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Figure 5 . Question 2 possible output.
aList = ['ab', 'cd', 'ef'] bList = ['The', 'old', 'list']cList = ['This old house'] dList = ['is falling down']Call listModifier In listModifierlistA = ['ab', 'cd', 'ef', 1.00001] listB = ['The', 'old', 'list', 2.00002]listC = ['C', 3.00003] listD = ['D', 4.00004]Back from listModifier aList = ['ab', 'cd', 'ef', 1.00001]bList = ['The', 'old', 'list', 2.00002] cList = ['This old house']dList = ['is falling down']

Go to answer 2

Question 3

True or False? The code in Figure 7 produces the output shown in Figure 8 .

Figure 7 . Question 3 program code.
def listModifier(cList,dList=["B"],aList=["C"],bList=["D"]):"""Illustrates keyword arguments""" print("In listModifier")cList.append(1.00001) print("cList = " + str(cList))dList.append(2.00002) print("dList = " + str(dList))aList.append(3.00003) print("aList = " + str(aList))bList.append(4.00004) print("bList = " + str(bList))return #End function definitionlistV = ["ab","cd","ef"] listW = ["The","old","list"]listT = ["This old house"] listU = ["is falling down"]print("listV = " + str(listV)) print("listW = " + str(listW))print("listT = " + str(listT)) print("listU = " + str(listU))print("Call listModifier") listModifier(listV,bList=listU,dList=listW)print("Back from listModifier") print("listV = " + str(listV))print("listW = " + str(listW)) print("listT = " + str(listT))print("listU = " + str(listU))
Figure 8 . Question 3 possible output.
listV = ['ab', 'cd', 'ef'] listW = ['The', 'old', 'list']listT = ['This old house'] listU = ['is falling down']Call listModifier In listModifiercList = ['ab', 'cd', 'ef', 1.00001] dList = ['The', 'old', 'list', 2.00002]aList = ['C', 3.00003] bList = ['is falling down', 4.00004]Back from listModifier listV = ['ab', 'cd', 'ef', 1.00001]listW = ['The', 'old', 'list', 2.00002] listT = ['This old house']listU = ['is falling down', 4.00004]

Go to answer 3

Question 4

True or False? The code in Figure 9 produces the output shown in Figure 10 .

Figure 9 . Question 4 program code.
def listModifier(listA,*args): print(listA)for arg in args: print(arg)#End function definition aList = ["ab","cd","ef"]bList = ["The","old","list"] cList = [1,2,3]listModifier(aList,bList,cList)
Figure 10 . Question 4 possible output.
['ab', 'cd', 'ef'] ['The', 'old', 'list'][1, 2, 3]

Go to answer 4

Figure index

What is the meaning of the following two images?

These images were inserted here simply to insert some space between the questions and the answers to keep them from being visible on the screen at the same time.

Spacer image of a rabbit and a penguin.

This image was also inserted for the purpose of inserting space between the questions and the answers.

Spacer image of a penguin and some houses.


Answer 4


Go back to Question 4

Answer 3


Go back to Question 3

Answer 2

False. The actual output is shown in Figure 6 .

Figure 6 . Question 2 actual output.
aList = ['ab', 'cd', 'ef'] bList = ['The', 'old', 'list']cList = ['This old house'] dList = ['is falling down']Call listModifier In listModifierlistA = ['ab', 'cd', 'ef', 1.00001] listB = ['The', 'old', 'list', 2.00002]listC = ['is falling down', 3.00003] listD = ['D', 4.00004]Back from listModifier aList = ['ab', 'cd', 'ef', 1.00001]bList = ['The', 'old', 'list', 2.00002] cList = ['This old house']dList = ['is falling down', 3.00003]

Go back to Question 2

Answer 1

False. The actual output is shown in Figure 3 .

Figure 3 . Question 1 actual output.
Create two lists aList = ['ab', 'cd', 'ef']bList = ['The', 'old', 'list'] Call listModifierIn listModifier Use incoming parameter to append to listANew listA = ['ab', 'cd', 'ef', 3.14159] Assign a new list to listBNew listB = ['A', 'new', 'list'] Back from listModifieraList = ['ab', 'cd', 'ef', 3.14159] bList = ['The', 'old', 'list']

Go back to Question 1


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  • Published: 10/26/14
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Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
I'm not sure why it wrote it the other way
I got X =-6
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
is it a question of log
I rally confuse this number And equations too I need exactly help
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
Commplementary angles
Idrissa Reply
im all ears I need to learn
right! what he said ⤴⤴⤴
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
many many of nanotubes
what is the k.e before it land
what is the function of carbon nanotubes?
I'm interested in nanotube
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
what is system testing
what is the application of nanotechnology?
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
anybody can imagine what will be happen after 100 years from now in nano tech world
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
silver nanoparticles could handle the job?
not now but maybe in future only AgNP maybe any other nanomaterials
I'm interested in Nanotube
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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