<em>The answer is Crimper.
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<em>Crimper is a networking device that would allow RJ-45 pins to get attached to UTP(Unshielded Twister Pair), CATs (Category Ns) and STPs (Shielded Twisted Pair) Cables. These are common type of cables used in computer networking. A crimper looks like a pliers only it has narrow ends with a hole where you would put RJ-45 devices together with the cables. In order to make the cable hold on to the RJ-45, the two hands must be squeezed together with an exact force (not too much force). Once the crimper made a sound, that means, your RJ-45 is now locked with the cables inside. </em>
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<em>You must be very careful when using this device. Sometimes crimpers has blades that could peel off cables or cut them for alignment and proper attachments. A pair of gloves and googles are also advisable to wear when you use this device for safety measures.</em>
Answer:
Think of some ideas of how their gonna create the new spaceship
Explanation:
Because I'm Smort (typo intended)
"Comparing the cost of care provided to Medicare beneficiaries assigned to primary care nurse practitioners and physicians" is a study that
<h3>What is the cost of Medicare according to the above study?</h3>
The study shows that after controlling for demographics, geography, comorbidities, and proclivity to visit an NP, Medicare evaluation and management payments for patients allocated to an NP were $207, or 29%, less than PCMD assigned beneficiaries.
The similar tendency was seen for inpatient and total office visit payment amounts, with NP-assigned beneficiaries receiving 11 and 18 percent less, respectively. The work component of relative value units yields similar results.
Learn more about Medicare:
brainly.com/question/1960701
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Answer:
See explaination for the code
Explanation:
def wordsOfFrequency(words, freq):
d = {}
res = []
for i in range(len(words)):
if(words[i].lower() in d):
d[words[i].lower()] = d[words[i].lower()] + 1
else:
d[words[i].lower()] = 1
for word in words:
if d[word.lower()]==freq:
res.append(word)
return res
Note:
First a dictionary is created to keep the count of the lowercase form of each word.
Then, using another for loop, each word count is matched with the freq, if it matches, the word is appended to the result list res.
Finally the res list is appended.
