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Lê Anh Duy
12 hour ago (10:59)Lê Quốc Trần Anh Coodinator
15 hour ago (07:58)Lê Quốc Trần Anh Coodinator
15 hour ago (07:45)
As we can see in this problem, we have 3 situations that may happen:
Door 1 Door 2 Door 3 Monkey Monkey Car Monkey Car Monkey Car Monkey Money Assume that you choose door 1, you can see that the probability to win the car is onethird. If the MC shows door 2 or 3, you may think that now the chance is fiftyfifty, but in fact not. If you stay with door 1, the probability to win is still onethird; but if you change the door, the fact is that now the probability to win is twothirds, so if you change, you are more ly to win.
Note: You can see the same problem on Google, 'The Monty Hall Problem', for an illustrated explanation.
Lê Quốc Trần Anh selected this answer. 
Can you explain why?

Lê Anh Duy 12 hour ago (11:01)
Of course change the door. What is in your mind???????
Lê Quốc Trần Anh Coodinator
15 hour ago (07:40)
Lê Anh Duy 12 hour ago (10:41)
Call Amos is A, Butch is B and Cody is 3
C_{1: }  If A is the one to shoot first and shoot C, he will be killed by B
=> SO A will shoot B first
Then it's time for C to shoot A
=> The percentage of A alive is 50% and C is always alive (out of shot)
C_{2: } If B is the one to shoot first and shoot C, he will be killed by A
=> SO B will shoot A first
Then it's time for C to shoot B
=> The percentage of B alive is 50% and C is always alive (out of shot)
 If C is the one to shoot first
+ He shoots A:
C_{3}. If shoot to target, B will kill C => B is alive
C_{4}. If shoot miss, B will kill A or A will kill B (reason above) and then C and A (or B) are alive (out of shot)
+ He shoots B:
C_{5}. If shoot to target, A will kill C
C_{6}. If shoot miss, B will kill A or A will kill B (reason above) and then C and A (or B) are alive (out of shot)
SO WE HAVE:
ALIVE 50% DEAD
C_{1}: C A B
C_{2}: C B A
C_{3: }B A , C
C_{4: }C, A (or B) B (or A)
C_{5}: A B, C
C_{6}: C, A (or B) B (or A)
SO C HAS THE BIGGEST CHANGE TO LIVE (\(\dfrac{4}{6}=66.66\%\) )
A (or B)' s probability to live;\(\dfrac{1\dfrac{4}{6}}{2}=\dfrac{1}{6}=16.66\%\)
Lê Quốc Trần Anh Coodinator
15 hour ago (07:31)
Alone 15 hour ago (08:20)
We have:\(V_{hinhcau}=3S_{hinhcau}\)
\(\Rightarrow\dfrac{4}{3}.\pi.r^3=3.4.\pi.r^2\) with \(r\) is the radius of this certain
\(\Rightarrow\dfrac{r^3}{r^2}=\dfrac{3.4.\pi}{\dfrac{4}{3}.\pi}=\dfrac{3.4}{\dfrac{4}{3}}=\dfrac{3.4.3}{4}=9\)
\(\Rightarrow r=9\)
Lê Quốc Trần Anh selected this answer.
Lê Quốc Trần Anh Coodinator
15 hour ago (07:30)
Lê Anh Duy 12 hour ago (11:07)
Sorry for OLD answer
First ask a simpler question: What is the maximum number of points of intersection for a line and a square? Because the square is convex, the answer is 2.
A Triangle consists of 3 line segments, each of which is a subset of a line. So each segment can clearly not intersect the square more than twice. This gives 6 as the upper bound, even if you extend each edge of the triangle into an infinite line.
Further note that if both endpoints of an edge are inside the square, then the whole edge is inside the square and doesn’t intersect it at all. And if one endpoint is inside and one is outside, then there can only be one point of intersection.
Therefore, you can only get 6 points of intersection if all 3 corners of the triangle are outside the square. If one is inside, the max is 4; if two are inside, the max is 2; and if 3 are inside, the max is 0.
Lê Quốc Trần Anh selected this answer. 
Lê Anh Duy 12 hour ago (10:42)
We might start by examining the number of ways that ONE SIDE of a triangle can intersect a square.
In other words, in how many ways can a LINE intersect a square?
After a bit of mental imagery, we might conclude that a SINGLE LINE can intersect a square in at MOST 2 ways
A triangle is composed of THREE LINE SEGMENTS.
If each SINGLE LINE can intersect a square in at MOST 2 ways, then the 3sided triangle can intersect a square in AT MOST 6 ways (with 2 intersections per line)
So, the correct answer must be 6 or less
At that point, if we're able to sketch a scenario in which there are 6 intersections, we can be certain that this is, indeed, the GREATEST number of intersections.
FC Alan Walker
22 hour ago (00:40)
Alone 15 hour ago (08:10)
We have:\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y3}{z}=\dfrac{1}{x+y+z}\)
\(=\dfrac{y+z+1+x+z+2+x+y3}{x+y+z}=\dfrac{2x+2y+2z}{x+y+z}=2\)
\(\Rightarrow\left\{{}\begin{matrix}y+z+1=2x\left(\cdot\right)\\x+z+2=2y\left(\cdot\cdot\right)\\x+y3=2z\left(\cdot\cdot\cdot\right)\\x+y+z=\dfrac{1}{2}\left(\cdot\cdot\cdot\cdot\right)\end{matrix}\right.\)
\(\left(\cdot\cdot\cdot\cdot\right)\left(\cdot\right)\Rightarrow x1=\dfrac{1}{2}2x\Rightarrow3x=\dfrac{3}{2}\Rightarrow x=\dfrac{1}{2}\)
\(\left(\cdot\cdot\cdot\cdot\right)\left(\cdot\cdot\right)\Rightarrow y2=\dfrac{1}{2}2y\Rightarrow3y=\dfrac{5}{2}\Rightarrow y=\dfrac{5}{6}\)
\(\left(\cdot\cdot\cdot\cdot\right)\left(\cdot\cdot\cdot\right)\Rightarrow z+3=\dfrac{1}{2}2z\Rightarrow3z=\dfrac{5}{2}\Rightarrow z=\dfrac{5}{6}\)
\(\Rightarrow P=2016x+y^{2017}+z^{2017}=2016.\dfrac{1}{2}+\left(\dfrac{5}{6}\right)^{2017}+\left(\dfrac{5}{6}\right)^{2017}\)
\(=1008+\left(\dfrac{5}{6}\right)^{2017}\left(\dfrac{5}{6}\right)^{2017}=1008\)
FC Alan Walker selected this answer.
Lê Anh Duy
Yesterday at 15:21Lê Quốc Trần Anh Coodinator
Yesterday at 03:04
Fixida Yesterday at 09:44
đúng rồi admin nên xóa người dùng này mãi mãi
Lê Quốc Trần Anh Coodinator
16/03 at 11:33Lê Quốc Trần Anh Coodinator
16/03 at 11:27
Tell the pencil is a, the paper clips is b and the eraser is c
=> a + 5b = 2c and a = 29b
=> 29b + 5b = 2c
=> 34b = 2c
=> c = 17b
So 17 paper clips weigh the same as an eraser.
Lê Quốc Trần Anh selected this answer.
Lê Quốc Trần Anh Coodinator
16/03 at 11:26
Tell these number to find is x
=> 100 < x < 1000
=> x has 3 digits.
The hundreddigit has 3 choice
The tendigit has 3 choice
The unitsdigit has 3 choice too
So there are 3.3.3 = 27 intergers satisfy
Lê Quốc Trần Anh selected this answer.
Lê Quốc Trần Anh Coodinator
16/03 at 11:25
2^{2017} . 7^{2017} = 2^{504.4+1}.7^{504.4+1} = ...2*...7 = ...4
So the units digit of 2^{2017}.7^{2017} is 4
Lê Quốc Trần Anh selected this answer.
Nhi Hạ Băng
16/03 at 08:21
We have: 4 apples + 6 bananas = 1.56
9 apples + 7 bananas = 2.60
=> 4 apples + 6 bananas + 9 apples + 7 bananas = 1.56 + 2.60
13 apples + 13 bananas = 4.16
=> 1 apple + 1 banana = \(4.16\div13=0.32\)
So the answer is: 0.32
P/s: Sorry I don't know how to do the euro sign.
Nhi Hạ Băng selected this answer.
Lê Quốc Trần Anh Coodinator
16/03 at 07:48
We have: a^{m} : a^{m} = a^{0} = 1
So 0^{5} : 0^{5} = 0^{5} : 0 undetermined
So 0^{0} undetermined
Selected by MathYouLike
Lê Quốc Trần Anh Coodinator
16/03 at 07:44
Convert: 6 feet = 72 inch ; 8 feet = 96 inch
So the area of this wall is: 72.96 = 6912 (squareinch)
The area of each tile is: 4.4 = 16 (squareinch)
So there are 6912: 16 = 432 (tiles) are neede to cover all area of this wall
Lê Quốc Trần Anh selected this answer.
Lê Quốc Trần Anh Coodinator
16/03 at 07:44
\(\widehat{ABC}=90^o \Rightarrow O\in AC \\\Rightarrow The\hskip0.1cmlength\hskip0.1cm of\hskip0.1cm arc \hskip0.1cmABC\hskip0.1cmis\\ l= \dfrac12\times2\pi\times3=3\pi \hskip0.1cm(meters)\)
Lê Quốc Trần Anh selected this answer.
Lê Quốc Trần Anh Coodinator
16/03 at 07:42
From the question, we have some equations:
\(\left \{ \begin{array}{l} A+B=150\times 2=300\quad(1)\\ B+C=127\times 2=254\quad (2)\\ C+D=168\times 2 =336\quad(3) \end{array}\right.\)
We do the following operation \((1)+(3)(2)\). after that, we have \(A+D=300+336254=382\).
Therefore, the average weight of sheep A and sheep D is: \(382\div 2= 191\)(pounds)
Note that \(A,B,C,D\) is the weight of each sheep A, B, C, D, respectively.
Lê Quốc Trần Anh selected this answer.
Lê Quốc Trần Anh Coodinator
16/03 at 07:42
From the question, we have 3 kinds of meat and 4 kinds of vegatables.
To get a twodistinctmeat pizza, we need to choose the first meat and the second meat.
From 3 kinds of meat, to choose the first meat, we have 3 options. After choosing the first meat, we have 2 kinds left (because the first and the second meat need to be distinct), so we have 2 options for the second meat.
So, we have \(3\times2=6\) twodistinctmeat pizzas.
To get a twodistinctvegatable pizza, we need to choose the first vegatable and the second vegatable.
From 4 kinds of vegatables, to choose the first vegatable, we have 4 options. After choosing the first vegatable, we have 3 kinds left (because the first and the second vegatable need to be distinct), so we have 3 options for the second vegatable.
So, we have \(4\times3=12\) twodistinctvegatable pizzas.
To get a onemeatonevegatable pizza, we need to choose the meat and the vegatable.
From 3 kinds of meat, we have 3 options. From 4 kinds of vegatables, we have 4 options.
So, we have \(3\times4=12\) onemeatonevegatable pizzas.
After all, we have \(6+12+12=30\) combinations of pizza.
Lê Quốc Trần Anh selected this answer.