₦airaland Forum

agentofchange1:
hey budy,am no chief oo, its ok glad ur cleared now

Lools. Thanks once again.

Mathematicians, pls help me solve this problem.

Tongsman:
Mathematicians, pls help me solve this problem.

The answer is -7 just looking at it.

Tongsman:
Mathematicians, pls help me solve this problem.

As x tends to infinity the second term of the numerator and the denominator dominates i.e.(7^(x+1)>>5^(x+1) and 7^(x)>>5^(x)) hence the function approaches (7^(x+1)/-7^(x)) = -7 asymptotically (that is as x tends to infinity). I have also used mathematica, hence find below the attached print screen of the mathematica script.

I have also plotted it on matlab so you can see the asymptotic behaviour of the function as x becomes large... In case you want to tackle a similar question in future.

masperano:
The answer is -7 just looking at it.

masperano:
As x tends to infinity the second term of the numerator and the denominator dominates i.e.(7^(x+1)>>5^(x+1) and 7^(x)>>5^(x)) hence the function approaches (7^(x+1)/-7^(x)) = -7 asymptotically (that is as x tends to infinity). I have also used mathematica, hence find below the attached print screen of the mathematica script.

Thanks very much, really appreciate.

Two hunters P and Q aim at a target at the same time. The probability that P hit it is 2/9, while thr probability that Q missess the target is 6/7. What is the probability that: (a) only 1 of them hits it.(b) one of them misses it?
Agentofchange1, karmanaut, masperano,Umartins1, Laplacian,and other gurus. Thanks

ayokunlei:
Two hunters P and Q aim at a target at the same time. The probability that P hit it is 2/9, while thr probability that Q missess the target is 6/7. What is the probability that: (a) only 1 of them hits it.(b) one of them misses it?
Agentofchange1, karmanaut, masperano,Umartins1, Laplacian,and other gurus. Thanks

Boss, I'm not good at probability ooo but let me try. I'm sure agentofchange1 will correct me.

pr that P hits= 2/9

Pr that P misses= 1- 2/9 = 7/9


Pr that Q misses= 6/7

Pr that Q hits= 1-6/7 = 1/7

(a) pr that only one of them hits

= (pr that P hits + pr that Q misses) * (pr that P missed + pr that Q hits)







****************************


Please, let me wait a little and see if the bosses will confirm if I'm right so far.

Please kindly help me how the Integral of sin square x(dx) equals to 1/2Integral(1-cosx)dx.
Cc : mathefaro, Umartins1, agentofchange1, and other gurus.

please kindly help me show how the Integral of sin square x(dx) equals 1/2Integral of (1-cos2x)dx.

Cc: agentofchange1, laplacian, kamarnaut,Umartins1, mathefaro, and other gurus

ayokunlei:
please kindly help me show how the Integral of sin square x(dx) equals 1/2Integral of (1-cos2x)dx.

Cc: agentofchange1, laplacian, kamarnaut,Umartins1, mathefaro, and other gurus

Hi,
To avoid stories lol
From R. Formula,
∫ sin^n(dx)=-1/n.sin^(n-1)xcosx+(n-1)/n ∫ sin^(n-2)xdx+.........«»
∫ sin^2x=-1/2sinxcosx+{x/2}+k.
In 1/2 ∫ 1-cos2xdx.. This is pretty simple.
∫ 1=x;
∫ cos2x=1/2sin2x.
Thus we have ,
1/2∫ (1-cos2x)dx=1/2(x-1/2sin2x)
But 1/2sin2x=sinxcosx.
Hence I=x/2-1/2.sinxcosx+k

Got it?

Alternatively @ayokunlei, rather than using the reduction formula.
You can use your day to day trigonometry. See how:

SOLUTION
$Sin²xdx

Recall from trig. that Cos2x = 1- 2Sin²x, then,

Sin²x = 1/2(1-Cos2x);

=> $Sin²xdx = 1/2$(1-Cos2x)dx.

Which is what you requested for.

Ans.: 1/2[x-1/2(Sin2x)] + c.

Arithmetic:
Alternatively @ayokunlei, rather than using the reduction formula.
You can use your day to day trigonometry. See how:

SOLUTION
$Sin²xdx

Recall from trig. that Cos2x = 1- 2Sin²x, then,

Sin²x = 1/2(1-Cos2x);

=> $Sin²xdx = 1/2$(1-Cos2x)dx.

Which is what you requested for.

Ans.: 1/2[x-1/2(Sin2x)] + c.

thanks alot.

bolkay47:
Hi, To avoid stories lol From R. Formula, ∫ sin^n(dx)=-1/n.sin^(n-1)xcosx+(n-1)/n ∫ sin^(n-2)xdx+.........«» ∫ sin^2x=-1/2sinxcosx+{x/2}+k. In 1/2 ∫ 1-cos2xdx.. This is pretty simple. ∫ 1=x; ∫ cos2x=1/2sin2x. Thus we have , 1/2∫ (1-cos2x)dx=1/2(x-1/2sin2x) But 1/2sin2x=sinxcosx. Hence I=x/2-1/2.sinxcosx+k
Got it?

yes. Thanks alot.

Umartins1:
Boss, I'm not good at probability ooo but let me try. I'm sure agentofchange1 will correct me.

pr that P hits= 2/9

Pr that P misses= 1- 2/9 = 7/9


Pr that Q misses= 6/7

Pr that Q hits= 1-6/7 = 1/7

(a) pr that only one of them hits

= (pr that P hits + pr that Q misses) * (pr that P missed + pr that Q hits)







****************************


Please, let me wait a little and see if the bosses will confirm if I'm right so far.

you try bro,

1) S= (PnQ') u (P'nQ)

=> pr(S) = pr(PnQ') + pr(P'nQ)

=> (2/9 *6/7 ) + ( 7/9 * 1/7)

= 12/63 + 7/63 = 19/63

Anyone with a step by step solution to the problem below?

There is a substance which can be diluted to a number of solutions. The concentration of the substance is 35%, which can be diluted to 12% and 3% concentration.
If one part of the 35% concentration is added to eleven parts of distilled water, it will give a solution of 3% concentration and one part of 12% concentration added to three parts of distilled water will also give a solution of 3% concentration.
The problem is how to get the solution of 12% concentration from the 35% concentration. So, how many parts of distilled water will be added to one part of the 35% concentration to give the solution of 12% concentration?

elmajor:
Anyone with a step by step solution to the problem below?

There is a substance which can be diluted to a number of solutions. The concentration of the substance is 35%, which can be diluted to 12% and 3% concentration.
If one part of the 35% concentration is added to eleven parts of distilled water, it will give a solution of 3% concentration and one part of 12% concentration added to three parts of distilled water will also give a solution of 3% concentration.
The problem is how to get the solution of 12% concentration from the 35% concentration. So, how many parts of distilled water will added to one part of the 35% concentration to give the solution of 12% concentration?

looks like chemistry

agentofchange1:
you try bro,
1) S= (PnQ') u (P'nQ)
=> pr(S) = pr(PnQ') + pr(P'nQ)
=> (2/9 *6/7 ) + ( 7/9 * 1/7)
= 12/63 + 7/63 = 19/63

thanks. Pls Help with the other part (b) pr that one of them misses it?

elmajor:
Anyone with a step by step solution to the problem below?

There is a substance which can be diluted to a number of solutions. The concentration of the substance is 35%, which can be diluted to 12% and 3% concentration.
If one part of the 35% concentration is added to eleven parts of distilled water, it will give a solution of 3% concentration and one part of 12% concentration added to three parts of distilled water will also give a solution of 3% concentration.
The problem is how to get the solution of 12% concentration from the 35% concentration. So, how many parts of distilled water will added to one part of the 35% concentration to give the solution of 12% concentration?

your question is so weird! lol looks like chemistry but i think it can be interpreted mathematically,lera!

elmajor:
Anyone with a step by step solution to the problem below?

There is a substance which can be diluted to a number of solutions. The concentration of the substance is 35%, which can be diluted to 12% and 3% concentration.
If one part of the 35% concentration is added to eleven parts of distilled water, it will give a solution of 3% concentration and one part of 12% concentration added to three parts of distilled water will also give a solution of 3% concentration.
The problem is how to get the solution of 12% concentration from the 35% concentration. So, how many parts of distilled water will added to one part of the 35% concentration to give the solution of 12% concentration?

i got 0.4%..how true?

I feel like solving mathematics, great to be back after a long vacation summer in United States of Dream.

Hi great NL mathematicians.

bolkay47:
i got 0.4%..how true?

Incorrect mister. U are to look for number of part(s) and not the concentration of the solution.

bolkay47:
your question is so weird! lol looks like chemistry but i think it can be interpreted mathematically,lera!

This is a mathematical problem; not chemistry.

Uncle Richiez it's like this problem is a hard nut to crack.

elmajor:
Anyone with a step by step solution to the problem below?

There is a substance which can be diluted to a number of solutions. The concentration of the substance is 35%, which can be diluted to 12% and 3% concentration.
If one part of the 35% concentration is added to eleven parts of distilled water, it will give a solution of 3% concentration and one part of 12% concentration added to three parts of distilled water will also give a solution of 3% concentration.
The problem is how to get the solution of 12% concentration from the 35% concentration. So, how many parts of distilled water will be added to one part of the 35% concentration to give the solution of 12% concentration?

I think it should be moved to the front page for everyone to see.

try this guys..

ayokunlei:
thanks. Pls Help with the other part (b) pr that one
of them misses it?

is it. at least one them? Re-check the question n tell me..

ayokunlei:
thanks. Pls Help with the other part (b) pr that one
of them misses it?

agentofchange1:
is it. at least one them? Re-check the question n tell me..

If question (b) is correct then it has the same solution as question(a) as provided by agentofchange1. That is if one of them misses it then it implies one of the must get it which is the same as question(a) 19/63.

agentofchange1:
try this guys..

It is like the double integral doesn't converge.

elmajor:
Anyone with a step by step solution to the problem below?

There is a substance which can be diluted to a number of solutions. The concentration of the substance is 35%, which can be diluted to 12% and 3% concentration.
If one part of the 35% concentration is added to eleven parts of distilled water, it will give a solution of 3% concentration and one part of 12% concentration added to three parts of distilled water will also give a solution of 3% concentration.
The problem is how to get the solution of 12% concentration from the 35% concentration. So, how many parts of distilled water will be added to one part of the 35% concentration to give the solution of 12% concentration?

Looks pretty straightforward to me.



First, we should note: if, for example, 200ml of 30% conc is added to 200ml of water, it gives 400ml of x% conc (more diluted)
200/400 = x/30; x=15% conc

1 part of 35% conc added to 11 parts of water gives 3% conc.
, i.e. 1 part can be any volume, but we can assume 1 ml for now.
1/(x+1) = 3/35
x = 32
for 11 parts of water; we can have
3/(3 + (32/11).11) = 3/35
where 32/11 = ~3, which makes sense for 1 part of 3ml of solution and 11 parts of 3ml of water

1 part of 12% conc added to 3 parts of water gives 3% conc.
1:4= 3:12
where 4 = 3 parts of water + 1 part of solution

how to get 12% conc from 35% conc?
1 part of 35% conc, how many parts of water to get 12%?

1/(x+1) = 12/35

x = (35-12)/12 = 23/12 = ~2 parts of water

Ans: ~2 parts of water.

Anyways, I love maths and this was a good one to do!

yea bro, never mind av solved it, its actually a problem on bi-variate Pdf

masperano:
It is like the double integral doesn't converge.

tanx anyway

seems we gat a female math killer in the house , nice1 dear , welcome on board

Vixxie:
Looks pretty straightforward to me.



First, we should note: if, for example, 200ml of 30% conc is added to 200ml of water, it gives 400ml of x% conc (more diluted)
200/400 = x/30; x=15% conc

1 part of 35% conc added to 11 parts of water gives 3% conc.
, i.e. 1 part can be any volume, but we can assume 1 ml for now.
1/(x+1) = 3/35
x = 32
for 11 parts of water; we can have
3/(3 + (32/11).11) = 3/35
where 32/11 = ~3, which makes sense for 1 part of 3ml of solution and 11 parts of 3ml of water

1 part of 12% conc added to 3 parts of water gives 3% conc.
1:4= 3:12
where 4 = 3 parts of water + 1 part of solution

how to get 12% conc from 35% conc?
1 part of 35% conc, how many parts of water to get 12%?

1/(x+1) = 12/35

x = (35-12)/12 = 23/12 = ~2 parts of water

Ans: ~2 parts of water.

Anyways, I love maths and this was a good one to do!