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Rectangle of X and Y distribution:
(1,1) (3,1) (3,4) (1,4)
(X,Y) is equally distributed in this rectangle with f(x,y)=1/6=1/2*1/3, on the assumption that X and Y are independent.
draw a line y=x, any (x,y) above this obeys y>x, whereas any (x,y) below this obeys y<x
y=x cut the rectangle at (1,1) and (3,3)
The area of the rectangular above the line is 4
Hence the probability that Y>X is
1/6*4=2/3
Ricky's result in Post 9 is 2a² times as large as my result, because what he evaluates is 2a² times as large as the original question, too.
So the answer is:
Uhh, +/- mistake corrected. Now my result is not 0 but a²/3, consistent with your result in Post 19.
My answer is 0. Can anyone evaluate it by software?
---my new answer is a²/3
This is correct instead:
George, your first step seems to be saying that:
Which does seem to be the case.
Uhh... you are right! I'm correcting it.
hmmm.... great!
Hmm, I prefer using metal pieces with holes to Lego in building my designs.
Hmm, but I like brown hair the most.
The essence of counting is identity. Though the last scoop is one ball, but it cannot be considered equivalent to either of the original two scoops.
It's like tossing n dices, and to get the chance of getting the sum of dice values
less than a(sounding ei) times of the least dice value.
But it's so hard !!!!!!!!!
when x=0, x/2=0; when x=π/2, x/2=π/4
Thus the result is
Next, Define S(θ)
Then
First, using trig formula cos(2x)=cos²(x)-sin²(x)
The original problem equals to
Awlsome!!
After going through so many evasive, imaginary pages of psychology, I come across some truth here!
Truth is truth- no matter how hard you try to ignore it, it will relentless show you are wrong, it is right.
are u and v both 3 dimensional?
BTW 1.5D1 can be generated by sliding the middle bar, using C
C 1 1.5
D D[sub]1[/sub] 1.5D[sub]1[/sub]
Hey, guys! Sorry for the delay of the last post on powers:
First, let me explain the basics:
Shifting Bases:
Suppose you have got a table of almost all of powers of 3,
..., 3^-0.1, 3^0, 3^0.1, 3^0.2,...
Now, however, you are to calculate a power of 9 instead of 3.
9^1.5
You cannot multiply
However, you know from the table of powers of 3 that 9=3^2
Then 9= (3[sup]2[/sup])[sup]1.5[/sup]=3[sup](2*1.5)[/sup]
Find 2*1.5 and by the table you can immediately get the answer, the power of 9
So with multiplication, a table of powers of a particular number can serve to find powers of any positive number.
That's how a slide rule works to help you guage a power.
LL scale, the power tables of e, e^0.1, e^0.01, e^-1, e^-0.1, e^-0.01 calculate more than powers of the only 5 e related bases.
Now suppose you are calculate 9^1.5 by the scale of e^x and multiplication on exponetial.
Use the simulated slide rule (SSR) above
Flip the SSR, find LL3+ at the bottom, e to 20M (meaning 20,000,000)
LL3+=e^D
Now you should find D[sub]1[/sub]
which satisfy
e[sup]D1[/sup]=9
D D[sub]1[/sub]
LL3+ 9
and then find D[sub]2[/sub]=1.5D[sub]1[/sub] on D scale, by multiplication of slide rule.
then you can get e^D[sub]2[/sub]=9^1.5
D D[sub]2[/sub]
LL3+ e^D[sub]2[/sub]
Next topic is shifting LL scales, solving too large and too small problems.
Who is gonna be the successor of Netscape or ICQ?
Who is gonna be the next victim of all-mighty Microsoft?
Wolfram, producer of Mathematica?
Or Maple?
PS: I use as little Microsoft as possible: Maxthon Explorer, Foxmail Mailreceiver, Skype communicator.
However this formula almost only makes it harder to solve.
I prefer the original one.:/
Beating maple, mathematica all over???
Microsoft Empire???
Recently I came across an assertion that brush your teeth with lemon can make them whiter...
The total distance of red line segments together is S.