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5.1.4ncuth#eredanemeOMd 100 Ll...

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5.1.4ncuth#eredanemeOMd 100 Ll

5.1.4 ncuth #eredaneme OMd 100 Ll



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(a) 200 (b) $-200$ (c) 0 (d) $-2001$

You want a subtract sits one is there he is. Three minutes. 503 minus two is one for minus three is one in one minutes. One is here, so our answers that

Lou. Today we will be determining if we have a problem destruction or a distribution today. But before doing that, we will be discussing or reviewing what a probability distribution is. And definition of probable distribution is a table or an equation that links each outcome of statistical experiment with its probability of occurrence. And, of course, the two rules that accompany that quick definition is that one all probabilities in the probability distribution must be between zero and one we can't have in negative probabilities and no probabilities. Great of the one crystal. That's not a probability anymore. And number two, some of off probabilities must evil one if we have more or less than one than were either accounting for some event more times than actually curse. Or if it's less the one them we're missing some event were missing some of our possible outcomes, and we're just not including them now. Our example for today as I will draw up now Sorry, dropping a pin is this follows. We have X, our event on top and our probability when bottom no plus where that's next. So our excess today are 20 30 40 and 50. The probabilities are 0.50 point 35 0.4 0.2. Now start with. First thing we can see is that we have no negative probabilities or any that exceed one. So we know that rule number one is followed. Now our next check will be the total of profile are probabilities. So we got to calculate that up. So point far 0.0.5 point 35 become 0.4 0.4 plus 0.20 point six. And if we add those together or total is one So number two rule number two is also followed. So now we know that the distribution went not here is indeed a probability Distribution is a table that can link each outcome of us. This experiment with the probability of occurrence for that outcome because all of the requirements were to be probability distribution are met

Right here. We need to distribute this. First of all, so we have 0.3 x when we distribute this point. 04 times 1000. Remember, we just have to move the decimal place three times. So 123 We end up getting 40. We have 40 point of four minus x negative 0.4 eps combining or like terms here, 0.3 Monet's point of four will give a negative 0.1 x Bring down the positive 40 and that, sir, Finally. And you're right there.

In this problem. We're going to solve the linear equation, and we're going to do this by eliminating decimals and then solved for the variable V m. So I am the longest decimals air 0.18 point 21. They have the most decimal places, so we're gonna mult. There's two in each one of those through the multi by two, tens of 10 times 10 is 100. We're gonna multiply everything by 100 100 times. 1000.3 is 30 a m. 100 has 1000.18 is 18, 100 has 1000.21 is 21. So let's go and distribute the 18 into the parentheses. So let me grab my calculator here and let's do UM 18 times 5000 18 times 5000 clear. The Saudi year 18 times 5000, and we get 90,000 and then 18 times. The minus M is a minus 18 m and 21 times 5000 on the other side is going to be, um, 100 5000. Let's combine the 30 EMS on this side. We have 30 minus 18, which would be 12 m plus. The 90,000 equals 105,000. So track 90,000 off both sides. And in this case, um, what's left over is going to be, um, let's check it out. We have 12 m equals R 15,000 divide by 12 on both sides. So let's take, um 15,000 and divide by 12 and we get 1000 200 50.


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