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kherson [118]
3 years ago
11

Which of these components is a part of the Central Processing Unit (CPU) of a computer?

Computers and Technology
2 answers:
WARRIOR [948]3 years ago
7 0

Answer: CONTROL

Explanation: PLATO

ladessa [460]3 years ago
5 0

Answer:

Control unit

Explanation:

central Processing unit has two components which include: Arithmetic and Logic Unit (ALU) that perform arithmetic and logical operation and Control Unit (CU) which fetches instructions from memory and decodes as well executes them

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Write a program with a function that accepts a string as an argument and returns the number of uppercase, lowercase, vowel, cons
diamong [38]

Answer:

Explanation:

The following code is written in Python. It is a function called checkString that takes in a string as an argument and loops through each char in that string and checking to see if it is lowercase, uppercase, vowel, consonant, or punctuations. It adds 1 to the correct variable. At the end of the loop it prints out all of the variables. The picture below shows a test output with the string "Brainly, Question."

def checkString(word):

   uppercase = 0

   lowercase = 0

   vowel = 0

   consonants = 0

   punctuation = 0

   vowelArray = ['a', 'e', 'i', 'o','u', 'y' ]

   consonantArray = ['b', 'c', 'd', 'f', 'g', 'h', 'j', 'k', 'l', 'm', 'n', 'p', 'q', 'r', 's', 't', 'v', 'w', 'x', 'y', 'z']

   punctuationArray = ['.', '!', '?', ',', ';', '(', ')']

   for char in word:

       if char.isupper():

           uppercase += 1

       else:

           lowercase += 1

       if char.lower() in vowelArray:

           vowel += 1

       elif char.lower() in consonantArray:

           consonants += 1

       if char in punctuationArray:

           punctuation += 1

   print('Uppercase: ' + str(uppercase))

   print('Lowercase: ' + str(lowercase))

   print('Vowels: ' + str(vowel))

   print('Consonants: ' + str(consonants))

   print('Punctuations: ' + str(punctuation))

8 0
3 years ago
. When would one use the analytic application fraud detection?
vaieri [72.5K]

Answer:Fraud detection through analytical method is used for detection of the fraud transactions,bribe activity etc in companies, business,etc. This techniques helps in the reduction of financial frauds in the organization, have the control over company to protect it,decrease in the fraud associated costs etc.

It has the capability of identifying the fraud which has happened or going to happen through the analytical ways and human interference. The organizations or companies require efficient processing and detection system for identification of such false happening.

4 0
3 years ago
Anyone use zoom<br><br>code:- 2574030731<br>pass:- HELLO<br>Z●●M ​
r-ruslan [8.4K]

Answer:

ok be there in a sec

Explanation:

3 0
3 years ago
Read 2 more answers
Which Operating System is used most often in households?
pshichka [43]

Answer:

Windows

Explanation:

Windows are one of the most popular household os.

3 0
2 years ago
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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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