of hops when transmitting a packet is k² - 3k + 3

if k = 0 then number of hops is 3

if k = 1 then number of hops is (1 - 3 + 3) = 1

multiple transmissions can be needed if K is between 0 and 1

The probability of successful transmissions through the entire path is (1 - k)²

for one transmission, the probility of success is (1 - k)²

for two transmissions, the probility of success is 2(1 - k)²(1 - (1-k)²)

for three transmissions, the probility of success is 3(1 - k)²(1 - (1-k)²)² and so on

∴ for transmitting a single packet, it makes:

     ∞                             n-1

T = ∑ n(1 - k)²(1 - (1 - k)²)

    n-1

   = 1/(1 - k)²

c.) Mean number of required packet = ( mean number of hops when transmitting a packet × mean number of transmissions by a packet)

from (a) above, mean number of hops when transmitting a packet =  k² - 3k + 3

from (b) above, mean number of transmissions by a packet = 1/(1 - k)²

substituting: mean number of required packet =  k^{2}  - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }

6 0
3 years ago
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