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Cool. Thanks Ricky.
Identity, Ricky
Eucludian Method - Is that method demonstrated by Identity? Pretty simple. I don't remember ever being taught that method in school. I was taught to compute the prime factorization which then could also be used to find the LCM (Least common multiple).
2% = .02
3800 * .02 = 76
The prime factorizations are:
2241 = 3^3 * 83
1411 = 17 * 83
The only factor they have in common is 83 so that is the greatest common factor.
12 1/2 % = 12.5%
You can multiply the orginal price by 12.5% (.125) to find out the amount of the discount and then subtract that from the original price. Or you can multiply the original price by 87.5% (100% - 12.5% = 87.5% = .875).
You would be right if the question was what are the chances of flipping 2 heads followed by a tails. What about THH and HTH? They also have a 1/8 chance of happening so add these together to get 3/8.
And when the problems involve a relatively limited set of possibilities, it's often easy to list them all to verify/reject your solution.
HHH
HHT good
HTH good
THH good
HTT
THT
TTH
TTT
I think I see what you were doing in post #4 now. It looks like you are adding in the requirement that 3 girls are sitting in the the 3 seats between the 2 guys. That wasn't stated in the original question but if you have the same books as mathsux, maybe that is what he/she meant.
I used a little brute force to double-check my answer as the question was originally stated. Consider the number ways you can arrange the letters BBBBGGGG with B standing for boy and G for girl. That would be 70 (8! / (4! * 4!). Using Excel, I generated those 70 possible combinations. I then counted how many of the seating arrangements had 2 boys sitting 3 seats apart. 54 of the 70 (=~ 77.1%) met the requirements.
I don't really understand your most recent post. If the original question is asking for the odds of 2 specific boys sitting 3 seats apart, then your answer is correct. If the question is what are the odds of any 2 of the 4 boys sitting 3 seats apart, then I think my answer is correct.
I think the answer is a bit more complicated than that. Your solution only takes into account when Morgan and Tom are 3 seats apart. What about when Bob and Fred are 3 seats apart? Or Morgan or Fred? There are 6 combinations of 2 boys which could be seated in the required arrangement.
But we just can't multiply the your answer by 6 because there will be duplicates when 2 differents sets of boys have 3 seats between them. Such as... M B _ _ T F _ _.
It may be easier to calculate the number of arrangements were no 2 boys are three seats apart. Let's have the boys pick their seats. Boy 1 can pick any of 8 seats. Let's say he picks seat 3. Boy 2 can then pick any of 6 seats because he can't pick seat 3 (boy 1 is already there) or seat 7 (he would be 3 seats from boy 1). Likewise, boy 3 could pick from 4 seats and boy 4 could pick from 2 seats. That's 8 * 6 * 4 * 2 = 384. The 4 girls could be arranged in 4! =24 ways. 384 * 24 =9216. And there are 8!=40320 total ways the kids would be arranged so 9216 / 40320 ~= 22.9%. That's the chances of no 2 boys being 3 seats apart. Subtract that from 1 to get a 77.1% chance to get the odds of 2 boys sitting 3 seats apart.
That sounds a little high to me. Maybe not though because if we multiplied Identity's answer by 6 we would get around 86%. Can someone else confirm or point out a mistake in my "solution"?
X=1 will work
#3. Again, I'm not sure if I'm reading this correctly but...
You must select 4 bowlers from a group of 5. That's (5 choose 4) and is equal to 5.
You must select an additional 7 players from the remaining 12. (12 choose 7) = 792.
5*792=3960 is what I come up with.
I'm reading question #1 differently than Jane. I think shivusuja is asking how many ways can you select 9 balls with the requirement that you have 3 of each color. But I've been wrong before!
If 3 balls of each color are right, then they are 20 ways to select 3 red balls (6 choose 3) and 10 ways to choose blue and 10 ways to choose white (5 choose 3). So that would mean 2000 different ways. This doesn't take into consideration the order of which the balls drawn.
The difference between p and q is 3. Therefore one of the primes has to be even and the other has to be odd. The only even prime is 2. And 5 is also a prime so the only pair of primes with a difference of 3 is 2 and 5.
My best guess is that it is a typo and it is supposed to be "/", i.e. division.
(21 x 8) / 4 + 60 - 17 =
= 168 / 4 + 60 - 17
= 42 + 60 - 17
= 102 - 17
= 85
Let N = the number of nickles. Then 20-N = the number of dimes.
.05N + .10 (20-N) = 1.40 (5 cents * the number of nickles plus 10 cents * the number of dimes)
.05N + 2.0 -.10N = 1.40
.60 = .05N
N = 12
Therefore you have 12 nickles (60 cents) and 8 dimes (80 cents).
If you make y=x^2, then your original equation becomes:
Speed of the fast train = F
Speed of the slow train = S
Time it takes for the trains to meet (pass each other) = T
Since both trains travel the same total distance and distance=time*speed:
F(T+1) = S(T+4)
We're trying to figure out F/S which is equal to (T+4) / (T+1) from the equation above. So we need to figure out the value of T.
After the meet, the fast train travels one more hour at speed F and covers the same distance the slow train covered in T hours
F*1 = S*T or F=S*T
After they meet, the slow train travels for 4 more hours and covers the same distance the fast train covered in T hours.
S*4 = F*T
Substituting S*T from the first equation in for F in the 2nd equation:
4S = S*T*T
4 = T*T
2 = T
Substitute 2 in for T in the (T+4) / (T+1) equation to get 6/3 or 2. The fast train is going twice as fast as the slow train.
Jane - I came up with 60000 also if you allow orders other than LLNN but I was wondering if there was an easier way to calculate, i.e. a simple formula. I calculated the number of ways where neither number or letter were repeated such as AB12 (48600), if either the letter or the number was repeated such as AA12 or AB11 (5400 each), and if both the letter and the number was repeated such as AA11 (600).
To get the percentage of orange juice, you divide the volume of orange juice by the total volume. The total volume is simply the amount of water (w) plus the amount of Tang-E (t). That's where the (t+w) comes from.
The orange juice comes only from the Tang-E portion of the mixture. Specifically, it's 45% of the amount of Tang-E or .45t . That gives you .45t / (t+w). Since you want the answer in terms of percentage, you multiply by 100 which gives you 45t / (t+w) %.
Break B into 4 groups:
Those members which are only in B
Those members which are only in A and B
Those members which are only in A and C
Those members which are in A, B and C.
If you can find out how many members are in each of those groups, you can add them up and get the total.
You were told there were 0 members which are only in B and 6 members which are in A, B and C. A and B have 17 members in common but 6 are also in C so there are 11 members which are in A and B but not C. B and C have 10 members in common but 6 of those are also in A. So there are 4 members in B and C but not A.
0 + 11 + 4 + 6 = 21
If the area is 49, then each side must be 7 since the figure is a square. The length of side WZ (and XY) is determined by calculating the difference of the X coordinates, in this case s and 3-s.
s - (3 - s) = 7
s - 3 + s = 7
2s = 10
s =5
Likewise, the lengths of sides XW and YX can be calculated by finding the different of the y coordinates.
t - (5 - t) = 7
t - 5 + t = 7
2t = 12
t = 6
That would be (5 choose 2). The general formula for (X choose Y) is X! / (N! * (X-N)!).
(5 choose 2)= 5! / (2! * (5-2)!) = 5! / (2! * 3!) = 5*4*3*2*1 / (2*1 * 3*2*1) = 10
If order was important (the team of Jane and Joe would be considered different from the team of Joe and Jane), then then answer would be 5 * 4 = 20.
Using Internet Explorer 7. Refresh, logging out and back in, rebooting... none of those worked. I deleted Temporary Internet Files and it's working again. I'm surprised that fixed it. If anything, I would have thought deleting cookies might have fixed it but I didn't delete those.
Anyway, it's no longer an issue. For me at least. Thanks.
For the last couple of days, whenever I click on the "show new topics since last visit" option, I get the same 4 posts. They all have todays date listed as the last post. The 4 topics are "tough one I;'m stuck on", "let me hear what you think about this", "I love maths so much" and "unbelievably fast mental calculations". Anyone else having problems? I'll try deleting my temporary files and cookies but I'm not confident that will work since I have the same problem on multiple computers.