## Maths & statistics exercises / French and English lessons

Started by scarface, June 16, 2013, 11:58 PM

#### humbert

#10
How far up in math did you go when you were in school? I also like math, but sadly all I could do is Calculus I.

#### scarface

#11
to humber...baccalaureate +5.
This exercise was in fact not very difficult but the notation can seem weird for a beginner.
Now Im proposing a new exercise of probability that is probably a little easier:
The antitheft device of a car radio is made up of 4 numbers, each number taking a value between 0 and 9 (10 possible values)
1)a)What is the number of all the possible codes ?
b)What is the number of codes made up of 4 distinct numbers (distinct 2 by 2).
2)After a power cut, the owner of the car radio has to reintroduce its code to use it.
He knows the four number of his code are 1,9,9,5, but he has forgotten the order of these numbers.
a)How many codes can he enter with these 4 numbers?
b)If the first code entered is not good, the owner has to wait 2 minutes before entering a new one. The waiting time is doubling between two successive attempts.
How many codes can the owner test to the maximum in 24 hours?

#### humbert

#12
@scarface - I envy your level of education in math. Sadly for me that wasn't possible - and not due to lack of interest or not understanding it.

The number of possible codes is 104, or 10,000. Give me an example of what you mean by distinct codes, I'm lost on that one.

Oh... I forgot. Did you post that you were now working? Do you finally have a job?

#### scarface

#13
Its a good answer...perhaps Ill give a correction tonight after a beer.
As for the level of these exercises, probably you can understand the correction if you understand the notation. If they were really difficult, you couldn't.
As for the job, yes, Im now working. I worked today but I took 3 weeks leave, Im tired.

#### scarface

#14
correction of the exercise:
1)
a) the number of possible codes is a 4-lists of 10 numbers. so they are 10^4=10000
b)The number of different codes (0987 has four different numbers, 1223 has not) is composed
by 10numbers, then 9, then 8 then 7, every chosen number cant be taken again. So this number is 10*9*8*7=5040.
2)
a) If the number were all different, the number of different codes possible would be a permutation of 4 elements:
it would be 4!=4x3x2X1=24. (the first number can be one of the four numbers, then the second the 3 following numbers and the third the 2 following numbers...) But we have 1,9,9,5 and the number 9 is repeated. there is 2 possibility to permute 2 different numbers in a code so the numers of possibility is to divide by 2,
so the numbers of different possible codes with the numbers 1,9,9,5 is (4*3*2)/2=12
b)24hours is 24*60 minutes = 1440 minutes
if we do a chart to represent the numbers of attempt, the time to add between 2 attempts and the total waiting time, we have this:
attempt-waiting time-total time
1           0           0
2           2           2
3           4           6
4           8           14
5           16         30
6           32         62
7           64         126
8           128       254
9           256       510
10         512       1022
11         1024     2046

after 10 attempts 1022 minutes have passed, after 11, 2046. So the answer is 10.

#### scarface

#15
Perhaps I'm going to come back to the first question of the exercise with vectors to give more explanations to humbert.
He seemed abashed, and I would not want him to stay traumatized.
I guess he didn't understand the terms "system of parametric equations"
In a normal plane, (Oij), we usually have reduced equation like y=ax+b. probably shadow97 knows what Im talking about, its something we can see at school.
for this kind of equation, we could use a Cartesian equation like y-ax-b=0 but its rare.
In the space,its different, we use a plane (Oijk) and a line is either defined with a system of parametric equations or a system of Cartesian equations because a line is considered as an intersection of 2 planes. if you look at the vectors you see 3 numbers, we are in the space.
so let's take the first question:
we have the point A(8,0,8) and the vector AB(2;3;2), a direction vector of delta.
Then,
x = 8+2t
y = 3t    t=IR
z = 8+2t
This is the system of parametric equations. This result is obvious and can be found rapidly.
now we are going to look for a system of reduced equations using this system, probably more understandable for you.
So we have this system. We are going to isolate t:
t=(x-8)/2
t=y/3
t=(z-8)/2
therefore:
(x-8)/2=y/3
y/3=(z-8)/2
then
3x-24=2y
2y=3z-24
so we have the system of cartesian equations:
3x-2y-24=0
2y-3z+24=0
if you dislike this notation we can write:
y=3/2x-12
y=3/2z-12
both equations are necessary unlike in a plane Oij where there is no z.

#### scarface

#16
The last time we have seen that humbert was happy to do some mathematics. This time we are going to do some statistics.
I chose a somewhat difficult exercise, but for those who know the distribution of Poisson, it's certainly not that complex.
So we have 2 variables, X and Y based on the Poisson distribution with the parameters l and u: X->P(l), Y->P(u).
We are going to demonstrate that the Sum S=X+Y is based on the parameters l+u: (X+Y)->P(l+u)

Here is what I did: #17
Impressive, I've got an A in maths but I have close to no idea what's going on. I think I'm going to further study Maths next year. It's a good thing to know.

#### scarface

#18
The last exercise in statistics is at an academic level. Probably You wont study the distribution of Poisson anytime soon. The exercises in maths are easier.

#### katana

#19
This is the part of Math that I'm glad I already finished my undergrad studies. Hopefully no complicated Maths in Masteral subjects! My big bro will probably be happy about this, he's really into math... I, on the other hand, choose to vow out of this situation. I'll leave you guys to it... I'm a WOMAN. Like a man but with a WO. Important distinction that. Can be misunderstood if not made properly.