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Peccary_NCA code review: code explanation, Shiny Integration and unit tests

hi this is a video explaining the pkwe,npf function so for the programmer so me,and everyone,who wants to improve,the good speaker a lot of things can be,improved of course so this is more,precisely a description of the file,which is in pica analysis,air folder of course and then the pkre,nca dot air,files so it contains the functions that,allow you to quickly perform,ncga analysis and as many functions,inside big array by simply,adding output expressions are true you,will not provide produce directly the,output but first the code that is,becoming dependent,which once evaluated will produce um,the output so there is a video,explaining how to use it as a function,from the user perspective here it's a,video from the developer perspective,so we will explain the rule code,of the function so first,in the console,for the console applications,then we will speak about the unit test,and finally we will show secretly the,code inside the shiny app,to make to use this function,installation app,so basically the world function is,back is meta programmed,so there is some by default,uh,value that can be used if you want to,manipulate the functions so uh and you,have of course the descriptions here of,all the arguments so basically it is all,metal programs so which means the code,we compute a code that is strictly,independent of p carry the one we have,seen here and at the end you have the,choice of either uh returning the,expressions or to evaluate the,expressions and to make that works you,need to,we need that block,so here you just basically capture the,expression that has been answered by the,users,and here so if you,want the code we need to substitute it,uh,so here is when you want to return the,code you need to substitute the names to,give you the kind of success or if you,want directly have the results you need,just to capture the expression data sets,but not the names because datasets exist,inside the scope,of the functions,so this is it so nothing special is just,here the group so you have speed dots so,here the dots allow you to give you,uh the groups that you want so you will,we will perform one ncaa pair uh all,groups that you enter into those results,so here for instance the filling it with,subjects,but you can have subjects and occasions,if you want to have a once a pair such,as any occasions so this is how you can,express uh capture that,uh so here you have directly the,expression actually you better have this,in plural,and so in for example we directly,capture dataset 2 as expression of,android material because,we want andromedacine,when we work that will allow us at any,time to evaluate uh where we are so far,to see that it works so there is many,possibilities and you will see one after,the other so if someone give avid colors,very easy we will just capture basically,um just a filter to only keep the evid,colon is equal to zero but we don't have,uh ones here,then we had another colon which is,uh let's do now,that so with the new colon in fact for

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CAPE MATHEMATICS UNIT 2 PAST PAPER

welcome back guys today we'll be doing,a cape unit 2 past paper question,coming from the year 2019 it is question,2,which contains partial fractions,integration by substitution,and integration by parts let's get,started,so this question we need to determine,the integral,x to the fifth times cos x cubed,dx now the x cube that we have within,here we have to get rid of it,we're going to replace x cubed to be,another letter let's,simplify it with t because if i replace,x cubed to be t it becomes cos t and i,know how to,integrate that now i'm going to,differentiate t with respect to x so i,have d,t d x then differentiate,x cubed so i get 3 x,square now when i do this it means i'm,doing a little bit of substitution to,simplify this,so i'm going to cross multiply so the 3x,squared,will be in the denominator so i have dt,over three x square,is equal to dx,now the next part now i'll be replacing,dx,with dt over 3x squared,so rewriting x to the fifth then,remember,x cubed is now t so i'm gonna replace,x cubed with the letter t,and then dx will be replaced with dt,over 3x square now,we can cancel x squared into x to the,fifth that will be left with x,cubed no x cubed it's the same thing as,t initially that's my substitution,so i can replace here with t,so i have the integral of t,instead of x cubed then cos,t,and then i have d t over,three so the three is not being,cancelled,no dt over 3 i can rewrite it as,1 3 times,dt so that means,here i can rewrite this as the integral,of t,cos t times,one third d t and i'll put this in my,bracket so the cost is together,and then times one third dt,now i'm gonna carry the one third once,you have a constant you can carry it,in front of the integral sign so here,we have to integrate this using,by parts because we still have two,functions,side by side but it's simple so i'm,going to let,u to be t,so let u be t and,dv being cos t,so dv being cos t so integration by,price i differentiate,this so d u is equal to,just like how you differentiate x to get,one you differentiate t to get,one and d v to get,v we have to integrate cos t so,integrating cos t,i will have sine t so this will give me,sine t just a reminder what is the,integration by parts formula,it's a u d v,being integrated is equal to,u v minus,the integral of v,d u that's by price that's what we're,gonna apply,so integrating this which is one third,integral of t cos t,applying the by part,rule i will substitute what is u,what is v so i have one third,open bracket my u which is,t my v which is,sine t,minus,then i have the integral,what is my v my v is sine t,and d u is one,then i'm integrating with respect to t,now one times sine t is just sine t so,i don't have to write the one i could,just have the integral of,sine t dt now after doing this we're,going to,integrate sine t but before i do that,remember this is,within the bracket 1 3 is outside,so we can distribute the one-third so,one-third,times this first brackets t,sine t number of the one-third is coming,from th

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CSC1141 Introduction to Computer Systems

hello,i'm now going to tell you about,introduction to computer systems in,general we start by,explaining some of the terms,the first one is information technology,which is iit in acronym so what is it,well it refers to an entire industry,that uses computers networking,software and the other equipment to,manage,information,so what is ict ict can be seen as an,integration of it,with the media broadcasting technologies,audio video processing and transmission,so ict is an extended acronym for,information technology,so ict versus iit what is the difference,between them and somehow,similar,so the term ict has widely been used in,the context of education,while the term i.t has been widely used,in the industry,so ict covers any products that will,store,data or information or programs um,example here is the hard disk drive,which is used in desktop computers,the second example is,solid state drive these are the hard,drives that are used in smart devices,like laptops and so on,so,that product of course will store,information or data or problems,then ict covers any product that will,retrieve that will enable you to get,back your data,also ict include any product that will,transmit,so ict has got different areas that it,covers some of them include,education,communication,data processing,and so on,so let us now attack,the first part um,among the two main parts of the computer,it is called computer hardware,this is the part that puts together all,the physical components of a computer,that you can see and detach as well,so there are many parts inside the,system unit,what do they exactly do,let us see some of the computer,components that make a computer function,whether it is a desktop computer or a,laptop,so every component,in a computer,is somehow immediately or,indirectly,connected to a stuck with the board,which is called motherboard,as you see displayed on the screen,so every computer has a large circuit,board which is called motherboard this,contains some of the most parts of the,computer such as the cpu,also known as the processor,the cpu you see it,you can also call it central processing,unit,so the cpu processes information and,executes our commands or instructions,so when the cpu is working it generates,a hit,that's why it is covered by the heat,sink you see the spray which draws the,heat away from the processor,it tries to reduce the heat generated by,the processor,so the motherboard also contains,the computer's ram or random access,memory so ram is the short term memory,that the computer uses when it is,performing calculations,however,you cannot store your files,or install the programs in the ram,because ram deletes everything in it,whenever the computer is shut down,the next component we are talking about,among the ones that are found inside the,cpu unity is the hard drive,so the hard drive provides the long term,storage keeping all of the computers,that even when it is turned off,so on many desktop computers,the motherboard has got,expansion slots,t

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Volume of a particular cone using The disk method.

okay,so now i'm gonna use both techniques to,calculate the volume of a cone,and in particular i'm gonna use a cone,where all the values are known to us the,radius the height and everything,now we're gonna we know that the volume,of a cone is 1 3 pi r squared h,and so we already know what result we're,going to get,but i'm going to use the cone because,it'll allow me to show you both,techniques in a natural,situation how we set up the problem,and and that we get the right answer,okay so let's begin,so okay so first of all i want to remind,you of course i know you already know,this because i just blurted it out and,probably all of you know this because,you guys,know it by heart in any case and if not,false like,it's a beautiful thing to learn a new,thing every day,and so this is the a cone,of radius r and of height h and i know,that the volume of said cone,is one third pi r squared,h okay okay so that's number one number,two,what cone are we gonna talk about now i,could do this one,in full generality um okay,uh um but what i wanted to do to make,this discussion even more concrete,is that i want to calculate the volume,of the following cone,which is the following one let me show,it to you,it's going to be oops that's a that's,not a cone that's a triangle,let me give it like some circular base,oh much better,and i want it to have radius two okay,and i want its height to be nine like,this,like this the height is 9 the radius of,the bottom is 2,and so we know that using this formula,we know the volume of this cone is,1 3 pi times r squared,times the height which is nine two,squared is four,four times nine is thirty-six thirty-six,divided by three is twelve,we know that the volume is twelve fine,oops,right,so we know that but now,i'm going to show you how we would do,this with calculus,using first the disk method,and then the cylindrical shell method,okay,so the first thing we have to do,is we have to figure out how we're going,to set this up,okay so let me begin,in that regard as follows i'm gonna,cons i'm gonna make my x y axis like,this okay,and now what i want you to realize is,the following,i could for example imagine to myself,that i'm going to make the line from,here,0 0 all the way to notice,if i was to put the x axis through here,the x will go through zero zero the x y,would be zero zero here and then what,would it be over here,well i would have had to go nine down to,reach the bottom,and two up so i'm gonna have to go nine,this way,and two up okay so it'll be like there,so so if this is my function,f of x right,then that cone is nothing but the,surface of revolution of this function,but what is this function,this function is the line,that is going through the points 0 0,and uh 9 2.,okay now of course i know that you all,probably know what this,equation is but let's do it together i'm,going to first calculate the slope,using my familiar formula y2 minus y1,over x2 minus x1,my y2 is 2. 2 minus 0 is 2,9 minus 0 is 9. and then i need to find,the eq

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Topic 11 Practice Assessment

hello everybody this is mr regner and,this is our practice,test for chapter 11. if you look at your,book,it says volume 2 page 481,and 482. in this video my hope is that,you can use this to help yourself out so,i'm going to write down the number with,a circle around it and that way you can,kind of skim ahead if you need to,so maybe you only need to look at some,of these problems maybe you watch,the entire video and that's totally up,to you as well so let's get started with,number one,it says julio used unit cubes to make a,rectangular prism,what is the volume of the prism now,there's a couple options here that i,could do i could count up the cubes,right there's some we can't see some we,can i could count the cubes just in like,the front face,right and then i could figure out how,many total rows there are that may be,another way to do that,or i could figure out the length the,width and the height,by calculating up all those so if i look,at this the height of this,is three high the width of it one two,three four five six wide,and it is one two three four deep,so i'm going to use that volume formula,length times width,times height and just plug in those,numbers so i think i said it was 6,wide are 6 long and 4 wide,and 3 high if i just solve this,problem i've got my answer so i can do,this in any order that i want to,it doesn't really matter i'll do 6 times,4 which is 24 and 24 times 3,which equals 72. i know i did that like,super super quickly,but part of that is because i'm trying,to make this video a little bit shorter,than it needs to be,my answer here is letter c and i can,move on now to number two,now number two we've got our little grid,here,we've seen this so many times before,this time we're matching up,the total volume on the left side with,the length width and height on the top,so i'm going to go ahead and do this,from the top and match it with the,volumes,so volume equals length times width,times height that's a formula we should,know,and have devoted to our brains because,we know that's the formula for volume of,rectangular prism,for the first one i have let's see,three times four times five,i'm just gonna solve that three times,four is twelve,and twelve times five 5 is,60. so i'm gonna find,60 and it says 60 centimeters cubed kind,of that third one down let's see there,it is right there,so i can mark that one on this,problem for the next one it says three,times three times five,so volume equals three times three,times five three times three is nine and,nine times five,is 45 so i can see 45 right here this,first option,there's my answer 45. the next one is 2,4 and 9. so volume equals 2,times 4 times 9. now just to mix things,up a little bit yeah,i mean i suppose i could do it any way i,want to i'll go 2 times 4 which is 8 and,8 times 9,which is 72 and if you look at your,options down here is 72.,now there's one left and just to make,sure that we've got it down,i might as well do the math for this,it's two times four,times seven well my h

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BCD and two's complement addition lesson 12

hi this is Chris Schaumburg and we're,going to talk a little bit about two's,complement arithmetic and adding and,subtracting so really when you add a,two's complement number you you do just,you just add like a regular binary,addition now the key thing here is,subtraction is really just what you do,is you convert that binary number into a,negative two's complement and then add,so in a sense you're just adding again,you're not really subtracting so that's,kind of a key point,so regular binary addition and then when,you do that when you add in binary using,two's complement addition you ignore the,most significant bit if it goes beyond,8-bit level that we're looking at okay,so let's look at this example in two's,complement and then we'll actually work,an example as well so let's just look at,this example to start out with we see,here we have our true binary for 35 so,35 is our number we have true binary for,that and that is going to be 32 plus 3,is 35 so that is our true binary now our,two's complement is going to be the,identical thing because we have a,positive sign here okay so because it's,positive 35 in true binary is the same,as two's complement okay now if we were,to do they get a 35 we look at the the,two's complement which is the same as,the 35 before and then we complement it,we deal with ones complement which means,that anytime there's a zero there's a,one zero there's a one a one there's a,zero a zero there's a one zero there's a,one zero there's a one a one there's a,zero and along there's a zero so then,essentially what we do is we have a,complimented number here then we add one,to that and then that is our two's,complement number it is really not easy,to see but really from this point it's,really not easy to see that it's 35 but,we can tell that there's a 1 in front so,we know it's a negative value to convert,it back it's really you just have to,the same steps that you did so you you,um let's go ahead and look at this one,so if we if we were wanting to convert,it back we would would make this a zero,we'd make this zero this a 1 this is 0,then 0 0 1 0 okay and guess what that,gets us back to our 35 okay so see this,right here is the 32 this right here,always still have to add 1 so we add one,right here and help me erase that real,quick here okay so we still have to add,one just like we did before,and we get 0 0 1 0 0 0 1 1 ok and now,this is our 35 if you want me to prove,it I have this is the 32 bit this right,here is the two bit okay it's going too,fast for my computer that's the two bit,that's the one bit when we add 32 plus 3,we get 35,I should get 35 okay now we have to,denote that we noted that this is this,right here is a 1 so that must mean that,we have a negative 35 here and that's,actually our answer okay okay so what,we're going to look at is we're actually,looking at converting 90 negative 98 to,decimal to in decimal to a two's,complement number so the first thing,we're going to do is we're going to,confirm that this is a 9

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Math140Sp22 Lec31 DifferenceEquations

okay so welcome to,math 140 this is i believe lecture 31 so,we are at the home stretch now in terms,of the,semester you know as always if there's,material you want me to cover and go,over just let me know,what i want to do is i want to use you,know some of the last few classes to,give you a little bit of a sense of what,can you do with the material that we've,been learning,and what might come next if you continue,most of you are not going to become,professional mathematicians that's,absolutely fine,but hopefully you'll have a sense of,what math can do and how it can help you,attack value problems,one of the most important things in the,21st century is to have some kind of,pattern recognition hey this looks like,something i've seen before,which means either,you know where to go to find out how to,do it or you know what kind of questions,to ask somebody who might then know how,to do it,one of my greatest successes is a,physics professor here years ago bill,ruders,i came to meet with one of his students,with a project that they were doing and,they needed help with certain,exponential songs and i was able to do a,little bit but you know what they needed,was extremely complicated and technical,well beyond my ability but i actually,know somebody who wrote one of the,classic posts in the subject said i'll,make an introduction,so i send it to him he goes oh no no no,no no you need god for this you need,young cats he's the only one,and so he passed it on to nick counts,who then wrote a 50-page paper title on,a question of buddhism,i know this is how the network goes you,don't need to know how to do something,you need to know enough to recognize,what is this,how can i push this to,whoever has the knowledge to do this,and that's why it's so important to get,a sense of what can and cannot be done,and so i want to talk today a little bit,about difference equations which are,discrete versions of differential,equations and thus not surprisingly the,next class will be on,differential liberations right,and so we will analyze one of the most,important battles in world history the,battle trafalgar from the napoleonic,wars so many of you have seen that in,the history class the first time i,taught this i actually had somebody who,was doing it in history class at,williams we will look at the battle from,a slightly different perspective than,you would in a history class,but it's interesting to see how can math,be applied,right so i want to talk a little bit,about fibonacci that numbers today talk,a little bit about generating functions,and then as an application,and you watch a video on why you should,not use the knowledge i'm about to give,you to go to las vegas to gamble,you know again i love teaching,but if it was guaranteed to make that,much money i did it,so yeah there's going to be a flaw and,so many times people are convinced they,have the solution occasionally yes,occasionally they find something that's,been missed by people for centuries but,odds are there's a

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Essential Mathematics, Chapter 2: Long Division #3

let's work a couple of challenging long,division problems,a 30 154,divided by 82 and b,174 000 divided by sixty,we'll do a first let's set up the long,division,box for our first problem,so thirty thousand one hundred fifty,four,goes inside the long division box,three zero one,five four,out in front goes the eighty-two,we can also read this division problem,as eighty-two divided into,thirty thousand one hundred fifty-four,now let's think about our four steps,that we're going,to repeat over and over so we need to,get started,with the divide into step we want to,think about how many times the number in,front,the 82 goes into,first the first digit three well it,doesn't even go into three once,and we look at the first two digits 30,doesn't go into 30 even once so we need,to look at the first three,and think of that as 301 how many times,does 82,go into 301,now this is complicated enough that i'm,going to have to make some guesses,and sometimes when i have no,idea on where to start i just,check by taking 82 times 5.,and see what happens now 5 times 2 is,10 write the zero carry the one,five times eight is forty plus one,gives me 410 well,too big right way too big what if i try,82 times four,that will be smaller four times two is,eight,four times eight is 32 so i have,328 still too big,in fact it is kind of interesting to,notice here,that if you if we,think of 4 times 8 as 32 4,times 80 is going to be 320,so 4 times 82 is going to be,a little bit more than 320 so,just by observing that 4 times 80 is 320,we can see that this 82 times 4 is too,big,let's go ahead and try three times,now before we go through the,multiplication 3 times 8 is,24 so 3 times 80 should be about,240 we should be close to 240 let's see,3 times 2 is 6. 3 times 8 is 24,we get 246 in fact we're a little bit,larger than 240.,now 246 is smaller than 301,and so we know 82 goes into 301,three times i write the 3,above the 1 in the 301,then i multiply 3 times 82,and we have that that's 246.,after the multiply step comes the,subtract step,we need to subtract 1,minus 6 i can't do so i need to borrow,can't get anything from the zero so i,borrow from the 3.,3 becomes 2 the place to the right,becomes a 10.,then i borrow from the 10 the 10 becomes,the 9,the place to the right i had 10 on so,that becomes 11.,11 minus 6 is 5.,9 minus 4 is 5,and 2 minus 2 is 0 out front,so i get 55 after i subtract,after the subtract step comes the bring,down step,i bring down the next place i bring down,a 5.,after the bring down step i go back up,to divide into,how many times does 82 go into,555 now from what we've seen,so far what we figured out so far,i know it's probably going to be more,than 5,times because 5 times 82 is just 410,what if i try 7 times here,82 times 7 seven times,two is fourteen seven times eight is,fifty-six,plus one is five seventy-four,well that's a little bit too big,so i need to back up and try eighty-two,times six six,times two is going to be twelve,right the two carry the one six times,ei

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u substitutions pt2

for I guess my final video I would like,to do you substitutions with definite,intervals and the way I do definite,integrals is slightly different than I,do,indefinite integrals but not the initial,steps I'll show you what I mean I'm,gonna start with my first example an,integral from 0 to PI over 4 sine of X,over cosine squared of X DX now when,you're investigating U and D you,possibilities what other things you,might try is U equals sine of X because,the derivative of sine of X is so easy,it's just cosine of X DX,and so when you'll do this then here's,what you'll get for an integral you'll,get au there you'll get a you there but,see the problem is down here you you do,not have cosine of X DX in the,denominator so and then even if you did,have a DX in the number you would have,something weird like this and we do not,simply have anything that deals with the,D U squared,and unless I guess we're doing something,like a double integral or not and having,the D u down here is their boat in any,way so don't do it it's not gonna work,for you usually you're gonna find what,you need in the denominator because when,you let u equal something from the,nominee D you always make sure that the,DX stays a top of any types of fractions,so what I'm gonna do this I don't take,away from your trial and error learning,I mean that's where you will one of the,most I'll get rid of those and I'll show,you what I would like equal to be equal,to u NB u equals cosine of X naught,cosine squared because if I did cosine,squared I'd have to deal with that -,coming down cosine of X and then take,the derivative of cosine and sine X and,I don't have cosine X sine X as,product anywhere in there so I'm not,gonna do it just cosine of X you'll,you'll learn how much you can get away,with later on we take the derivative,derivative cosine of X is negative sine,of X DX and of course I have sine of X,sine of X DX all in the numerator not,negative though so what I'll do is I'll,take the negative out of here and I will,put it in front of my D you then it's,all a matter of substituting everything,so let's go ahead and put it in okay,where I do things different is in the,limits of integration it will deal with,that in a second,right now let me go ahead and rewrite,the problem in a way that you'll,understand how I'm doing the,substitution remember cosine squared of,X is the same as cosine of x squared so,since I'm still dealing with X I'm not,gonna monkey with these things just yet,so and we'll go to the next step that's,what I'm gonna start putting in my U so,down here I'm gonna have au squared in,the denominator in the numerator I'm,gonna have negative well negative D u,this is an extremely simple integral now,here's where I do things a little bit,differently you can go ahead and do your,integration now but what I recommend,doing right at this moment is changing,the limits of integration because when,you switch to you while it's still,integrating inside of this rectangle per,se it's a different r

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CoSc 425 Lecture 10

okay,uh what we want to do today is finish,discussing the design,the design of the pep8 arithmetic logic,unit the alu,so last time we we saw this over this,overview the alu performed 16 different,functions,and in figure 10.54,is a block diagram of the alu the,arithmetic logic unit we see that,how many bits how many data bits for a,eight and how many data bits for b,obviously eight and how many bits for,the result,eight so those fat pipes represent eight,lines then we have the four lines coming,in from the right,that specify the function that the alu,is to perform and we have a carry in,that goes into the aou,and then the the four outputs besides,the result,are the n z out v and c out,okay and then we saw,in figure 10.55 we looked at the 16,different functions that this aou,performs,either a can go through straight or we,can do a plus b or we can do a plus b,plus c n etc etc and the last one was a,little bit strange,it was what it did is it it funneled,a4 a5 a6 and a7 to the nzvc,output lines and then we looked at,figure 10.56 which is,a little bit lower level of abstraction,which means that more details are become,visible,because after all abstraction is hiding,details,and so here's more details and so we saw,that those four lines coming in from the,alu,coming into the alu from the right,go into a 4x16 decoder and what is,always the output of a decoder,when it's enabled what is the output of,a decoder,it just generally speaking all of the,what,all of the output lines are what except,for,except which is one except for one which,is one and how do we know which one is,one,well it depends on the what the alu the,the number of,the aou the combination of the alu,number,going in right is everybody with me on,this,okay so then and so the first,15 of those lines go into the,computation unit,and the 16th line which is number 15,goes into these 12 two input,multiplexers and we,talked we resolved the little question,and we did,say that it was indeed this bottom box,is indeed,12 two input multiplexers as opposed to,what did we say 10 wait is that the one,where we,we had the little confusion of oh no,that was a later that was a different,one,sorry yeah these are 12 two input,multiplexers,anyway and then we see and we see the,individual lines for a and b,on the top and we see that a 4 5 6 and 7,are tapped off and they go into the nzvc,of the,uh of the the bank of 12 two input,multiplexers,and we understand now that we have,we have um the n,on the on the left group of the nvc,nzvc corresponds to the,n of the right group nzvc,and depending on whether 15 is one or,zero one of those ends or the other,will go to the n output,now do you remember how that worked,okay and then then here this describes,which one if line 15 is one then the,result of the nzbc from the left,are routed to the output if line 15 is,zero the result from the right are,routed to the output and then oh this is,the one,figure 10.57 at the end of class last,time was the one where we're trying to,ma

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