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timama [110]
3 years ago
5

When you are almost done with your soup and want to get at the last little bit, how should you do this 2

Computers and Technology
1 answer:
pav-90 [236]3 years ago
7 0

Answer:

<u>B. Tip the bowl slightly, then spoon up the last bit</u>

Explanation:

Generally speaking none of the other answers made much of sense.. "tip the bowl from side to side" doesn't help you at all, neither does "Keep spooning as much as you can, then stop eating". Neither of these will help you spoon up the last little bit, <u>the most logical answer is B. "Tip the bowl slightly, then spoon up the last bit"</u>. If this is not the correct answer then it'd be D, but I don't believe/feel that it's "poor etiquette" to "leave the last little bit".

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What operating system are you using? On most computers, you can answer this question by right-clicking on the "My Computer" icon
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Answer the question ASAP please
RUDIKE [14]

Answer:

Shows the programming checking if num1 is greater than num2

Explanation:

So num1 and num2 are inputs

for you to code this you would need to put

num1=int(input("What is your first number? ))

and the same for num2 except change num1 for num 2 and first for second

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it will do this by using a code something like:

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But in this example if it greater it just ends

But if it was less than you would put

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elif num1<num2:

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What is the HIPAA Privacy rule, and why does it affect IT professionals?
CaHeK987 [17]

Solution:

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IT affects it by these ways:

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This takes for patient.

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6 0
3 years ago
Write a program that generates 1,000 random integers between 0 and 9 and displays the count for each number. (Hint: Use a list o
masya89 [10]

Answer:

import random

count0, count1, count2, count3,

count4, count5, count6, count7,

count8, count9, i = [0 for _ in range(11)]

while i < 1000:

   number = random.randint(0,9)

   if number == 0:

       count0 = count0 + 1

   if number == 1:

       count1 = count1 + 1

   if number == 2:

       count2 = count2 + 1

   if number == 3:

       count3 = count3 + 1

   if number == 4:

       count4 = count4 + 1

   if number == 0:

       count5 = count5 + 1

   if number == 6:

       count6 = count6 + 1

   if number == 7:

       count7 = count7 + 1

   if number == 0:

       count8 = count8 + 1

   if number == 9:

       count9 = count9 + 1

   

   i = i+1

print("0's: "+ str(count0) + "\n"+ "1's: "+ str(count1) + "\n"+

"2's: "+ str(count2) + "\n"+ "3's: "+ str(count3) + "\n"+

"4's: "+ str(count4) + "\n"+ "5's: "+ str(count5) + "\n"+

"6's: "+ str(count6) + "\n"+ "7's: "+ str(count7) + "\n"+

"8's: "+ str(count8) + "\n"+ "9's: "+ str(count9) + "\n")

Explanation:

- Initialize variables to hold the count for each number

- Initialize <em>i</em> to control the while loop

- Inside the while loop, check for the numbers and increment the count values when matched.

- Print the result

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