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The IQ scores of ten students randomly selected from an elementary school are given. $$ \begin{array}{ccccc} 108 & 100 & 33 & 125 & 87 \\ 105 & 107 & 105 & 113 & 118 \end{array} $$ Grouping the measures in the 80s, the 90 s, and so on, construct a stem and leaf diagram, a frequency histogram, and a relative frequency histogram.

Hello everyone. So in this question we will be learning about the basic concept of the polarization late. So the given question has four parts and the first parties were to describe in detail the politicians. States of age of this. Right? So the stock rector Representations of the two beams would be B. One. Yes it was too 1100 and B two would be equal to T. C. You go see you know 38. To answer the first part of the question. The B one is linearly polarized horizontally. All of the intensity completely horizontally polarized. Right? And the beetle has intensity of today and completely polarized in the right handed circular polarization late so be one is to linear polarized horizontally late and B two is light ended. Second liberalized. Mhm. Yeah. And has the intensity of the In the 2nd part of the question. The stock vector of the combination of these members of these beams would be B one Plus B2 which would be equal to fall 103. So the beam is mostly right circularly polarized but there's also an unpublicized part of the beam which has an intensity of I. N. That is given by as not minus square root of as one square plus has to be square plus has taste square. So this would be equal to four a squared order of one plus zero plus nine. That is equals two 4 -2 sq Root of 10. It is approximately equal to 0.84. In the same part of the question we have to find the degree of the politician. So degree of position we would be equal to I. P. Divided by I people. S I N. Can also be readiness. It's square load off As one is square plus has to be square plus has three square divided by it's not Right. So this would be equal to the square root of 10 divided by four Which would be approximately equal to 0.791. Now coming to the 4th part of the question. The talks vector representation of the beam is 1100 plus 1 -100. Which will be equal to 2000. Right? These are horizontally polarized and the vertically polarized beam, the result is an an polarized beam of double the intensity. So this comes from the definition of these dogs parameters which is to go in this condition. Right? So b. s. equals two. Yeah they

So this question were given a contour diagram that were asked to find partial derivative with respect to X. Oh, see that one zero then departed derivative with respect that X at 01 the partial derivative with respect. Why, All right, so let's start with the first point. So we're gonna goto 10 So which is this former here? But then I wouldn't draw horizontal. And then now so this is not really a horizontal, but I imagine that it's horizontal. So as we notice and this Linus read lying right here, this red horizontal line right here start at X equals one and ex people's 1.5 and then ours e at our final points to undersea our initial 0.1. So now we find a difference of that. We do to minus ones of the conflict. Final Contour Point is to initial going to appoint this one there by that by the final counterpoint, the final explain this 1.5 the initial list one we subtract that we get one divided by 0.5. That's just all right. So now let's do the 2nd 1 So it z derivative Bobbsey, with respect to X. That means we're gonna hold. Why constant that 0.0. But so x zero, Why is once over right here and then we're gonna hold by. Constance are only gonna find the difference. Uh, sport, So we're moving with respected ex, while keeping likely. If you notice this point, it's like a political. It's like they're top of the mountain, so that means we have a maximum of them now, since we have a maximum at that point, the derivative is zero respect. All right, my mouth. Finally. Where? The last point we have er ze subway at 01 So again, 01 right over here. But right now, it's not flat because we're going up. All right, so the point that I chose is one and 1.1. So at one one and one point wants us the little horizontal line. Right? So, uh, one are contour are contour is the equals one, and at 1.1 or contour is too. So basically, what we're gonna do is a difference of the contours, which is Thio two minus one, and divide that by the difference in the life value. So we start at one and we end up 1.1, take a distance of there. And then what we have is one divided by 0.1, which is 10 again. All these values air approximate because, uh, yeah, we don't know the exact coordinates for these, but they're around.

Here for the solution step one. The expression for the spectral with is that is omega according to pissy by lambda not hair. Lambda note is the prevalent and she is the velocity of the light in vacuum. Now for the next step the spectral width is omega equals two pi C by lambda north. Now we substitute one new M. For London north and three multiplied by gentle about it. Emperors foresee. So we get omega not equal to in Bracket 3.14. And bracket 3.82 deep three multiplied by two and three power eight m per second Divided by one. and from here we get at 1.8 at five multiplied by 10 to the power 15 hours At Lambda, not equal to one. Be square. And by the λ North Square which is equal to 0.01- zero mule. I am university square. No gamma equal to the square K. By the omega square. Which is equal to the lambda, not by two pisces square In Beckett. Lambda North square D square N. By D. Lambda North square. Now from the given table, substitute the values in the above equation. Okay. Carmichael to one um In Beckett and Power minus six M by one. Um All divided by two pi in Beckett three multiplied by 10 to report it Emperors. Holy Square in record one. Holy Square in Baghdad. 0.0 120. University square. And from here we get 2.1 to multiply by 10 to depart -26 am universe as a square. And we know that question that equal to two by gamma to buy town or square. Now we substitute 2.12 multiplied by 22 by minus 26 M. Universe as the square for gamma, one km for dessert and 100 PS four town halls. So and by substituting the values in the equation we get called to in brackets 2.12 multiplied by 10 to the power minor statistics and in words X square and back at one camp in Beckett and Power three per one km whole divided by 100 PS. In Beckett tend to depart. Minister has per one PS all square. And from here we get bigger to 4.24 multiplied by 20 power -3 Step three. So the values of frequency sleep are yet delta omega by omega nautical to buy omega not north square one plus p square and baggy t minus two by E. Or by substituting the values, we get 4.5 multiplied by 10 to the power minus stand in Becky t minus that by we therefore delta omega by omega note it will be quoted to be by omega no teen town old square in Beckett one plus P square in Beckett minus 100 PS. And from here we get it minus 4.5 10 to the power minus eight. Now omega delta omega by omega North Dakota to be by omega. North town North square and back at one plus P square and back at 50 minus 50 PS, which is equal to minus 2.25 10, 20 power minus eight. Or delta omega by omega not equal to to be by omega, Note Town Hall Square and Beckett one plus P square in record plus 100 PS. And from here we get it 4.5, multiplied by 10 20 power minus eight has the required values of the frequencies three par minus 4.5, multiplied by 10 to depart minus eight minus 2.25, multiplied by 10 to depart minus 82.25 multiplied by 10 to depart minus eight and 4.5 multiplied by 10 to depart minus eight. So this is a compared solution step by step and you tell about the explanation concept, please go through this. Thank you.

Hello, everyone and welcome. We want to estimate the limit of export. Zero of ever vex limit as exports zero off our backs. So for a limit to exist at a certain ex value, um, it has to be the same from the left and the right hand side. So if we look it from the left hand side or the negative side, um so as we get from negative, want a negative 0.1 negative 0.1 Our values seem to getting way closer and closer to three. Goes from 2.9222 point 998 And it seems to be going the limit as exporters. Urinal negative side fffx seems to be approaching three or very close to a 2.992 point and an eight. Um, you know, just to make sure that this limit exists, we have to look at it from the right hand side. So the limit as X no space there, Ato limit as X approaches zero from the positive side of FX is going to be equal to. And now let's take a look at our table once again, and we see that It's also getting closer and closer to three as Ex gets closer to zero. And so this also three. Since, uh, this is equal to this, we know that the limit as experts zero off F F X is equal to three. So thank you for watching, and I hope this helped.


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