Recent content by YODA0311

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    Auto Insurance monthly interest rate. Having trouble

    [b]1. If you took out a $11,250 car loan at 3.840 % for 5 yrs (60 months), (a) what would be the monthly payment? (b) what is the difference between paying the $11,250 up front versus taking a loan out for it? [b]2. Interest formula: I= P r T [b]3. (a) I= (11,250)(.03840)(5)...
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    Post Measurements for Golden Ratio Study

    thanks, ill just take the (1)'s out :)
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    What is the probability of drawing a full house with three aces and two kings?

    To make it more clearer for you: (a) in how many ways can 5 cards be drawn from a whole deck?= 52!/(52-5)!=311875200 (b)how many ways can 3 aces be drawn from the total 4 possible aces?=4!/(4-3)!=24 (c)how many ways can 2 kings be drawn from the total 4 possible kings?=4!/(4-2)!=12 (d)how...
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    What is the probability of drawing a full house with three aces and two kings?

    Yes part (a) is from a whole deck of cards, however I already checked the answers with previous tutors. However maybe you see something wrong. I was using the permutation formula= n!/(n-r)!
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    Post Measurements for Golden Ratio Study

    Post your measurments pleasezz I am answering a statistics problem and it asks me to answer questions referring to human measurements. Trying to reveal "the golden ratio" base on the info below: I am gathering TOTAL HEIGHTS(inches) and HEIGHTS OF A PERSON'S NAVEL, MEASURED FROM THE...
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    What is the probability of drawing a full house with three aces and two kings?

    I understand where both you are coming from now. I guess it was easier to read from a textbook and follow the examples. However the actual question is "in how many ways can three aces and two kings be drawn?" So i figured out how 1 hand of 3 aces and 2 kings can be arranged=10 I found out how...
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    What is the probability of drawing a full house with three aces and two kings?

    Actually that answer only applies to 1 hand of 5 cards. Therefore I must multiply it by 288. in how many ways can three aces be drawn=24, in how many ways can two kings be drawn=12...12x24=288
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    What is the probability of drawing a full house with three aces and two kings?

    I think i figured out problem "in how many ways can three aces and two kings be drawn?" I used the permutations rule(when some items are identical to others). 3 aces and 2 kings are different, but are all considered cards. In the book they used an example of how many was can 11 girls and 3...
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    Maximizing Expected Profit: A Contractor's Dilemma and How to Calculate It

    Expected profit, help please 1. A contractor is considering a sale that promises a profit of 38,000 with a probability of 0.7 or a loss (due to bad weather, strikes, and such) of 16,00 with a probability of .3. What is the expected profit...
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    What is the probability of drawing a full house with three aces and two kings?

    in that example you used 3 kings and 3 aces, 3 of each
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    What is the probability of drawing a full house with three aces and two kings?

    you chose 3 because that is how many spots the aces are taking up
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    What is the probability of drawing a full house with three aces and two kings?

    in my problem, n=288, r=5. I came up with that based on there are 288 ways, three aces can be drawn out of four aces, and two kings out of four kings. Then i used 5, because we want to know within a hand how many different arrangements can we get
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    What is the probability of drawing a full house with three aces and two kings?

    Vela I appreciate your help and I will start fresh so you can better understand the problem: a. in how many ways can five cards be drawn from the deck? I answered= 311875200 b. in how many ways can three aces be drawn? i answered= 24 c. in how many ways can two kings be drawn? i answered= 12...
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    What is the probability of drawing a full house with three aces and two kings?

    so to answer the question "in how many ways can three aces and two kings be drawn? is 3!/(3-2)!2!=3!/1!2!=6/2=3= there are 3 ways, three aces and two kings can be drawn
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    What is the probability of drawing a full house with three aces and two kings?

    Okay if i was using the combination method(order does not count) I am having a hard time incorporating the factors. the equation is = n!/(n-r)!r!, n=different items available, we select "r" of the "n" items(without replacement) I do not understand how you got n=3, r=2?
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