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I thought about the same thing Mathsperson so I did a little more research. Here in the United States we have several multistate lotteries. The top prize grows each time there is a drawing with no winner. I believe the largest jackpot ever was $340M (million US dollars). The odds of picking the winning combination of numbers is over 146 million to one! Sounds like a good deal in this case where they're paying $340M for odds that are only 146M to 1.
But the jackpot prize is paid over 30 years. Or you can just take the cash equivalent lump sum right away. For the $340 million, that drops the prize to $165 million. Then you have to pay taxes and now you're down to $110 million. If you bought every ticket possible, you would also have another $29 million worth of non-jackpot winning tickets. That's around $20 million after taxes. So you would have buy 146 million to win 130 million. And you better hope that you had the ONLY jackpot winning ticket!
Is your lottery paid in a lump sum or over so many years? Is it taxable?
reference: www.cnn.com/2005/US/11/08/powerball/index.html?eref=sitesearch
Plus you have to have the cooperation of whoever runs the lottery. Or a lot of people helping you. If you could purchase 5 tickets every second, it would take 32 full days to purchase all 14M tickets!
Let's call the angle X. The complement angle would be equal to 90 - X and the supplement would be 180 - X. If I understand you correctly:
5*(90-x) = 180 - X + 150
450 - 5X = 30 - X
4X = 420
X = 105
That means the angle is an obtuse angel (90 < x < 180) and I don't think obtuse angle have complementary angles.
Multiply both sides of the 1st equation by 2:
Now subtract your new first equation from your second:
Multiply both sides by b:
You can use the quadratic formula to solve this now.
It's not a shortcut but more of a verification. If a number is evenly divisible by 3, the sum of it's digits is divisble by 3.
123: 1 + 2 + 3 = 6, therefore divisible by 3 (41)
943: 9 + 4 + 3 = 16, therefore not divisble by 3 (314.333)
123456789: 1+2+3+4+5+6+7+8+9 = 45, therefore divisible by 3 (41152263)
No, don't change it. I wasn't suggesting that. I think it's cool!
Hey Toast,
I visit this website while at work and I had a visitor who asked "why is that piece of toast wearing a bikini?"! Once I told him it was sunglasses he agreed but I could certainly see his original point of view. Those are sunglasses, right?
#117
49! can be evenly divided by 14 only 8 times. To be divisible by 14, it must also be divisible by 7. When calculating N = 49!, the only time a 7 is factored in is when multiply by 7, 14, 21, 28, 35, 42, and 49 (two 7's in that one).
It looks good to me except all of your 16x should be 16x^2.
I'm not sure if this will result in the same answer as Ricky's but here's some general formulas.
The number of ways to arrange 7 unique letters (abcdefg) is 7!
For each duplicate letter, divide by 2!. abcdeff gives you 7! / 2! permutations and abcddee gives you 7! / (2! * 2!).
For a triplicate (don't know if that's a word!), divide by 3!. So abcdeee gives you 7! / 3! permutations.
SUCCESS is a seven letter word with one duplicate and one triplicate. So you have 7! / (3!*2!) permutations. Thats 420. And that does give you the same results as Ricky.
First, remember that:
edited typo mistake... Thanks RickyWe need to figure out what to multiply both sides of your first equation. We want to get the exponents equal:
So multiply both sides of the first equation by 2 raised to the 2x - 1 power:
Maybe they aren't looking for a specific amount of time, like 2 hours and 3 minutes. Maybe the answer should be expressed in terms of the variables you have?
His average speed would be:
Then to figure out how long to run 20 miles, you divide 20 by the average:
I believe the above equation is the answer.
Let's say he run 8 miles in 1 hour and then 10 files in 2 hours. His average speed is 6 hours (18/3), so it should take 3 1/3 hours.
The first one should be:
The second is
The way I read the orginal question is that books on the same subject must be grouped together. So you could have the 2 math books sitting to the left of the law books or the 2 math books sitting to the right of the law books:
MM LLLLLL or LLLLLL MM.
There are only 2 ways to arrange the math books: M1 M2 or M2 M1. Thats 2!, by the way.
There are 6! ways to arrange the law books. That's 720.
You have to multiply these 2 numbers (2 and 720) together to get the numbers of ways to arrange them in MMLLLLLL fashion. That's 1440 ways.
There are an equal number of ways to arrange them in LLLLLLMM fashion. So multiply the 1440 by 2 to get a total of 2880. A
Bah, pi man got there first. Well, at least it means that we've confirmed each other's answers.
Mathsyperson - That's happened to me more than once also. And I still have less than 100 posts. Maybe the next time Mathisfun is doing someweb redesign work, there could be another option added next to "post reply" that's something like "Will post reply". And then the thread could be flagged somehow with something like "Pi Man is working on a reply". The flag would be cleared when I do post my answer or after some time limit (1-2 hours?). Being flagged wouldn't prevent anyone else from replying, just informing them that someone(s) else is already working on it.
Or I guess for solutions which we might be spending a few minutes working out and typing up, we could just post a reply which states we're working on it? Too simple!
N + D + S = 35
0.05N + 0.1D + S = 12.25
subtract -0.05N both sides
0.1D + S = 12.2
subtract 0.1D from both sides
S = 12.1
You were doing fine until you started subtracting. When you subtract 0.05N from both sides, it does "cancel out" the existing 0.05N on the left side of the equation. But on the right side you would have 12.25 - 0.05N. You simply subtracted .05 rather than subtracting .05N.
You need to use the statement that there are twice as many silver dollars as nickles.
S = 2N
Going back and doing some substitution:
N + D + 2N = 35
3N + D = 35 (A)
0.05N + 0.1D + S = 12.25
0.05N + 0.1D + 2N = 12.25
2.05N + 0.1D = 12.25 (B)
Let's mutliple both sides of the equation I denoted as (B) by 10
20.5N + D = 122.5 (B1)
Now subtract A from B1
17.5N = 87.5
N = 5
So you have 5 nickles ($.25) and 10 silver dollars ($10). Since you have 35 coins total, you must have 20 dimes ($2). And that works out: $10 + $2 + .25 = $12.25
10. Since any digit can be repeated, there are 5 possibilities for the first digit, 5 for the second, and 5 for the third. Or 5*5*5=125 different numbers
There are 12! ways to arrange the 12 kids. That's 479,001,600. diffent ways.
For all of the girls to be sitting in a row, they could be occupying seats 1-6, 2-7, 3-8, 4-9, 5-10, 6-11 or 7-12. That's 7 different grops of seats that could be occupying. Within each of those groups, they could be arranged 6! different ways (720) for a total of 7*720=5040 different ways. 5040 ways out of 479001600 = .001052%
6. Find the probability of getting 1 or 2 heads exactly when flipping a coin 3 times. It's easier to consider the opposite: what's the odds of getting 0 or 3 heads exactly? Theres only 1 way each to get those two results (TTT, HHH). There's 8 differents ways total (2*2*2) so that leaves 6 of 8 chances (75%) of getting 1 or 2 heads.
5. Ways of throwing a particular total when throwing 2 dices:
2 - 1 (1,1)
3 - 2 (1,2), (2,1)
4 - 3 (1,3), (2,2), (3,1)
5 - 4 (1,4) (2,3), (3,2), (4,1)
6 - 5 (1,5), (2,4), (3,3), (4,2), (5,1)
7 - 6 (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
8 - 5 (2,6), (3,5), (4,4), (5,3), (6,2)
9 - 4 (3,6), (4,5), (5,4), (6,3)
10 - 3 (4,6), (5,5), (6,4)
11 - 2 (5,6), (6,5)
12 - 1 (6,6)
So there are 15 (5+4+3+2+1) ways of throwing a total of 8 or better. 8 of 36 is 22.22%
4. There are 366 days in a leap year. That's 52 full weeks plus 2 extra days. If the year starts on a Sunday, there's going to be 52 full weeks of Sunday through Saturday, plus an extra Sunday and a Monday. Likewise, if it beings on a Monday, there's going to be 53 Monday's and 53 Tuesday's. There's going to be 53 Tuesdays whenever the year starts on a Monday or a Tuesday. So that's 2 of the possible 7 days the year could start on. 2 / 7 = 28.57%
3. There are 52 choose 3 ways of picking 3 cards from a deck of 52.
You want to choose 1 of the four possible jacks, one of the four possible queens and one of four possible kings (4 choose 1 for each).
There are 64 different ways to choose 1 jack, 1 queen and 1 king. That's out of 22100 possibilites, so the probabily is 64 / 22100 or 2.896%
2. There are 24 different ways (4!) to put the 4 letters (a, b, c and d) in the 4 envelopes (A, B, C and D).
There is only 1 chance out of the 24 combinations of getting all 4 letters in the right envelope.
How many ways of getting exactly 3 of the 4 letters right? Zero. If three of them are right, the 4th would have to be right also.
How many ways of getting exactly 2 of the letters right? Lets say we get the first two correct (Aa, Bb). That means c has to go in D and d has to go in C: Aa, Bb, Cd, Dc. So there's only 1 way to get only letters a and b in the correct envelope. There's also other ways to get 2 letters right - a and c, a and d, b and c, b and d, and c and d (4 choose 2). So there are 6 different ways to get exactly 2 letters in the correct envelope.
How many ways of getting exactly 1 letter right? Let's say D is correct. The possible combinations for the first 3 are ABC, ACB, BAC, BCA, CAB and CBA. Only 2 of these combinations have the none of the letters in the right envelope - BCA and CAB. So there are 2 ways to get only letter D in the correct envelope. Similarily there are 2 ways each to get only A, B or C in the correct envelope. That gives a total of 8 ways to get just one right.
How many ways of getting exactly 0 letters right? THere are 24 combinations total. 1 way to get 4 right, 6 ways to get 2 right, 8 ways to get one right. That leaves 9 ways to get 0 right (24 - 1 - 6 - 8)
1. There are 12 marbles in all. You want to randomy pick 3 of them. So there are (12 choose 3) ways to do that:
You want to know what the odds of drawing 2 red balls plus either a green or black ball. There are 10 different combinations of 2 red balls (5 choose 2) and there are 7 different greeen or black balls (7 choose 1) to choose from. So that's 10*7 different ways out of the possible 220 ways. 70/220 ~= 32%.
Unless you don't count the row and column which goes through the center square? If that's the case, then there is a solution. Label the houses as follows:
A B C
D E
F G H
The top row (A, B, & C) and the bottom row (F, G & H) add up to 15 each for a total of 30. That means houses D and E must add up to 6. Likewise. houses B plus G have 6 people total. Using each number 1 through 8 once and only once, there are only 2 sets of numbers that add up to 6 ==> 1+5 and 2+4. That leaves 3, 6, 7 and 8 for the corner squares. Let's go ahead and put 5 in spot B. That means A+C = 10 and therefore A and C must be 3 and 7.
3 5 7
D E
F 1 H
Now look at the column with the 3. We know D and E are going to be filled in with 2 and 4. D can't be the 2 since that would make F=9. So D=4, E = 2, F = 8 and H = 6.
3 5 7
4 2
8 1 6.
So there's really only 1 solution. Any other "solution" is just some rotation or mirror image of the above.