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## #1 Help Me ! » stats urgent! » 2006-03-28 19:54:05

Flowers4Carlos
Replies: 1

1) a car insurance company has determined that 8% of all ddrivers were involved in a car accident last year. among the 15 drivers living on one particular street, 3 were involved in a car accident last year. if 15 drivers are randomly selected, what is the porobabiltity of getting 3 or more who were involved in a car accident?

now my question is when it sais "if 15 drivers are randomly selected" does this mean the 15 drivers living on the particular street?? i assumed it is so i worked out the problem as follows:

P(x>=3) = P(3) + P(4) + ... + P(15) but there were only three drivers that were involved in an accident so P(4) = P(5) = ... = P(15) = 0
P(x>=3) = .08565 + 0 + 0 + ... + 0 = .08565

is my assumption and work correct???

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2) find the probabilty that 2 randomly selected people all have the same birthday. ignore leap years.

P(2 randomly selected people all have the same birthday) = 1/365
or should it be (1/365)^2???

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3) how many ways can an IRS auditor select 4 of 10 tax returns for an audit?

i'm not sure how they audit tax returns so i assumed they audit the first one, then the second one, then the third, and fourth, meaning order matters, so the total combinations is 10P4 = 5040
however, if they audit all the tax returns at the same time, meaning that order does not matter, then it should be 10C4. so which one is it???

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identify which of these types of sampling is used:
4) a market researcher selects 500 people from each of 10 cities.

a)systematic
b)stratified
c)convenience
d)cluster
e)random

there were many math tutors in my school that said it should be cluster. now the definition of cluster is to divide population into sections, then randomly select some of the sections and then choose ALL members from those sections. but the question doesn't say if those 500 people are ALL of the members in those 10 cities. can i make the assumption that it is???

## #2 Re: Help Me ! » some problems » 2006-02-26 15:40:49

oops.  i read question 4 all wrong.  quarters should be 16.

my mistake.

## #3 Re: Help Me ! » some problems » 2006-02-26 15:00:28

hi yaz!

4)  30coins = quarters + dimes
we are given that: quarters > \$5.20
one quarter is .25, so 20 quarters is: 20(.25) = \$5.  however, 20 quarters < \$5.20, so we must have an additional quarter to make that statement true giving us a total of 21 quarters.  knowing this, we can now solve for dimes:
30 = 21 + dimes
30 - 21 = dimes
9 = dimes

8) let n be the small integer
let (n+1) be the consequitive larger integer
we are given:
5n < 4(n+1)
solving for n:
5n - 4n < 4
n < 4

11)
A = \$4(maintenance fee) + .10(check cashed)
B = \$6(maintenance fee) + .06(check cashed)
so we wanna find out when is A > B:
\$4 + .10x > \$6 + .06x     solve for x
4 - 6 > .06x - .10x
-2 > -(1/25)x
-2(-25) < x
50 < x

## #4 Re: Help Me ! » helpz!! » 2006-02-26 06:36:48

john what happened to your avatar??  i'm gonna break my neck trying to view it upright!

and yes, y' is dy/dx.  got any ideas??

## #5 Help Me ! » helpz!! » 2006-02-24 16:34:23

Flowers4Carlos
Replies: 2

hi yaz!

i'm having some difficulty understanding the "existence and uniqueness of solution" theorem.  if someone could be so nice as to work this problem out and explain it to me.

1)a. show that xy' + 2y = 3x has only one solution defined at x=0.  then show that the initial value problem for this equation with initial condition y(0) = y0 has a unique solution when y0 = 0 and no solution when y0≠0.

b. show that xy'-2y=3x has an infinite number of solutions defined at x=0.  then show that the initial value problem for this equation with initial condition y(0)=0 has an infinite number of solutions.

also this other problem has me stumped!!!

dy/dx + P(x)y = x,          y(0) = 1

where P(x) = {1, 0≤x≤2
{3, x>2

1)general solution for 0≤x≤2 is  y=x-1+ce^(-x)  and c=2
2)general solution for x>2 is  y=(1/3)x-(1/9)+ce^(-3x)

question is: choose the constant in the general solution from part 2 so that the solution from part 1 and the solution from part 2 agree at x=2.

## #6 Re: Introductions » Happy new year! » 2006-01-01 11:20:25

happy new year everyone!!

i "celebrated" new years a couple of hours ago.  it wasn't the best way to start the year but atleast the food was real good.  who else is going to the gym w/ me???

## #7 Re: Introductions » VR Hawks - New Member of Math Is Fun Forums » 2006-01-01 11:10:52

griever is easy to defeat only if you are prepared.  stupid me never bothered to learn the junction system so none of my characters were properly equipped.  even worse, i fought all my battles using the GFs exclusively.  As soon as battle began i would whip them out and hope for the best.  not once did i attack, used an item, nor cast magic  the GFs did all the fighting.  i never upgraded my weapons nor stocked up on potions as a result.

my avatar is also a game character.  you know what game he's from??  zach seems to know a lot about games he's never played so may be he might just know.

## #8 Re: Help Me ! » D(x^x) » 2006-01-01 10:46:41

johny!!!!!!!!!!!!!!
hehe.. sorry bout that.

think of y as a function in terms of x, that is, y=f(x).  can you say [d/dx]f(x) = 1???  no, because f(x) might be x² or x³ or it might not be differentiable.  in our problem, we can rewrite it like this:

f(x) = x^x
lnf(x) = xlnx
[d/dx]lnf(x) = [d/dx](xlnx)
f'(x)/f(x) = lnx + x/x

you don't have to replace y=f(x), it just makes the problem look much easier, which is what i should have done in the first place.  this is also called "implicit differentiation".  go here for more examples and a thorough explanation

http://archives.math.utk.edu/visual.calculus/3/implicit.7/

krassi:  oh geee... i tried using integration by parts on that creatuer but i ended up w/ an uglier looking monster.  i'm baffled!!

## #9 Re: Help Me ! » D(x^x) » 2005-12-30 17:42:52

hey john!

there is a similar problem like this here:
http://www.mathsisfun.com/forum/viewtopic.php?id=2310

i'll work it out for ya!

y = x^x                   apply log to both sides

lny= lnx^x

lny = xlnx                now differentiate both sides

y'/y = lnx + x/x

y' = y(lnx + 1)         remember that y=x^x

y' = (x^x)(lnx + 1)

y' = (x^x)lnx + x^x

y' = (x^x)lnx + x*x^(x-1)

that last part is typical calculator manipulation.  they always spit out some funky looking solutions like x*x^(x-1) when it could just be x^x.

## #10 Re: Introductions » VR Hawks - New Member of Math Is Fun Forums » 2005-12-30 17:16:32

the only final fantasy game i've played is ff8.  it was a real fun game but i will mostly remember it as the first game i could not complete.  griever was my stopping point.  his shockwave pulsar made all my characters drop dead like bowling pins.

oh and welcome VR!  you can call me F4C

## #11 Re: Help Me ! » Help » 2005-12-30 08:04:36

siva:  that is a very interesting approach and one that i'm not too sure if it's true, however, it looks like you found the x-intercept and not the summation.

## #12 Re: Help Me ! » Factor Theorem? » 2005-12-30 07:46:19

hello katy!

aren't you given a lovely divisor like (x-c) to try out??  otherwise we need to figure out where this function equals to zero and that requires using the rational zeros theorem.

## #13 Re: Help Me ! » derivative of an exponential function » 2005-12-26 19:43:12

it's not that difficult really.  just apply log to both sides and differentiate.

y = (cos x )^0.7x

lny = ln(cos x )^0.7x

lny = .7xln(cox)

give it a try!  oh and happy holidays everyone!

## #14 Help Me ! » unit normal vector » 2005-12-19 22:11:46

Flowers4Carlos
Replies: 2

i have a quick question regarding the unit normal vector on the following integral:

∫∫FdS = ∫∫FndS
s             s

if the surface S is given by z=g(x,y) then the unit normal vector is f(x,y,z)= z - g(x,y)  ⇒

n= -----------------

my question is if i'm given the surface 3z=g(x,y), then what should the unit normal vector be???

is it f(x,y,z) = 3z - g(x,y)     or     f(x,y,z) = z - g(x,y)/3

## #15 Re: Help Me ! » logarithims, remind me again? » 2005-12-13 21:18:24

hooray!!!  i solved it!!!

(e^x - e^-x)/2 = 7

e^x - e^-x = 14                             multiply everything by e^x

e^x(e^x - e^-x) = 14e^x

e^2x - 1 = 14e^x

(e^x)² - 14e^x - 1 = 0                   let w=e^x

w² - 14w - 1 = 0                            using the quadratic formula, i obtain

14±(200)^1/2)
w= ------------------                        which simplifies to
2

w= 7±(50)^(1/2)                           plug w back in w=e^x

7±(50)^(1/2) = e^x

ln(7±(50)^(1/2)) = x                     but lnx > 0

ln(7+(50)^(1/2)) = x

## #16 Re: Help Me ! » logarithims, remind me again? » 2005-12-13 20:28:22

hi yaz mikau!!!

i'm sorry but i'm having trouble solving (e^x - e^-x)/2 = 7

also i can't seem to find a value for x to make that statement true!!

however, you did make one mistake:

e^x = 14 + e^-x

ln e^x = ln 14 + ln e^-x

you cannot distribute the log because that would imply: ln 14 + ln e^-x ⇒ 14e^-x which is not true.

the correct way of applying logs would be:

e^x = 14 + e^-x

ln e^x = ln (14 + e^-x)

## #17 Re: Help Me ! » Double/Half Angle Difficulty. » 2005-12-12 20:17:10

what is "Maths at A2 Level, (C3)"???

## #18 Re: Help Me ! » Integration » 2005-12-12 20:12:10

i tried working out ∫e^(-x^2)dx but i wasn't able to go n e where w/ it.  i guess we need a super computer for this one.  *shrugs*

your half way through calculus??  wow... ur pretty fast.  are you learning it on your own or in an accelerated course???

## #19 Re: Help Me ! » Integration » 2005-12-12 19:37:59

hi yaz chemist!!

(1+x)^(1/2)
∫---------------dx          and you are trying to solve this integral using x=cosθ ⇒ dx=-sinθdθ
(1-x)^(1/2)

(-sinθ)(1+cosθ)^(1/2)
∫-------------------------dθ     multiply everything by (1+cosθ)^(1/2)
(1-cosθ)^(1/2)

(-sinθ)(1+cosθ)^(1/2)*(1+cosθ)^(1/2)
∫--------------------------------------------dθ
(1-cosθ)^(1/2)*(1+cosθ)^(1/2)

(-sinθ)(1+cosθ)
∫--------------------------dθ
(1-cos²θ)^(1/2)

(-sinθ)(1+cosθ)
∫------------------dθ
sinθ

i guess you can take it from here!!  i may have done something wrong so please √√

## #20 Re: Help Me ! » Curvy line length calculus? » 2005-12-11 10:07:51

you are absolutely correct john!  the integral for finding the distance on a curve is how you explained it!  although it's a common formula, i find it smart of you (and sooo cute!!!!) that you were able to figure it out on your own!!  so smart!!!!

as for ∫(1+4x²)^(1/2)dx you either gotta use trigonometric substitution or a table chart (which is also fine).  if you want i can solve it "manually" for you... it's not gonna be pretty though.

## #21 Re: Introductions » Здрасти by krassi_holmz » 2005-12-09 18:02:14

Rod: hello in russian is "привет"

и не, я не говорю по-русски, но я могу притвориться!

http://www.freetranslation.com/

i was working on problem one in degree mode and my calculator kept spitting out positiive results even for very large values of x.  then i switched over to radian mode and it worked!  sometimes a lil trial and error will get you on the right track!

2)

f(x) = x² + sin(x/2)                f(0) = 0
f'(x) = 2x + (1/2)cos(x/2)       f'(0) = 1/2
f''(x) = 2 - (1/4)sin(x/2)          f''(0) = 0
f'''(x) = -(1/8)cos(x/2)            f'''(0) = -1/8
f''''(x) = (1/16)sin(x/2)            f''''(0) = 0
f'''''(x) = (1/32)cos(x/2)          f'''''(0) = 1/32

0 + [(1/2)x]/1! + 0x²/2! - [(1/8)x³]/3! + [0x^(4)]/4! + [(1/32)x^5]/5!

## #23 Re: Help Me ! » 4; 6; 8 and 9 » 2005-12-02 16:26:04

uh oh!  you are not allowed to post your email.  you gotta send in your birth certificate to rod proving you are of a certain age first

and as for your query.... i guess it would be (4!)^(6!)^(8!)^(9!)

my calculator can't even handel 2^9!

## #24 Re: Help Me ! » Hard task! Need help! » 2005-12-02 10:33:06

it reads "I can't love her because just by looking at her i forget what i'm about to say...."

## #25 Re: Help Me ! » What is a partial derivative? » 2005-12-02 10:19:15

one use is in optimization problems.  if you want to find the minimum cost of constructing, oh i don't don't, say a can, you would probably use partial derivaties.

i'm sure someone in the field can give you a better use for it.