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Harut

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  • 2 weeks later...

Mike, unfortunatly i haven't taken Statistics yet. i'm going to take it on next term.

i hope i'm going to like it because i like everything else i learned, especially Calculas.

 

Armo, where do you take Calculas? in college or high school(AP class)? and what kind?

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here, i'll make up one Calculas question, even though this kind of problems are comman.

-----

 

a helicaptor is flying horizontally over road with amplitude of 50 meters. it's speed is 25 meters per second.

there is a car coming toward it on the road.

the radar on the helicaptor shows 50 meters per second and 150 meters of distance between the car and helic.

what is the real speed of the car???

------

here, i'll try to draw a picture to help you.

 

 

.....H ----> 25 m/sec

....|.

5..|.. 150 m

0..|...

m..|.............pic is not drawn to scale

....|.............pls, discart all the dots

....|......

______

 

the rest of the action figure out yourselves.

----

 

have a good time with this one.

 

sorry, for any inconvinance with numbers. i made them up myself

 

[ August 30, 2001: Message edited by: Harut ]

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  • 2 weeks later...

sorry Mos, i didn't see you have an answer.

 

asum es 75? hmmmm

 

Mos, ko matemi dasatun ova eghel, Mos???

 

sorry Mos, it's not the answer. it's not that simple. you have to do some calculation.

 

i see nobody else is interested, so i'll just go ahead and try to solve it myself.

 

i'll have the answer shortly.

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here we go. let me try.

-----

a^2+b^2=c^2

 

50^2+b^2=150^2

 

b^2=22500-2500

 

b~=141.42

 

(a(t))^2+(b(t))^2=(c(t))^2

 

taking derivative

2a(t)*da/dt+2b(t)*db/dt=2c(t)*dc/dt

2's cancel out

 

a(t)=50

b(t)=141.42

c(t)=150

da/dt=0

db/dt=25+?(car's speed)

dc/dt=50

 

50*0+141.42*(25+?)=150*50

 

3535.5+141.42*?=7500

 

141.42*?=3964.5

 

?~=28

 

therefore the speed of car was, approxematly

28 meter per second.

-----

 

how was it.

i hope your were confused and my colculations were correct.

 

have a good time.

 

[ September 12, 2001: Message edited by: Harut ]

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  • 2 weeks later...

Harout Jan,

Here is a simple(kind of fun) question for you?

 

A billiard ball is struck so that it strikes all four cushions and returns to its starting position. In what direction is it struck, and how far does it travel?

 

I will post the answer tomorrow.

Azat

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May be I can help. let me try, though i don't know much about the game Just by imagination

 

On 3 corners you place balls (one each). On your left hand side you place 2. You start the game by striking one ball onto the other, which has to travel straight forward and strikes the ball in the very next corner(vertical), continously travels straight line onto the parallel corner, strikes the ball standing there and vertically to you right hand side, where the lonely ball stands, strikes that too and comes to the starting point: Jlump! into the hole!

 

If you know the curcumference (?) of the table, then you will know how many meters ball has travelled I guess

 

Is my calculation make sense Azat

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not bad Dragon.

here is mine.

 

______________________

|41|

||

|5 2|

||

|63|

-------------------------

the ball is in the center(0). you hit it so it goes to 1, bounces to 2, then 3,4,5,6, and back to 0.

 

this might not be correct because it hits the longer sides twice. but if there is not limit then this works.

 

if we suppose the play area is 2x4, then the ball has traveled 8 times squre root of 2.

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Harout,

I was thinking much simpler and I did confuse you by not telling you that the ball must hit each cushion once.

 

You were on a good thought process though.

 

Here is the answer that I was thinking.

It must be stuck parallel to one of the diagonals, and its total path back to its starting point is double the length of a diagonal.

 

Azat

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Okay, but this one is too simple.

 

We each have a watch and it reads 12 o'clock on it right now. My watch is 1 minute too fast per hour, and your watch is 1 minute too slow per hour.

 

When will our watches read the same time again?

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Good.

 

Here are 2 more

 

1. Two strangers from different parts of America both build similar apartment buildings in their home towns. By chance, they both forget an important part of their project. They each, once again by chance, call the same national hardware store and order the missing items. The prices they are quoted are as follows:

 

one will cost them $2.

two will cost them $2.

twelve will cost them $4.

and a hundred and forty-four will cost $6.

 

What was the item they needed?

 

 

2. You have fifty bags, each with approximately one hundred balls inside. You have a weighing machine with a digital readout, with 3 places of decimals on the display, calibrated in grams, it is capable of weighing an infinite amount (but it won't have to). All of the balls in all of the bags weigh 1 gram, except in one bag in which all of the balls weigh 1.001 grams. You cannot tell by look or feel which bag has the heavier balls.

 

How do you determine which bag contains the heavier balls?

You can use the weighing machine only once (which means you can obtain only one reading from it before it self destructs).

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Okay one more.

 

The following multiplication example uses all the digits from 0 to 9 once and once only (not counting the intermediate steps). Finish the problem. One number has been filled in to get you started.

 

xxx

x5

------

xxxxx

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okay last one for the day.

 

There are two lengths of rope.

Each one can burn in exactly one hour.

They are not necessarily of the same length or width as each other. They also are not of uniform width (may be wider in middle than on the end), thus burning half of the rope is not necessarily 1/2 hour.

 

By burning the ropes, how do you measure exactly 45 minutes worth of time?

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Baranin megu gu varem yev jamatsuyts gu prnem: 45 vargyan yev gu marem gragu. Baranin mnatsyal masu gu chapem yergrort baranin yev mnatsyal adkan mu gu gdrem chvaradz baranen. Arachinu gu vari jamatsuytsov, yergrortu arants jamatuytsi: yergukn al 45 vargyanen

 

Khelatsi em

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quote:
Originally posted by Azat:
2. You have fifty bags, each with approximately one hundred balls inside. You have a weighing machine with a digital readout, with 3 places of decimals on the display, calibrated in grams, it is capable of weighing an infinite amount (but it won't have to). All of the balls in all of the bags weigh 1 gram, except in one bag in which all of the balls weigh 1.001 grams. You cannot tell by look or feel which bag has the heavier balls.

How do you determine which bag contains the heavier balls?
You can use the weighing machine only once (which means you can obtain only one reading from it before it self destructs).



Ok. i think i got this one. here it is.
===
first you number all of the bags from 1 to 50.
then you take 1 ball from bag #1, 2 balls from bag #2, 3 from #3, and so on.
you put all of those balls together and wight them.
then you look to the result.
if it says (i don't want to calculate so i'll use *s) ****.001 then #1 bag is the one with heavier balls
if it says ****.002 then it's #2.
****.003 then it's #3.
and so on.

and THEN you hide so you won't be hurt from the destruction. but be sure to read what it says first. even if you have to take high risk
=====

this one was easy. but the rest will take a while.
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