Recent content by hholzer

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    Physics problem involving sum of torques

    That is much easier, actually. Thanks for the reply.
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    Physics problem involving sum of torques

    Homework Statement Given a beam of 5 meters and uniform weight 100N position on a fulcrum such that it subtends an angle of 20 degrees and the length of the beam to the left of the fulcrum is 3.3m, how far could an 80N cat walk before the beam tips? Here is an image that I drew up in mspaint...
  3. H

    Set Associative Cache: 8-Way Determination

    If I have the addressing format: Tag: 31-12 Index: 11-6 Byte offset: 5-0 And if my cache size is 32KB = 32*2^10 = 2^5*2^10 = 2^15 then this would represent a how many-way set associative cache? i.e., an x-way associative cache? Since bits 11-6 are used for the index, then this...
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    Equating Truth Table Expressions: A NAND Solution

    Wasn't sure of where the most appropriate place would be for this post. If you have a truth table, say for inclusive OR, then you get a sums-of-products expression: (A * B^c) + (A^c * B) + (A * B) From this, how could I arrive at the following: A + B = ((A + B)^c)^c = (A^c * B^c)^c...
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    What is the probability of drawing a spade from a deck of 51 cards?

    Ah, that's what I was trying to determine. So we break it up into (Lost card was spade) and (Lose card not spade). The answer is indeed 1/4 but I was more concerned with how we partition the sample space. And on the word "event", "event" is a subset of your sample space, as you of...
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    What is the probability of drawing a spade from a deck of 51 cards?

    Suppose you had a normal deck of 52 playing cards and lost a card. You then decide to draw a card from the remaining 51 cards. What is the probability the drawn card is a spade? Would this be appropriately captured by the following events: A : event card was drawn from the deck S ...
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    Conditional probability question

    @VandeCarr: My comment was addressed to your and mathman's solution. Not statdad's. statdad responded after you presented a solution and my response preceded statdad's. So it isn't possible that I was referring to your's and statdad's. I know the probability of the event being a boy with...
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    Conditional probability question

    I was puzzling over my book's use of a reduced sample space with the format of a triplet listed above. Their sample space had 12 points in it and so I was not arrive at a probability of 1/7. Where the reduction in sample space is Q(A) = P(A|B) B our sample.
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    Conditional probability question

    It looks like you two reached different answers. I know that the event the middle child is a boy with an older brother and younger sister is 1/14. I was asking about why: P( {ggb*} ) = P( {gb*g } ) = P( {b*gg} ) = 1/7 With a reduced sample space of: (given a triplet (xyz) take it...
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    Conditional probability question

    Consider this scenario: "From families with three children, a family is selected at random and found to have a boy. What is the probability that the boy has an older brother and a younger sister? Assume that in a three-child family all gender distributions have equal probabilities."...
  11. H

    Model of parachute that doesn't include gravity

    It isn't clear to me why you would consider them the same. Could you elaborate on this?
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    Model of parachute that doesn't include gravity

    I'll post this question here because it isn't a homework question. I've already solved this problem that I have taken from a physics book. My question pertains to why would gravity not be accounted for? (The reason I think no gravity is account for is because the solution provided by the...
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    An expression for a sequence of events

    if you have a sequence of events {A_1, A_2, ...} then an expression for the event that "infinitely many A_i's occur" is: U(n = 1 to inf, U(m = n to inf, A_m) ) but wouldn't U(n = 1 to inf, A_n) also satisfy this?
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    Proof that P(A) = 1, P(B) = 1, then P(AB) = 1

    Theorem: If P(A) = 1, P(B) = 1, then P(AB) = 1 My book starts out with the proof as follows: P(A U B) >= P(A) = 1, so P(A U B) = 1 How do they reach such a conclusion? Things I know: P(A U B) = P(A) + P(B) - P(AB) How can I use that to be sure that P(A U B) = 1?
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    Solving Jack's Ticket Problem at Maryland University

    Thanks for your cogent reply. The key here is realizing, under the given assumption, that if P(A) is small, then it contradicts our assumption. Which, as you have pointed out, suggests that the police do not patrol the parking lot randomly.
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