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Luck E Iet Watuilprefetence UMpLMY Uu" MuH Lluu i Ux the Lalle for the fulluw ing pirl?.rrddIu- Iauna Liera 4ellDi Ia FunueANIeDutublg HouliALB DutupleTotalFealeKakTutalTwintI Hatiubability tae #UIIEULleFeuiueurelenDuwpling(4 IiJts) Whatpubalxlity that RuncuteNule @telthley Hrefer Din Tal Fung? ?VuatsalprulrbI ' that qumleutle dotan ureler Aati LI ? Dutuple Holle alld Eu t WAe-(4 fnints) te eeni Dumpling Bar: AreaenispenolMuiAHruIl prefer XLBitudepeulent? ShutTwinta Giveeveuts which ae

Luck E Iet Watuil prefetence UMpLMY Uu" MuH Lluu i Ux the Lalle for the fulluw ing pirl?. rrddIu- Iauna Liera 4ell Di Ia Funue ANIe Dutublg Houli ALB Dutuple Total Feale Kak Tutal TwintI Hat iubability tae #UIIEULle Feuiue urelen Duwpling (4 IiJts) What pubalxlity that Runcute Nule @tel thley Hrefer Din Tal Fung? ? Vuatsal prulrbI ' that qumleutle dotan ureler Aati LI ? Dutuple Holle alld Eu t WAe- (4 fnints) te eeni Dumpling Bar: Areaenis penol Mui A HruIl prefer XLB itudepeulent? Shut Twinta Give eveuts which ae disjoint aud explain #hy they ate disjuint .



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A candy company distributes boxes of chocolates with a mixture of creams, toffees, and cordials. Suppose that the weight of each box is 1 kilogram, but the individual weights of the creams, toffees, and cordials vary from box to box. For a randomly selected box, let $X$ and $Y$ represent the weights of the creams and the toffees, respectively, and suppose that the joint density function of these variables is $$ f(x, y)=\left\{\begin{array}{ll} 24 x y, & 0 \leq x \leq 1,0 \leq y \leq 1, x+y \leq 1, \\ 0, & \text { elsewhere. } \end{array}\right. $$ (a) Find the probability that in a given box the cordials account for more than $1 / 2$ of the weight. (b) Find the marginal density for the weight of the creams, (c) Find the probability that the weight of the toffees in a box is less than $1 / 8$ of a kilogram if it is known that creams constitute $3 / 4$ of the weight.

Okay, So we know that the horizontal and vertical areas from aiming at a point target are independent of each other and each have a normal distribution para meters. I mean off zero under standard deviation off one. Also, given that the density function F V is V over seating squared times E to the power monetary screwed over to signal skirt for one V is great in zero on, otherwise chicken to do so Basically for views. Liston culture. So no. Do we know that the standard norm alleviation distribution is? Trust me. Sure. She was one way consult be. I calculated the proportion of darts that land between 25 millimeters of the target. Okay, that's just given by the entry of 0 to 25. No FV TV. Did you go thio actual 0 25 of the over 20 squared. I am e to the minus fee squared over two times 20 squared TV. So if you succeed for V in terms of tea, get his tea. She was a negative V squared over two terms. 20 screamed, and we know the D key is therefore negative to v just differentiating Curator over two times 20 squared. This will then reduce this integral. You too. And to go who is you to 0.78 1 to 5 e to the power of negative Chief teach you. So if you're just playing the boundaries way, know that Did you bring this or just have the minus coming front? Something negative. Same boundaries. Oh, sure, which will give negative zero from 4578 plus one, which is equal to 0.542 which is the proportion of darts that land within 25 millimeters.

Hello. This is problem 92 And we were years the asan approximation to the binomial distribution. Its first the final random variable fax. But capital acts the note, the number uh cream technicality one items that film commission 25. Okay. Which is when the Challenger exploded. Okay, for now let's work on step one. We need to to find our sample science which is 748. And it's the number uh, criticality one. Right? And P is equal to 0.1 Which is the probability a failure for criticality one. Right to me. Okay. Now, we will use this in step two. First step to we need to satisfy some conditions. We have little and equal to 748 which is greater than or equal to 100 and N. Times P, which is equal to 748. Multiplied by 0.1 Which is equal to 0.7 It has to be listed and I go to 10. This is true. Now we can continue to step three. First step three. We have the probability that capital acts is equal to a little acts, Jessica. To lander, to the X. E. To the native lander or X. Factorial. So this is the Tucson probability mass function. Since we're using the Poisson approximation uh to the phenomenon distribution to approximate right, That's what we're going to approximate the normal decision. We have that land. How should we were two? End touch people which is 0.7 for part a probability that capital access through zero zika to 0.7 to zero E. To negative 0.7 over zero factorial. You plug this in the calculator. You get your 0.93 problem. Or for part B we have the probability the X is greater than or equal one. So that means that there's more, there is one or more criticality on him. So is the same thing as one minus rates as opposed to one. Ah Probably that there are no criticality ease, which is equal to and notice that this part is here. So show your work. You'd be one minus 0.93 which is 0.7

Part of discussions. We have two parts. Fifties for R one equals 23 multiplayer bait. And to the power eight kilometer, that is 11 m. So we have from the capital. Slow time period. Capital T equals two pi R to the power three by two divided by cheap and s. Okay. And four time period. Even this will be good. Uh huh. So, substituting values we get even equals two pi Mhm organ is 3 to 10 to the power 11 power three by two divided by and the root of G E 6.67 multiplayer beta into the par minus 11 and mass of the sun M s is equal to 1.99 will plummet into the power 30. Okay, so from here we get capital T V equals to 8.96 multiple aerator into the power 7th 2nd, which can be converted into years. So we get even equals to 2.84 years. So this is the answer for the part A. And the first part of the party. So for the part two in which we have our two equals two, 5 to 10 to the power eight kilometers, 11 m. So it will be equals two. Yeah. Uh, bullfight are you? So r s 5 to 10 to the power 11. Power three by two. Mhm. I never understood. 6.67 will carve it into the four minus 11, but played by 1.99 when applied by 10 to the power 30. Okay, so from here, T two comes out to be 1.928 molecular weight into the power eight seconds, which can be converted in two years. So we get to equals to six point development. Yes. So this is the second part of the part of the question. Okay? No, we will solve part B of the human question. Okay. Part B in this part, we have time, period. Capital T equals to 5.93 years and we have to determine the orbital radius 40 Kirkwood gap. So from the capital, so we have people equals equals two pi r to the power three by two. They were by G. M s under it, so substituting value. Here we get 5.9 ft years and years can be converted into seconds as 3 65 multiplied by 24 multiplied by 36 double zero seconds and go back R to the power three by two Development and the Group G A 6.67 will carve it into the par minus 11. And M s is 1.99 manipulate and to the power 30 k z. Okay, so now after solving it, we get r equals two four point 90 10 to the power 11 m. So this is the orbital radius for the gap. Okay, so this is our answer for the party. Okay? No moving to the party in which we have orbital time. Capital t was too. 0.4 molecular by 1.86 years. That is 4.7 double four years. Okay, So from the capital Oslo, we can write 4.7 double food multiplied by 3. 65 multiplied by 24 multiplied by 36 double G seconds equals two buh bye r to the power three by two they were by under root of 6.67 multiplied by 10 to the par minus 11, multiplied by 1.99 multiplied by 10 to the power 30 cases. So from here, after solving our comes out to be four point double, two multiplied by 10 to the power 11 m. So this is our little tedious for the car coat.


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