mr.fizwidget
Smash Champion
what is the probability that someone will have the matching birthdays in a room with 30 people?
-
Welcome to Smashboards, the world's largest Super Smash Brothers community! Over 250,000 Smash Bros. fans from around the world have come to discuss these great games in over 19 million posts!
You are currently viewing our boards as a visitor. Click here to sign up right now and start on your path in the Smash community!
Need Math Help?
- Thread starter AltF4
- Start date
Corpsecreate
Smash Lord
"A doctor interviewing a female patient asks her the ages of her three children. She tells him
that the product of the three children’s ages is 36. The doctor says he needs to know more so
the patient says that the sum of the ages is equal to the number of the house next door. The
doctor runs outside and returns. He is exasperated and says that he needs more information.
The conversation continues, with the woman saying that she thought he was a smart doctor
but she is beginning to change her mind. She also tells him that she is in a hurry because she
must take her oldest child to her violin lesson. The doctor jumps up and claims he knows the
ages of the children. What are they?"
Don't understand...seriously. abc = 36, are the only numbers your given.
EDIT: I solved it, I'll leave it up here for anyone who wants to try it.
that the product of the three children’s ages is 36. The doctor says he needs to know more so
the patient says that the sum of the ages is equal to the number of the house next door. The
doctor runs outside and returns. He is exasperated and says that he needs more information.
The conversation continues, with the woman saying that she thought he was a smart doctor
but she is beginning to change her mind. She also tells him that she is in a hurry because she
must take her oldest child to her violin lesson. The doctor jumps up and claims he knows the
ages of the children. What are they?"
Don't understand...seriously. abc = 36, are the only numbers your given.
EDIT: I solved it, I'll leave it up here for anyone who wants to try it.
That problem has two answers. It's also very annoying.
Corpsecreate
Smash Lord
I managed to get one unique solution 
_KuyaSombreo_
Smash Ace
My teacher said that Calculus is relatively easy, but it is the algebra involved that makes it difficult.
Is that true?
Is that true?
POKE40
Smash Lord
It is a lot of Algebraic steps.
That is what most of calculus is so far.
Calculus should not be that hard. It is just finding the answer takes quite a while.
Good luck!
cutter
Smash Champion
I managed to do quite well in calculus when I took it last year because I had a very solid understanding of all the previous fundamentals of math before I took it. If you are quite comfortable at doing virtually any algebraic problem and understanding fundamentals like trig identities, calculus won't be as hard as you think.
I pretty much agree with what POKE40 said. Some problems like integrating even powered trig equations are a huge pain the *** because of all the steps required to integrate it.
I pretty much agree with what POKE40 said. Some problems like integrating even powered trig equations are a huge pain the *** because of all the steps required to integrate it.
Corpsecreate
Smash Lord
Yep. Thats what I got.
Gosu_Engineer
Smash Ace
I don't know if anyone answered you or even if you still need the answer, but for a group of 30 people it's about 70%
Assume the total combination of people having overlapping birthdays is the sample space meaning all possibilities of no one having a birthday from everyone having a birthday is 100% or 1
so to get the probability that at least 2 people will have the same birthday is you subtract the possibility that no one has a birthday from the total (1)
To get that value you first calculate the probability that no one has the same birthday:
first person having the same birthday as themselves: 100% or 1 (here I'll denote it as 365/365; it'll be useful later)
second person not having a birthday with the first: 364/365
third person not having the same birthday as the second and first: 363/365
fourth person not having the same birthday as the others: 362/365
.
.
.
thirtieth person not having the same day as the others 335/365
now the probability that none of these people have the same birthday is all these values multiplies together:
(365*364*362*361*360***335)/(365^30)
that yields ~.3
so the probablility that at least 2 people do have the same birthday is the whole set of posibilities minus the possibility that no one has the same birthday:
1-.3 = .7
-or- ~70%
Velox
Smash Ace
prove Stoke's theorem!!!
on a serious note though, if any of you use MATLAB or Mathematica, I just bought Maple for 100$ and it's totally superior to those other programs if anyone is wondering (not a lot of people know about Maple I think..)
also, math may be the purest degree, but it's overly pure, everything in moderation :D
on a serious note though, if any of you use MATLAB or Mathematica, I just bought Maple for 100$ and it's totally superior to those other programs if anyone is wondering (not a lot of people know about Maple I think..)
also, math may be the purest degree, but it's overly pure, everything in moderation :D
- Joined
- Aug 31, 2005
- Messages
- 8,189
Alright guys, I could use some help.
I am having trouble with this problem.
F(x)= -6x^6*x^(1/2)+7/(x^2(x^1/2)
find F'(x)
I start out my problem by separating the two equations(just to keep it clean.)
So h(x)=-6x^6*x^(1/2)
and g(x)= 7/(x^2(x^1/2)
So I start working on h(x)
' of -6x^6 is -36x^5 and ' of x^(1/2) is 1/(2x^(1/2))
So using a'b + b'a I get h'(x)= -36x^5*x^(1/2)-3x^6/(x^(1/2)
then when I get to g'(x). I get confused. Am I supposed to use quotient rule, even though the numerator is a constant? And we aren't supposed to use chain rule yet(thats the next chapter.), so are we supposed to do product rule then? I am just a bit lost. :/
I am having trouble with this problem.
F(x)= -6x^6*x^(1/2)+7/(x^2(x^1/2)
find F'(x)
I start out my problem by separating the two equations(just to keep it clean.)
So h(x)=-6x^6*x^(1/2)
and g(x)= 7/(x^2(x^1/2)
So I start working on h(x)
' of -6x^6 is -36x^5 and ' of x^(1/2) is 1/(2x^(1/2))
So using a'b + b'a I get h'(x)= -36x^5*x^(1/2)-3x^6/(x^(1/2)
then when I get to g'(x). I get confused. Am I supposed to use quotient rule, even though the numerator is a constant? And we aren't supposed to use chain rule yet(thats the next chapter.), so are we supposed to do product rule then? I am just a bit lost. :/
ipitydatfu
Smash Master
why dont you combine powers it simplfies everything (that is, if im reading your problem correctly)
-6x^(13/2)+ 7x^(-5/2)
then
F'(x) = (13/2)(-6)x^(11/2) +(-5/2)(7)x^(-7/2)
note: when you have something like constant/variable^power, you can always write it as constant(variable^-power), and proceed with powerrule from there
-6x^(13/2)+ 7x^(-5/2)
then
F'(x) = (13/2)(-6)x^(11/2) +(-5/2)(7)x^(-7/2)
note: when you have something like constant/variable^power, you can always write it as constant(variable^-power), and proceed with powerrule from there
ipitydatfu
Smash Master
no prob.
Corpsecreate
Smash Lord
You learnt the quotient rule before you learnt the chain rule? I learnt it in the other way around hehe.
Healer
Smash Apprentice
- Joined
- Apr 6, 2009
- Messages
- 115
Does anyone know any tricks to do proofs in Geometry (the two column kind) real fast. I'm in a Geo Honors class and the teacher must be obsessed with proofs or something, because my whole week of hw comprises of proofs. ITS SO ANNOYING AND BORING!
cF=)
Smash Lord
- Joined
- Aug 22, 2005
- Messages
- 1,909
An infinitely long, uniform rod of mass µ per unit length is situated on the z axis.
a) Calculate the grav. force on a point mass m at a distance p (rho) from the z axis.
I did the EXACT same thing in my electromagnetism class 2 years ago using a charged rod and a charged point particle, yet I'm helpless now as to how to start this problem. Anybody can give me a pointer or a tip on how to write this in cylindrical coordinates?
EDIT: If you have John R. Taylor's classical mechanics book, it's on page 153, #4.24***
a) Calculate the grav. force on a point mass m at a distance p (rho) from the z axis.
I did the EXACT same thing in my electromagnetism class 2 years ago using a charged rod and a charged point particle, yet I'm helpless now as to how to start this problem. Anybody can give me a pointer or a tip on how to write this in cylindrical coordinates?
EDIT: If you have John R. Taylor's classical mechanics book, it's on page 153, #4.24***
ipitydatfu
Smash Master
F = MmG/r^2 (G = gravitation constant)
have you attempted to try something using gaussian surfaces?
edit: just spewing ideas. dont mind me.
have you attempted to try something using gaussian surfaces?
edit: just spewing ideas. dont mind me.
cF=)
Smash Lord
- Joined
- Aug 22, 2005
- Messages
- 1,909
I know what's the expression for gravitational attraction between 2 point particles. The mental block I'm having is about integrating a tiny piece of the rod's mass from -infinity to infinity to find the rod's total mass. If I consider my rod to be infinite, then every little force from a non perpendicular angle acting on my point particle should cancel out with the other side of the rod, and here I am again wondering how I can find the rod's mass!
GunmasterLombardi
Smash Champion
I'm taking Honors Algebra 1.
What all needs to be learned here before taking calculus?
What all needs to be learned here before taking calculus?
POKE40
Smash Lord
Needs to be learned in Algebra 1:I'm taking Honors Algebra 1.
What all needs to be learned here before taking calculus?
-Graphing
-Finding an equation for a line
-Rise over Run
-Factoring
-Different Properties of Foiling
-Collecting Like terms
-"T-Tables"
-Uhh.... there is more I forgot.
Then you move on to Geometry, Algebra 2, Pre-Calculus, and Trigonometry.
Good luck!
ipitydatfu
Smash Master
would it work if you tried something like this? dm = μdz
that would imply, r = sqrt( z^2 + x^2), or something in terms of cosine,
well dang, good luck with that physics problem. and if you find out the answer can you tell me too?
Dark Ryu
Smash Ace
Hi I need help! This problem is really confusing me... it's:
wool
Smash Ace
2+(1/2) = 5/2
1/(5/2) = 2/5
3+(2/5) = 17/5
1/(17/5)=
5/17
is that the answer idonno it makes sense to me
basically all i am doing is converting the denominator into an improper frac then doing 1/that since thats what the problem is to get a frac then adding the number (2, or 3, depending on what it says) then doing the 1/ again.
idonno if its right
1/(5/2) = 2/5
3+(2/5) = 17/5
1/(17/5)=
5/17
is that the answer idonno it makes sense to me
basically all i am doing is converting the denominator into an improper frac then doing 1/that since thats what the problem is to get a frac then adding the number (2, or 3, depending on what it says) then doing the 1/ again.
idonno if its right
Velox
Smash Ace
I got 2.5/8.5 or 5/17 by getting rid of the complex fractions one at a time, Maple also gives this answer which is pretty convincing lol. Just use wool's method, it's cleaner.
Velox
Smash Ace
I got 2.5/8.5 or 5/17 by getting rid of the complex fractions one at a time, Maple also gives this answer which is pretty convincing lol. Just use wool's method, it's cleaner.
hey, cF, about that physics problem, I'm about to be late for my modern optics class here at school, which I choose to take this semester instead of the mechanics course that I'm sure you've found that problem in, so although I'm probably no sure on how to do it, it seems to me that finding the total mass wouldn't be good because if you applied the inverse square law it would be wrong because the point masses along the rod farther from Ro are going to contribute less gravitational force than just assuming the entire rod is a point mass like you do with planets. I think it goes something like the integral from negative infinity to Ro of the perpendicular component of the gravitational force (you may have to draw this out, look at it geometrically and maybe use a trig substitution on the integral) + the integral of Ro to positive infinity of the perpendicular component of the gravitational force.. That seems like how they're going to do that problem, but like I said, I haven't taken Mechanics yet, all I do is listen to my lab partners b**** about how hard it's supposed to be..
oh, well one more thing, in the integrals, I think it's gunna be Mu dz for the mass you're integrating with (and you're integrating the inverse square law) and then somehow u need to find a way to slip the distance in the integral from the infinitely small length of mass along the z-axis to the object located a distance Ro away from the rod of mass.
Actually no, there's a lot more going into the problem than that, because then you would have just integrated the entire portion of the gravitational force produced by each point mass along 'z'. Well.. I hope I got you started at least lol..
hey, cF, about that physics problem, I'm about to be late for my modern optics class here at school, which I choose to take this semester instead of the mechanics course that I'm sure you've found that problem in, so although I'm probably no sure on how to do it, it seems to me that finding the total mass wouldn't be good because if you applied the inverse square law it would be wrong because the point masses along the rod farther from Ro are going to contribute less gravitational force than just assuming the entire rod is a point mass like you do with planets. I think it goes something like the integral from negative infinity to Ro of the perpendicular component of the gravitational force (you may have to draw this out, look at it geometrically and maybe use a trig substitution on the integral) + the integral of Ro to positive infinity of the perpendicular component of the gravitational force.. That seems like how they're going to do that problem, but like I said, I haven't taken Mechanics yet, all I do is listen to my lab partners b**** about how hard it's supposed to be..
oh, well one more thing, in the integrals, I think it's gunna be Mu dz for the mass you're integrating with (and you're integrating the inverse square law) and then somehow u need to find a way to slip the distance in the integral from the infinitely small length of mass along the z-axis to the object located a distance Ro away from the rod of mass.
Actually no, there's a lot more going into the problem than that, because then you would have just integrated the entire portion of the gravitational force produced by each point mass along 'z'. Well.. I hope I got you started at least lol..
GunmasterLombardi
Smash Champion
I'm also learning functions, rate of change, and lots of ******** fractions.
Thanks, just wanna be prepared if I get honors next year.
Dark Ryu
Smash Ace
Thanks guys really appreciate it :]
The answer is 5/17.
The answer is 5/17.
Zero Beat
Cognitive Scientist
Question..
Determine whether the graph of the equation is symmetric with respect to the x axis, y axis, and or the origin.
y= 4x^2-3
So I tried testing it for all:
x axis: -y= 4x^2-3 is not equivalent to the original equation
y axis: y= -4x^2-3 is not equivalent to the original equation
origin: -y= -4x^2-3
multiply by (-1) on both sides, gives me y=4x^2+ 3, is not equivalent to the original equation
And the answers are:
A) y axis
B) origin
C) x axis
D) x axis, y axis, origin
E) none
Now, I went with choice E(none) based on my work above, yet the answer key says it's A: y axis. Could it be that there's a mistake with my work or is there an error within the answer key?
Disregard this post, I made the co-efficient negative instead of the x. -x^2= x^2, thus y=4x^2-3 = y=4x^2-3
Determine whether the graph of the equation is symmetric with respect to the x axis, y axis, and or the origin.
y= 4x^2-3
So I tried testing it for all:
x axis: -y= 4x^2-3 is not equivalent to the original equation
y axis: y= -4x^2-3 is not equivalent to the original equation
origin: -y= -4x^2-3
multiply by (-1) on both sides, gives me y=4x^2+ 3, is not equivalent to the original equation
And the answers are:
A) y axis
B) origin
C) x axis
D) x axis, y axis, origin
E) none
Now, I went with choice E(none) based on my work above, yet the answer key says it's A: y axis. Could it be that there's a mistake with my work or is there an error within the answer key?
Disregard this post, I made the co-efficient negative instead of the x. -x^2= x^2, thus y=4x^2-3 = y=4x^2-3
ipitydatfu
Smash Master
hey cf=)
i thought of something
lets say this: we can turn it into terms of cosine
when its so far away, you can assume theta is 180, and when at its closest point is 90.
so technically, df = MmuG/r dr
r = zcostheta
then
dr= zsintheta d(theta)
integrate from 180 to 90
multiply by 2
* i think this way will give you an answer. im just not sure if i set up the integral for the trig correctly tho *
i thought of something
lets say this: we can turn it into terms of cosine
when its so far away, you can assume theta is 180, and when at its closest point is 90.
so technically, df = MmuG/r dr
r = zcostheta
then
dr= zsintheta d(theta)
integrate from 180 to 90
multiply by 2
* i think this way will give you an answer. im just not sure if i set up the integral for the trig correctly tho *
Fuelbi
Banned via Warnings
Oh god just looking at this makes me not want to keep taking math
*sees all the horrible and complicated equations in his future*
I dont want to take math anymore
*cries in a corner*
*sees all the horrible and complicated equations in his future*
I dont want to take math anymore
*cries in a corner*
cF=)
Smash Lord
- Joined
- Aug 22, 2005
- Messages
- 1,909
I didn't read your post yesterday, but I was able to solve this pretty neatly using a couple of identities. Specifically, I was right saying both sides of my infinite rod would cancel out, but I wasn't aware of the intrinsic relationship between rho, z and r. Here's how I solved it:
The following drawing sets up my variables:
And now, here's the magic:
Pretty sure it's good, you could still take a look and tell me if I messed up somewhere :D
ipitydatfu
Smash Master
oh dang, so thats how you do it.
i screwed up with the r^2 and put cosine in the wrong places lol. i forget its f dot r. no wonder i didnt do so well in electromagnetism
i screwed up with the r^2 and put cosine in the wrong places lol. i forget its f dot r. no wonder i didnt do so well in electromagnetism
wool
Smash Ace
Yo I am not looking forward to physics next year..
ipitydatfu
Smash Master
lol. you;re not gonna do anyting that crazy.... yet
hopefully you'll have a decent teacher who'll build a solid foundation from the ground up
hopefully you'll have a decent teacher who'll build a solid foundation from the ground up
A batter strikes a baseball moving horizontally towards him at 15 m/s. The ball leaves the bat horizontally at 24m/s, 40.0 degrees to the left of a line from the plate to the pitcher. The ball is in contact with the bat for 0.01 s.
a) What is the change in velocity of the ball?
I've set this up with vector components but I keep getting really low values. The answer should be 37 m/s [24.8 degrees left of the pitcher] but I don't know how to solve it.
a) What is the change in velocity of the ball?
I've set this up with vector components but I keep getting really low values. The answer should be 37 m/s [24.8 degrees left of the pitcher] but I don't know how to solve it.
ipitydatfu
Smash Master
change of velocity is just from (24- (-15)) divided by .01 = 3700 m/s^2
i dont remember anything about angles tho
i dont remember anything about angles tho
Super Touhey
Smash Cadet
You have to find change in horizontal velocity (Vx) and change in vertical velocity (Vy) separately.
Vx1 = -15 m/s
Vy1 = 0 m/s
Vx2 = 24*cos(40) m/s
Vy2 = 24*sin(40) m/s
Vx = Vx2 - Vx1 = 24*cos(40) m/s + 15 m/s = 33.39 m/s
Vy = Vy2 - Vy1 = 24*sin(40) m/s = 15.43 m/s
The change in velocity would be (Vx^2+Vy^2)^.5 which is about 36.78 m/s.
To get the angle, you just use arctan(Vy/Vx) which is about 24.8 degrees.
I don't think number of seconds in contact matters, since you're looking for the change in velocity, not the rate of change in velocity. Sometimes they throw extra information at you to try to mix you up.
Vx1 = -15 m/s
Vy1 = 0 m/s
Vx2 = 24*cos(40) m/s
Vy2 = 24*sin(40) m/s
Vx = Vx2 - Vx1 = 24*cos(40) m/s + 15 m/s = 33.39 m/s
Vy = Vy2 - Vy1 = 24*sin(40) m/s = 15.43 m/s
The change in velocity would be (Vx^2+Vy^2)^.5 which is about 36.78 m/s.
To get the angle, you just use arctan(Vy/Vx) which is about 24.8 degrees.
I don't think number of seconds in contact matters, since you're looking for the change in velocity, not the rate of change in velocity. Sometimes they throw extra information at you to try to mix you up.