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RPubs - Ryan Tillis - Data Science - R Programming - Quiz 1 - Coursera

RPubs – Ryan Tillis – Data Science – R Programming – Quiz 1 – Coursera – Indexing with a boolean. 11. Use the Week 1 Quiz Data Set to answer questions 11-20. In the dataset provided for this Quiz, what are the column names 18. Extract the subset of rows of the data frame where Ozone values are above 31 and Temp values are above 90. What is the mean of Solar.R in…Since the information in statement two does not provide any information regarding the value of x, statement two is not sufficient to answer the question. Statements One and Two TogetherAnswer: a Explanation: Since X(z) is a infinite power series, it is defined only at few values of z. The set of all values of z where X(z) converges to a finite value is called as Radius of Answer: c Explanation: A discrete time LTI is BIBO stable, if and only if its impulse response h(n) is absolutely summable.

If t and x are integers, what is the value of x? : Data Sufficiency (DS) – If h(x) = 6 – x, what is the value of ( h•h)(10)? Follow • 1. Add comment.The value of the variable for which the equation is true (4 in this example) is called the solution of the equation. Notice that x + 3 = 7 and x = 4 are equivalent equations since the solution is the same for both, namely 4. The next example shows how we can generate equivalent equations by first…rottenbanana0007 asked in Science & Mathematics. Chemistry · 1 decade ago. Calculate the concentration of OH- ions in a 1.4 x10^-3 M HCl solution? 1.0 x 10^-14 / 1.4x 10^-3 <—– concentration of hydronium.

If t and x are integers, what is the value of x? : Data Sufficiency (DS)

Z Transform Questions and Answers – Sanfoundry – The value that you want returned if the result of logical_test is FALSE. Following are examples of some common nested IF(AND()), IF(OR()) and IF(NOT()) statements. The AND and OR functions can support up to 255 individual conditions, but it's not good practice to use more than a few because…So we're graphing equations right now, and I need to find what the exact x value is on a graph for a large y value (too large for me to even estimate on the little calculator screen zoomed all the way out). Is there a way when graphing to have the ti-nspire cx tell me an x value if I tell it a y value (or just plot…Related questions. The domain of f(x)=x−2. +log(4−x)1 is.

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❖ LaGrange Multipliers – Finding Maximum or Minimum Values ❖ – All right quick shameless plug for my website there it is in this video.
We're going to talk about finding Max and mins using Lagrangian multipliers so The basic idea is you're again trying to find a maximum or minimum of a function? [I] wrote [it] Generically depending on three variables my actual first example I'm just going to use [two], but the idea is going to be the same it's going to be subject to another equation Involving x Y&Z that's going to equal some number And this is what's called your constraint so we'll make some more sense out [of] this I [think] in our actual concrete example The basic idea and you can do this stuff If you've seen gradient notation and things like that you can use all that, but I'm going to do it. I mean it's equivalent I'm going to do it without it though so what you do is you form a new function that depends on x y z and the symbol lambda and To get that new function you just take whatever you're trying to maximize or minimize And you subtract away the original constraint you set it equal to zero though, and then you also multiply that by Lambda so that's how you come up with this new function and Then you basically this new function you take partial derivatives of that with respect to x y z and lambda whatever your variables are set it equal to zero this is going to be the tricky part usually because you're going to get a big system of Equations that you have to solve and in general There's just no procedure to you know solve a generic system of equations If they're linear you can use matrices things like that, but in general that will not happen and then once you get your solutions from the system you just plug them back into your original function and That's either going to give you a maximum or a minimum All right So let's do a concrete example here to make some sense out of this so just [kind] of a little bit of a word problem to put it in context I guess so Suppose you make TV sets at two different factories will call them factory a and Factory B suppose You're making portable TVs and plasma TVs who cares Suppose you make x TVs at Factory a y TVs at Factory B and the cost giving by the production of x TVs and y TVs is this function 6x squared plus 12y squared? Now suppose you have to produce 90 TV sets [a] month and what you're trying to do is figure out well How many of x should you produce and how many of y should you produce? so that you meet that requirement of 90 exactly 90 TV sets but Obviously in this case you're going [to] want [to] keep your cost to the minimum. Okay [unless] [you're] a bad business guy or gal [we'll] assume though. We're not because hey we're smart people. We're taking calculus, so probably want to minimize this function okay, [so] what I'm going to do on this again is I'm basically going to Do like I said in the last part you take whatever function. You're trying to maximize or minimize and Then you subtract away, lambda times your constraint and notice in this case our constraint since we have to produce 90 TV sets Our constraint would be that well the number of TV sets from Factory a plus the number of TV sets from Factory B Turn on the light here. Sorry about that That would have to equal well 90 What you do with this? Constraint again though is you want to make one side [equal] to 0 so I'm going to write this as x plus y? minus 90 And now I'll have 0 on the right side and in this case. This is my function Well, it's not in terms of xy and z but it is in terms of x y and the symbol lambda [okay] So what I get to do now is I should take partial derivatives of all of this So maybe I'll just multiply it out first to make it a little clearer So I'll get minus lambda x minus Lambda Y plus 90 times Lambda Okay, and now I'm going to take partial derivatives So the partial derivative with respect to x remember, we're treating now y and lambda as constants I'll get 12x my 12 y squared will go away. I'll get a minus lambda and then everything else will disappear if I take the partial derivative with respect to y, I'll get 24 y minus Lambda everything else will disappear and then if I take the partial derivative whoops almost [over] [to] z with respect to Lambda The x squared the y squared's will disappear. It looks like I'm going to get a minus x term a minus y term plus 90 okay, [so] now what we [have] to do is we [actually] set each of these equal to 0 and we solve ok so if I set this first one equal to 0 if I solve this for x notice I could add Lambda to both sides and Get that x equals LAmbda over 12 If I do that to the middle one as well, I'll add lambda to both sides and [divide] by 24 And I'll get that y equals lambda over 24 now what I'm going to do is I'm going [to] take these two things and plug it [into] [my] Other partial derivative [ok] and what this is going to allow [me] [to] do is it will allow me to solve for Lambda [I'll] then be able to go back and figure out values for x and y okay, so if I plug in that x is negative Well, it's negative Lambda over 12 minus Lambda over 24 plus 90 equals 0 and now on the left side you could simply get Common denominators multiply top and bottom of the first one by two on top you would get negative three Lambda over 24 I'm going to subtract the 90 and get negative 90 on the right side if You divide both sides by negative three we'll get that lambda over 24 equals Looks like positive 30 to me whoops squeeze it in there and multiply both sides by 24 And you'll get that lambda equals 720 All right, so now at this point I'm going to plug lambda back into my x and y [values] because after all I'm trying to find values for x and y that maximize or minimize So if I plug [that] back in I'm going to get that x equals lambda Which is 720 over 12? I'll get that y equals okay. That was LAmbda over 24 So I'll get 720 over 24 and you can simplify both of these down it looks to me like x should equal 60 Y [should] equal 30 Okay, typically what you do once you have values for x and y you may find more than one pair. That works here We're only finding a single pair 60 comma 30 and what you need to do is you need to plug those back into the original equation [and] So the thing you're trying to minimize or maximize in this case, we're trying to Minimize this cost function 6x squared plus 12y squared Well the cost of making 60 TV sets times 30 and 30 of the other TV sets will be 6 times 60 Squared Plus 12 times 30 Squared and to justify that this in if this at this point does either has to be [a] Maximum or a minimum if you wanted to to justify that it's a that It's a minimum cost you could also take just some other random point Maybe you make [61] of the other TV sets from Factory a which means you only make 29 from Factory B Plug that in – this cost equation, and you'll see that the value you get out When you plug in 61 and 29 you'll actually find out that this cost is greater than the cost We found up here when you make when you plug in 60 and 30 So since you can actually make the cost greater that in fact now Justifies that well this it was either a maximum or a minimum. Well we can make it bigger. So okay We've [now] reduced it to the fact that this must be a minimum. Okay, so this is the basic [idea] with Lagrange's [the] only thing that's going [to] be trickier again is if you have more variables floating around you're going to get more partial derivatives You're going to have more variables floating around your equations more than likely and again solving the system of equations [to] me turns out to [be] the tricky part in general, so this one wasn't [so] bad I was trying to take a straightforward [example], but um sometimes you have to be pretty clever and definitely, there's times where I look at them and get stuck myself, so this is the tricky part again, whatever values you find for [X&y] you should plug those back into your whatever, you're trying to maximize or minimize, and then just justify that that is a maximum or a minimum, so Again, you can do this stuff using gradient notation and some other notation I prefer this notation myself But um you know use whatever you're comfortable with I'm going to do another one as well using Xy z– and z so take a look for that. [I] hope this helps if you have any questions, shoot me an email I'll be happy to get [to] them as soon as I can .

Finding Function Values – – FIND FUNCTION VALUES.
OUR GOAL IS TO EVALUATE A GIVEN
FUNCTION GIVEN DIFFERENT INPUTS. I'D LIKE TO RELATE THIS
TO SOME THAT WE'VE ALREADY DONE THIS SEMESTER. MANY TIMES FUNCTIONS
ARE WRITTEN UP AS EQUATIONS LIKE Y = 5X – 2
OR Y = X SQUARED + 4. BEFORE WHEN WE WERE FIND POINTS
ON THE GRAPHS OF THESE EQUATIONS WE PICKED VARIOUS X VALUES
AND FOUND CORRESPONDING Y VALUES. HERE'S A SIMPLE EXAMPLE. EVALUATE Y = 5X – 2 WHEN X = 4. WE KNOW THE PROCESS
IS ESSENTIALLY TO SUBSTITUTE 4 IN 4X AND EVALUATE. SO WE HAVE Y IS = TO 18. THERE ARE SEVERAL WAYS
TO EXPRESS THIS RELATIONSHIP. ONE WAY WOULD BE TO WRITE IT
AS A ORDERED PAIR; WHEN X IS 4 Y IS 18. WITH THE NEW LANGUAGE
OF INPUTS AND OUTPUTS THAT COME WITH FUNCTIONS
WE COULD SAY THAT WHEN THE INPUT IS = TO 4 THE OUTPUT IS = TO 18. WE SUBSTITUTED 4 IN FOR X,
X IS OUR INPUT AND Y WAS = TO 18, Y WAS OUR OUTPUT. LET'S COMBINE THIS THOUGHT
WITH FUNCTIONS. LET'S SAY THAT WE WERE GIVEN
F OF X IS = TO X SQUARED + 1 AND WE ARE ASKED TO EVALUATE
F OF -4, -4 IS OUR INPUT. TAKE A LOOK
AT THIS NOTATION THOUGH; NOTICE HOW THIS X
HAS BEEN REPLACED WITH -4 AND THAT IS ESSENTIALLY
HOW YOU EVALUATE F OF -4; YOU REPLACE OR SUBSTITUTE
X WITH -4. -4 SQUARED + 1 GIVES US
16 + 1 OR 17 SO F OF -4 IS = TO 17. AGAIN, -4 WOULD BE CONSIDERED
OUR INPUT, 17 WOULD BE CONSIDERED
OUR OUTPUT. I'D ALSO LIKE TO MAKE
THE CONNECTION THAT THIS REPRESENTS 1 POINT
ON THE GRAPH WHEN X IS -4, Y IS = TO 17. MOST OFTEN X IS REGARDED
AS THE INPUT, Y IS THE OUTPUT. OKAY, LET'S GO ON
WITH ANOTHER EXAMPLE. F OF 7 SO THIS BECOMES 7 SQUARED
+ 1, 49 + 1 IS 50 SO JUST TO CLEAN IT UP
F OF 7 IS = TO 50. ONE MORE EXAMPLE,
F OF K–NOW THIS MAY LOOK A LITTLE CONFUSING,
BUT ESSENTIALLY WE'RE GOING TO BE DOING
THE SAME THING. NOW WE'RE GOING TO REPLACE
X WITH K. SAME FUNCTION SO F OF K
IS = TO INSTEAD OF X SQUARE WE HAVE K SQUARED + 1. NOW WE CAN'T DO ANYTHING ELSE
BECAUSE THESE ARE NOT LIKE TERMS, NOTHING SIMPLIFIES
SO THAT WOULD BE OUR FINAL FORM. F OF K IS = TO K SQUARED + 1. LET'S TAKE A LOOK AT A COUPLE
MORE INVOLVED EXAMPLES OF EVALUATING FUNCTIONS. A FUNCTION SUBTRACTS THE INPUT
FROM THE SQUARE OF THE INPUT. A DESCRIPTION OF F IS GIVEN
BY F OF X IS = TO X SQUARED – X. OUR GOAL IS TO FIND F OF X + H
OR IN OTHER WORDS OUR INPUT IS X + H. THE QUESTION IS WHAT WOULD
THE OUTPUT BE? WELL, LETS REPLACE X WITH X + H. SO WE'RE GOING TO SQUARE IT
AND THEN WE'RE GOING TO SUBTRACT IT. OKAY AND REMEMBER THERE IS NO
SHORTCUTS TO SQUARING X + H. WE HAVE TO WRITE THIS OUT TWICE
AND FOIL IT; THAT WOULD GIVE US
X SQUARED + 2HX + H SQUARED, YOU CAN CHECK ME ON THAT
IF YOU'D LIKE BY WRITING IT OUT AND MULTIPLYING IT. NEXT WE WANT TO ELIMINATE
THE PARENTHESIS HERE; WE'RE GOING TO SUBTRACT
THIS QUANTITY. SOME OF YOU MAY HAVE BEEN TAUGHT
TO PUT A 1 THERE DISTRIBUTE A -1; WHICHEVER METHOD
YOU'RE USED TO DOING GO AHEAD AND MAKE SURE THAT
YOU GET – X – H. AND IN FACT THERE ARE NO
LIKE TERMS HERE SO WE ARE DONE. F OF X + H IS = TO THE SUM
AND DIFFERENCE OF ALL THESE TERMS. THE LAST EXAMPLE I WOULD LIKE
TO TAKE A LOOK AT; WE WANT TO FIND F OF X + H – F
OF X ALL OVER H OR DIVIDED BY H. OKAY, LET'S TAKE THIS
ONE STEP AT A TIME. NOW, THIS PREVIOUS PROBLEM
DOES GIVE US F OF X + H. IT GIVES US THIS PIECE
OF THIS FRACTION SO LET'S REWRITE THAT. WE HAVE X SQUARED
+ 2HX + H SQUARED – X – H. – F OF X WE HAVE TO SUBTRACT
THE QUANTITY X SQUARED – X. SO IF WE SUBTRACT BOTH OF THOSE
TERMS YOU WOULD GET A – X SQUARED. AND THEN IF WE SUBTRACT A – X,
WHICH THAT WOULD BE + X, ALL OF THIS WOULD B OVER H. NOTICE WE'RE NOT SUBSTITUTING
H IN BECAUSE THEY DIDN'T GIVE US F OF H, THEY JUST GAVE US
THE VARIABLE H. LET'S SEE IF WE CAN
SIMPLIFY THIS. FIRST OFF LET'S LOOK
AT THE NUMERATOR AND WE'LL NOTICE THERE ARE SOME
TERMS THAT ARE OPPOSITES. SO WE HAVE AN X SQUARED HERE
– X SQUARED THERE. WE HAVE – X + X. LET'S REWRITE
WHAT WE HAVE SO FAR. WE HAVE 2HX + H SQUARED – H,
NOW DIVIDING ALL OF THAT BY H. NOTICE THEY ALL HAVE AN H
IN COMMON SO WHAT I'M GOING TO DO IS DIVIDE THEM
ALL BY H INDIVIDUALLY. REMEMBER YOU ARE ALLOWED
TO DO THAT AS LONG AS YOU ARE DIVIDING BY A MONOMIAL,
ONE TERM. AND THE REASON I DID THAT
IT'S VERY EASY TO SIMPLIFY THESE NOW. NOTICE HOW THE COMMON FACTOR
OF H HERE AND HERE, ONE COMMON FACTOR OF H HERE
THAT WOULD TURN INTO AN H TO THE 1st POWER AND
H OVER H WOULD SIMPLIFY AS WELL. SO WHAT'S LEFT? WE HAVE 2X + H
AND THEN THIS IS – 1. SO THIS ENTIRE FRACTION
THAT INVOLVES THE FUNCTION NOTATION IS = TO 2X + H – 1. THIS IS ACTUALLY CALLED
THE DIFFERENCE QUOTIENT THAT WILL SHOW UP LATER ON
IN THE SEMESTER. I HOPE THAT HELPS EXPLAIN
HOW TO EVALUATE FUNCTIONS. .

❖ Limits Involving Absolute Value ❖ – .