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STatiana [176]
2 years ago
6

Which of the following is not a common input device?

Computers and Technology
1 answer:
Debora [2.8K]2 years ago
3 0

The device which provides any piece of information from a computer into human-readable form is known as the Output device.

Explanation:

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The purchase and subsequent sale of a securities position in a customer account solely to generate commissions is____________.
Citrus2011 [14]

Answer:

Churning

Explanation:

Churning is termed as an act of a broker conducting immoderate trading in the account of client solely to generate commissions. It is an illegal and deceptive practice. It violates security laws. The purchase and subsequent sale of a securities that are little or insignificant to meet the investment goals of client can be the evidence of churning. Consequently it causes considerable losses in client's account or can produce a tax liability.

Churning occurs due to over trading by a broker to generate commissions by buying and selling stocks excessively on the behalf of investor. This often happens when broker has permissive authority over client's account.

5 0
3 years ago
Where must virtualization be enabled for VM (virtual machine) software to work?
Lelu [443]

Answer:

b) BIOS/UEFI

Explanation:

Virtualization can be defined as a technique used for the creation of a virtual platform such as a storage device, operating system, server, desktop, infrastructure or computing resources so as to enable the sharing of resources among multiple end users. Virtualization is usually implemented on a computer which is referred to as the "host" machine.

Generally, virtualization must be enabled in the BIOS/UEFI for VM (virtual machine) software to work.

BIOS is an acronym for Basic Input/Output System while UEFI is an acronym for Unified Extensible Firmware Interface. BIOS/UEFI are low-level software that serves as an intermediary between the operating systems and the computer's firmware or hardware components. The UEFI is actually an improvement of the BIOS and as such is a modernized software.

Basically, the BIOS/UEFI is a software which is an essential tool or feature which must be enabled to link the virtual machine with the hardware components of the computer.

5 0
2 years ago
What does the word "e-business" mean?
Zepler [3.9K]
It stands for Electronic Business, a typical example of this is a online store.
4 0
3 years ago
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quester [9]

Answer:

Network administrator

5 0
3 years ago
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1. Scrieţi un program care citeşte un număr natural n şi determină produsul cifrelor impare ale lui n. De exemplu, pentru n = 23
saveliy_v [14]

Answer:

1.  

num1 = input("Enter the value of n")  

n = int(num1)  

def calc(n):  

Lst = []  

while n > 0:  

remainder = n % 10  

Lst.append(remainder)  

quotient = int(n / 10)  

n = quotient  

i = len(Lst) - 1  

sum = 1  

for i in range(0, len(Lst)):  

if(i % 2 != 0):  

sum *=Lst[i];  

else:  

continue  

print(sum)  

return(0)  

\ r2

num=input("Enter the value of n")

arr=[12,-12,13,15,-34,-35,35,42]

def div7positive():

  sum = 0

  m = 0

  i = 0

  while m <= 7:

      if(arr[i]>=0 and arr[i]%7 == 0):

          sum +=arr[i]

      i = i + 1

      m += 1

       

   

  print("sum of number divisible by 7 and positive is", +sum);

div7positive()

\ r

\ r3.Write an algorithm that reads a natural number n and calculates the sum:

\ rS = 1 / (1 * 2) + 1 / (2 * 3) + 1 / (3 * 4) +… + 1 / ((n-1) * n)

\ r

num=input("Enter the value of n")

def sumseries():

  sum = 0

  m= 0

  while m <= 7:

      sum +=1/((int(num)-1)*int(num))

      m += 1

  print("sum of series is", +sum)

sumseries()

\ R4. Read a natural number n. Calculate the sum of its own divisors n. For example, for n = 12, the sum of its own divisors is 2 + 3 + 4 + 6 = 15

num=input("Enter the value of n")

def sumdivisors():

  sum = 0

  m= 2

  while m <= int(num):

      if int(num) % m == 0:

          sum += m

      m += 1

  print("sum of divisors is", +sum)

sumdivisors()\ r

\ R5. We read a natural number n and then whole numbers. Calculate and display the sum of the natural numbers between 10 and 100. For example, if n = 5 and then read 30, –2, 14, 200, 122, then the sum will be 44 (that is, 30 + 14).

\ r

Lst = []  

num=input("Enter the value of n")

i = 0

while i<= int(num):

  num1=input("Enter the element of array")

  Lst.append(int(num1));

  i += 1

def numbet0and100():

  sum = 0

  m= 0

  while m <= int(num):

      if Lst[m] <= 100 or Lst[m] >=0:

          sum += Lst[m]

      m += 1

  print("sum of numbers between 0 and 100 is", +sum)

numbet0and100()

\ R6. A natural number n of maximum 4 digits is read. How many digits are in all numbers from 1 to n? For example, for n = 14 there are 19 digits, and for n = 9 there are 9 digits.

\ r

num = input("Enter the value of n")

n = int(num)

print(“sum as required is:” (n -9) + n)  

 

\ R7. Read the natural numbers n and S, where n can be 2, 3, 4 or 5. Show all the numbers of n digits that have the numbers in strictly ascending order, and the sum of the digits is S. For example, for n = 2 and S = 10, 19, 28, 37, 46 will be displayed.

num = input("Enter the value of n")

n = int(num)

Lst = []

def calc(n):

  i = 1

  total = pow(10,n)

  while i <= total:

      j = i

      while j > 0:

          remainder = j % 10  

          Lst.append(remainder)

          quotient = int(j / 10)

          if quotient > 0:

              j = quotient

          else:

              length = len(Lst) - 1

              sum = 0

              while length >= 0:

                  sum += Lst[length]

                  length = length - 1

              if sum == pow(10,n):

                  print(j)

              k = len(Lst)

              del Lst[0:k]

      i = i + 1

  return(0)

calc(n)

\ r

\ R8. We consider the row 1, 1, 2, 3, 5, 8, 13, ... in which the first two terms are 1, and any other term is obtained from the sum of the preceding two. Write an algorithm that reads a natural number n and displays the first n terms of this string. For example, for n = 6, 1, 1, 2, 3, 5, 8 will be displayed.

\ r

nterm = int(input("What number of terms do you want?"))

a, b = 0, 1

totalcount = 0

if nterm <= 0:

 print("Please input a positive number")

elif nterm == 1:

 print("Fibonacci number upto which",nterm,":")

 print(a)

else:

 print("Fibonacci series:")

 print(nterm)

 while totalcount < nterm:

     print(a)

     nth = a + b

     a = b

     b = nth

     totalcount += 1

 

\ 9. Write an algorithm that reads two natural numbers n1 and n2 and displays the message "yes" if the sum of the squares of the digits of n1 is equal to the sum of the numbers of n2 or "no" otherwise. For example, for n1 = 232 and n2 = 881, "yes" will be displayed, and for n1 = 45 and n2 = 12, "no" will be displayed.

num1 = input("Enter the value of n")

num3 = input("Enter the value of n")

n = int(num1)

num2 = int(num3)

def calc(n):

  Lst = []

  while n > 0:

      remainder = n % 10  

      Lst.append(remainder)

      quotient = int(n / 10)

      n = quotient

       

  i = len(Lst) - 1

  sum = 0

  while i >= 0:

      sum += Lst[i]* Lst[i]

      i = i - 1

  return(sum)

def calc1(num2):

  Lst = []

  while num2 > 0:

      remainder = num2 % 10  

      Lst.append(remainder)

      quotient = int(num2 / 10)

      num2 = quotient

       

  i = len(Lst) - 1

  sum = 0

  while i >= 0:

      sum = Lst[i] + Lst[i]

      i = i - 1

  return(sum)

a = calc(n)

b = calc1(num2)

if a == b:

  print("yes")

else:

  print("No")

\ r

\ R10. Write an algorithm that reads a natural number n and displays the message "yes" if all of its n numbers are distinct, or "no" if n does not have all the distinct digits. For example, for n = 37645 it will display "yes" and for 23414 it will show "no".

\ r

num1 = input("Enter the value of n")

n = int(num1)

def calc(n):

  Lst = []

  while n > 0:

      remainder = n % 10  

      Lst.append(remainder)

      print( remainder)

      quotient = int(n / 10)

      n = quotient

       

  flag = 1

  for i in range(0, len(Lst)):

      for j in range(i+1, len(Lst)):    

          if Lst[i] == Lst[j]:    

              flag = 0

              break

           

  if flag == 1:

      print("YES")

  else:

      print("NO")

  return(0)

Explanation:

Please check answer.  

3 0
2 years ago
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