Computer Science Illuminated
7th Edition
ISBN: 9781284155617
Author: Nell Dale, John Lewis
Publisher: Jones & Bartlett Learning
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Chapter 6, Problem 4TQ
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Software borrowed:
“Yes”, there are cha...
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Possibility of copying software:
- “Yes”, there are chances that the user’s friends would have taken a copy of software for their future use, but the user cannot be sure about it, because it is not possible to find when a software getting copied.
- This action is called as software stealing and using this type of copied software is illegal, because all software (other than open source) contains its own copyright; technically this can be named as software piracy...
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I need help!! Writing a long pseudocode for a modular program that will display the monthly payment on a mortgage. P=Principal amount borrowed (loan)R=Rate of interest computed for each monthN=Number of months to pay back the loan or mortgageThe help I need is creating a module that you can input the principle, rate of percentage, and years to repay the loan, and another module "displaypayment" that accepts the 3 values and calculates the monthly payment needed for the rates. Lastly 2 local variables needed!Equation:Monthly Payment=[(R*(1+R)^N)/((1+R)^N-1)]*P
Two pseudocode questions I need help with: How do I design a module called findSum that will display the sum of two integer passed by parameter, and a module called findArea that will display the area of a rectangle when passed 2 real values for the length and width of the rectangle?
For the pseudocode module, what is displayed with the call findValue(1, 4, 2)?Module findValue(Integer a, Integer b, Integer c)
Declare Integer value
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Chapter 6 Solutions
Computer Science Illuminated
Ch. 6 - Prob. 1ECh. 6 - Prob. 2ECh. 6 - Prob. 3ECh. 6 - Prob. 4ECh. 6 - Prob. 5ECh. 6 - Prob. 6ECh. 6 - Prob. 7ECh. 6 - Prob. 8ECh. 6 - Prob. 9ECh. 6 - Prob. 10E
Ch. 6 - Prob. 11ECh. 6 - Prob. 12ECh. 6 - Prob. 13ECh. 6 - Prob. 14ECh. 6 - Prob. 15ECh. 6 - Prob. 16ECh. 6 - Prob. 17ECh. 6 - Prob. 18ECh. 6 - Prob. 19ECh. 6 - Prob. 20ECh. 6 - Prob. 21ECh. 6 - Prob. 22ECh. 6 - Prob. 23ECh. 6 - Prob. 24ECh. 6 - Prob. 25ECh. 6 - Prob. 26ECh. 6 - Prob. 27ECh. 6 - Prob. 28ECh. 6 - Prob. 29ECh. 6 - Prob. 30ECh. 6 - Prob. 31ECh. 6 - Prob. 32ECh. 6 - Prob. 33ECh. 6 - Prob. 34ECh. 6 - Prob. 35ECh. 6 - Prob. 36ECh. 6 - Prob. 37ECh. 6 - Prob. 38ECh. 6 - Prob. 39ECh. 6 - Prob. 40ECh. 6 - Prob. 41ECh. 6 - Prob. 42ECh. 6 - Prob. 44ECh. 6 - Prob. 45ECh. 6 - Prob. 46ECh. 6 - Prob. 47ECh. 6 - Prob. 48ECh. 6 - Prob. 49ECh. 6 - Prob. 50ECh. 6 - Prob. 51ECh. 6 - Prob. 52ECh. 6 - Prob. 53ECh. 6 - Prob. 54ECh. 6 - Prob. 55ECh. 6 - Prob. 56ECh. 6 - Prob. 57ECh. 6 - Prob. 58ECh. 6 - Prob. 59ECh. 6 - Prob. 60ECh. 6 - Prob. 1TQCh. 6 - Prob. 2TQCh. 6 - Prob. 3TQCh. 6 - Prob. 4TQ
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- I need help finishing my python rock paper scissors code, where it asks the user "rock, paper, scissors", the user types either option (repeatedly asks until its one of the three), and for a random option from the code to select also rock paper scissors. The end result would be you winning, code winning, or a tie.My code:import randomOPTIONS = ("ROCK", "PAPER", "SCISSOR") def getInput(): """ Function getInput() will receive and validate user input to be the string ROCK, PAPER, or SCISSOR. Input must be spelled correctly and in all capital letters. Function will return the validated input @return user input as ROCK, PAPER, OR SCISSOR """ #need help here def checkWinner(user, computer): """ Function checkWinner will accept 2 string parameters that must be unique values of ROCK, PAPER, or SCISSOR. Functions should return true if the string in parameter user beats the string in parameter computer in a game of rock…arrow_forwardI need help understanding how could I got the IP Address for the 1st Host on this Subnet, the IP Address for the Last Host on this Subnet & the Broadcast Address for this Subnet My Explanation:8 bits --> 17 & 0 128 64 32 16 | 8 4 2 1 0 0 0 1 0 0 0 1 -->17+ 0 0 0 0 0 0 0 0 --> 0 0 0 0 0 0 0 0 0 --> 0 (The Major (Base) Network Address): 142.20.33.0 To find the broadcast address, we can set all the bits in the host portion of the address to 1, which gives us 142.20.33.255. Based on the subnet mask 255.255.255.0, no bits were borrowed from the host portion, meaning 8 bits remain in the host portion. Using the formula 2^h – 2 (where h represents the number of host bits), we get 2^8 - 2 = 254 hosts. 8 bits --> 17 & 252 128 64 32 16 | 8 4 2 1 0 0 0 1 0 0 0 1 --> 17+ 1 1 1 1 1 1 0 0 --> 252 0 0 0 1…arrow_forwardWhich of the following opens when you click the launcher in the Size group on the Ribbon? Question 19Select one: a. Size dialog box b. Layout dialog box c. Width and Height dialog box d. Format dialog boxarrow_forward
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