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Looks like I didn't go far enough. Thanks Mathsyperson. But why did you break down the (x^4 - 1) the way you did?
Looks good to me. Just one small mistake and then one more step. Your equation for calculating SIN(C) looks right but the answer is incorrect. It should be SIN(C) = .692320138.
You now know what the SIN(B) and SIN(C) are equal to so you can use the arcsin function to determine the angle.
B = arcsin(.945952038) = 71.07
C = arcsin(.692320138) = 43.81
Add those 2 together plus your 65 for Angle A and you get 179.88. Not quite 180 but I think that can be explained to rounding errors.
You made a couple of mistakes Neha. First, you can't just assume angle C is 65 degrees just because the they look like they are in the picture. Second, the Pythagoreom Theorem (a^2 + b^2 = c^2) only applies to right triangles.
I think you're going to need some trig functions to solve this. Here's what you know:
(Capital letters designate an angle, lowercase letters designate a side. Side c would be opposite angle C).
A= 65 dg
B = ?
C = ?
a = ?
b = 15
c = 11
I could be totally wrong here since its been a long time since I studied trig, but I believe the Law of Cosines could be used to find the value of a. Once you know that, the Law of Sines can be used to figure out B and C.
The Law of Cosines followed by the Law of Sines:
I stated the Law of Cosines as it's normally written which makes it confusing in this case because you only know A. So for you, it would be:
Hopefully this helps. And if I'm wrong, hopefully somebody will correct me.
So each number in the series is 2 greater than the number before the previous number.
How many different letters can be the first letter of the "word"? There are 8 letters in EQUATION and no letter appears twice, so 8 different letters can be the first letter of the word.
How many letters can be the 2nd letter of the word? All of them but the one that's being used as the first letter. So that's 7. We now have 56 combinations for the first 2 letters of the word.
Keep using this method and it will become apparent that the answer will be 8*7*6*5*4*3*2*1. That's 8 factorial, usually written as 8!. That's equal to 40320.
It's important to note that no letter appeared twice in EQUATION. If you were rearranging letters that had repeats, you have to take some additional steps. If one letter is repeated twice, you divide your answer by 2!. If one letter is repeated 3 times, divide by 3!. If 2 letters are repeated twice, divide by (2! *2!).
AABCDEFG = 8! / 2! different arrangements
AAABCDEF = 8! / 3! different arrangements
AABBCCDD = 8! / (2! * 2! * 2! * 2!)
The supervisor can complete 18 jobs in 3 minutes or 36 jobs in 6 minutes. So with the worker (24jobs / 6 minutes) they can complete 60 jobs in 6 minutes. Or 10 jobs per minute.
In the 2 minutes the worker worked by himself, he can complete 8 jobs. That leaves 80 jobs left to complete together. At a rate of 10 jobs/minute, they would need to work 8 minutes together.
Dividing by a fraction is the same as muiltiplying by its inverse:
You're right, Mathsyperson. Even at 80% (.8), I still want him on my team!
The probability of scoring 0 goals out of 6 shots = .000064. The probability of missing 6 shots in a row is equal to the probability of missing one shot raised to the 6th power. So in this case, the probability of missing a shot is .2 and the probability of making it is 98.8. I want him on my team!
The 3 digits numbers below represent the way the balls could end up in the boxes. For example, 0 in the first box, 0 in the second box, 4 in the 3rd box.
004, 013, 022, 031, 040, 103, 112, 121, 130, 202, 211, 220, 301, 310, 400
Since each ball is placed randomly, each of the above possibilities are equally likely. Only 3 of the 15 possible outcomes have at least 1 ball in all 3 boxes. Thats 20%.
It's amazing how these things get under your skin...
I probably shouldn't be surprised but it's a Fibonacci-based series!
# flips # Possible # with 2 heads # without 2 heads
(n) Outcomes in a row (Xn) in a row
_______________________________________________________________
1 2 0 2
2 4 1 2*X1+1 3
3 8 3 2*X2+1 5
4 16 8 2*X3+2 8
5 32 19 2*X4+3 13
6 64 43 2*X5+5 21
7 128 94 2*X6+8 34
8 256 201 2*X7+13 55
9 512 423 2*X8+21 89
10 1024 880 2*X9+34 144
So for 11 tosses it should be 2*880 + (21+34) = 1815 (or 2048-233)
And for 12 tosses it should be 2*1815 + (34+55) = 3719 (or 4096-377)
I'll apologize in advance because there has to be a better way. I'm fairly confident it's correct but it will be interesting to see if anyone else gets the same answer.
There are 1024 (2**10) possible outcomes of the 10 flips. The probability of each one of the outcomes are equal. For example, HHHHHHHHHH is just a likely to happen as is HTHTHTHTHT.
Of the 1024 possible outcomes there are:
1 way to get 10 Heads (10 choose 1)
10 ways to get 9 heads (10 choose 2)
45 ways to get 8 heads (10 choose 8)
120 ways to get 7 heads (10 choose 7)
210 ways to get 6 heads
252 ways to get 5 heads
210 ways to get 4 heads
120 ways to get 3 heads
45 ways to get 2 heads
10 ways to get 1 head
1 way to get 0 heads
Now we have to figure out how many of those ways have at least 2 heads in a row. If you get 6 or more heads, at least 2 of them have to be in a row. So thats 386 ways (1+10+45+120+210).
On the other side, there aren't any ways to get 2 heads in a row if you only rolled 0 or 1 heads. Of the 45 ways to roll 2 heads there are 9 ways to get them in a row (1 and 2, 2 and 3, ... 9 and 10). So add that to your 386 ways to wake it 395.
Now it gets tougher for the 3,4 and 5 heads. (Especially since its late and I'm suffering from too much Thanksgiving turkey!)
Let's see if we can figure out how many ways we can get 5 heads and NONE of them are in a row. For the sequences below, the numbers indicate on which flips the heads were tossed.
1, 3, 5, 7, 9
1, 3, 5, 7, 10
1, 3, 5, 8, 10
1, 3, 6, 8, 10
1, 4, 6, 8, 10
2, 4, 6, 8, 10
I think that's all. So 246 of the 252 ways to get 5 heads have 2 or more heads in a row. Add that to 395 for a total of 641.
***
4 heads... gotta be an easier way but here goes. Consider the 9 cases where only 2 heads were tossed and they were in a row (1 and 2, 2 and 3, etc.). For each of those 9 cases, you can add 2 more heads in any of the 8 other tosses. That's (8 choose 2) or 28 ways. 28 * 9 = 252. But that includes a lot of duplicates. Anytime 4 heads were tossed in a row, they were counted 3 times. For example, 1234 would be counted under 12, 23, and 34. There are 7 ways to get 4 in a row (1234, 2345, ... 78910) and each of those was counted 2 times too many. Subtract 14 from the 252 to get 238.
There are 42 ways to get exactly 3 in a row (1235, 1236, 1237, 1238, 1239, 123A, 2346, 2347, 2348, 2349, 234A, 1345, 3457, 3458, 3459, 345A, 1456, 2456, 4568, 4569, 456A, 1567, 2567, 3567, 5679, 567A, 1678, 2678, 3678, 4678, 678A, 1789, 2789, 3789, 4789, 5789, 189A, 289A, 389A, 489A, 589A, 689A - where A indicates the 10th roll). Each of those ways was counted twice (1235 was counted under 12 and 23). Subtract 42 from 238 to get 196.
Lastly there are the cases where you have 2 heads in a row being rolled twice without getting 3 or 4 in the row. For example 1256. There's 21 cases of these being counted twice (1245, 1256, 1267, 1278, 1289, 129A, 2356, 2367, 2378, 2389, 239A, 3467, 3478, 3489, 349A, 4578, 4589, 459A, 5689, 569A, 679A). Subtract 21 from 192 for a final total 175 ways to get 2 or more heads in a row if 4 heads were rolled. Previous total was 641 so it's now 816.
***
3 heads...
We have to count the number of ways to get 2 or 3 heads in a row if 3 heads were flipped. I brute forced it and counted 64 ways. So the final count is 816+64= 880. So 880 of the 1024 possibilities will have 2 or more heads in a row. That's around 86%.
The only other factors of 145 are 1 and 145.
The diagram for the 2nd child really helps. Do you have one for the 1st child?
2nd child: The kid's dad's mum is 3/4 Armenian and 1/4 greek. That means her dad is 7/8 Armenian and 1/8 Greek. That means she is 15/16 Armenian and 1/16 Greek.
The 1st child is a little more difficult because I'm not sure exactly what you mean. Sounds like his grandma's grandpa was French. So his grandma would be be 1/4 French and 3/4 Maltese. His mom would be 7/8 Maltese and 1/8 French. His Dad is full Armenian so he is 1/2 Armenian, 7/16 Maltese and 1/16 French.
Now if these 2 kids have a child together, he/she would be 15/32 Armenium and 1/32 Greek on his her mom's side and 1/4 Armenian, 7/32 Maltese and 1/32 French on his/her dad's side. That makes 1/32 Greek, 1/32 French, 23/32 Armenian and 7/32 Maltese.
I qualify this answer with a big "I think".
1101 0110
- 0100 0111
========
1000 1111
You need to borrow for the first 5 (rightmost) columns. Hopefully you can understand my notations.
1
1101 0100 You must borrow for the first subtraction which makes the 1 in the 2's
0100 0111 column a 0
=======
1
1
1101 0000 You must borrow for the second subtraction also, making the 1 in the 4's
0100 0111 column a 0
=======
11
1
1100 1000 You must borrow again, this time from the 16's column.
0100 0111
=======
111
1100 1000 Now it's easy with no more borrowing.
0100 0111
=======
1000 1111
Let X be the first number with a remainder of 3. So X = 7a + 3
Let Y be the second number with a remainder of 5. So Y = 7b + 5
XY = (7a+3) * (7b+5)
= 49ab + 35a + 21b + 15
49ab, 35a, and 21b are all divisible by 7. We only need to worry about the 15, which is 2*7 with a remainder of 1.
This may help:
http://www.algebra-online.com/relatively-prime-numbers-1.htm
I took it to mean that characters are allowed to be repeated. But you want to find the number of combinations for UP TO n characters. So for N=3 you need to find all the 3 character combinations plus the number of 2 character combinations plus the number of 1 character combinations. And to be technically correct, you should probably add in the number of 0 character combinations (1 - the null set).
Is this what you mean:
Like you Beanny, I came up with the answer of 533 but also suspected there must be a more efficient way. But now after putting a lot of thought and effort into it, I'm 90% convinced that 533 is the best you can do. This is a great problem - certainly one you can spend more than 10 minutes on! (Which also leads me to believe 533 is the correct answer. If 567 is the correct answer, I suspect it's a pretty complicated solution which would require more than 10 minutes of effort.) Were you interviewing for a job? If so, what type of job?
Some of my failed efforts to come up with 567:
Figuring the best way to maximize the use of the camel was to make sure he was carrying as close to 1000 bananas as possible, I tried making a whole lot of 1 km trips. Below is the start of that journey. The camel eats a banana at the end of each 1 km trip. The "*" marks where that segment of the journey ended and the "<" and ">" indicate the direction of travel for that segment.
3000
> 2000 999*
< 1999* 999
> 999 1998*
> 999 998 999*
< 999 997* 999
< 998* 997 999
> 1994* 999
> 0 994 1998*
> 994 998 999*
Working this all the way through still yields 533 banananas at the end.
I also tried making the first trip 200km but in addition to eating 1 banana every km, I dropped off 4 bananas at each 1 km marker for "snacks" on future trips. So after returning from the 200 km trip (each way), I still had 2000 bananas at the start and 3 bananas littering each 1km marker along the way up to 200km. That enabled the camel to carry 1000 bananas all the way to the 200 km marker on the next 2 trips. So I ended up having 2000 bananas at the 200 km marker - the same thing you have in your original answer.
A line has a formula of y = mx +b, where m is the slope (rise over run). In the case of a horizontal line, the slope is 0. So the equation is simply y = -2.
It takes 120 days of work (4 days * 30 workers) to do 6 jobs. So it takes 20 days of work to complete 1 job. 4 jobs takes 80 work days. If you get 20 workers on those 4 jobs, it would take 4 days.
Looks right to me.
My first inclination is to say no but it really depends on some other things. Am I staying in good physical and mental health or am I going to spend eternity in a nursing home (I'd chose hell first!)? And how about my family and friends? Are they living forever or are they dying at a normal age?