jackpot:
my left mind is wishing the guy drops Laplace Transform as the compulsory question

I can use laplace to some extent cos I apply it in chemistry very well (am a chemistry student)
But wouldn't expect it sha cos he never teach us that

jackpot:
Let me follow your argument to see if it holds water:
you gave your sample space as
S={(124),(125),(126),(245),(246),(145),
(156),(645),(641),(265)}.

Following your logic,
S={(124),(512),(126),(524),(246),(514),(156),(564),(164),(526)} because, for instance, 125 and 512 are the same numbers without repetition (whatever that means).

So, probability = 0/10=0.

Contradiction!

Class dismiss. Everybody go home.

hmmm,

i normally don't like arguing much, but let me try to explain this for the last time & if u guys think otherwise, that's Ok . so be it.

to explain this,

[b]lets say we have 5 students in a class ( named S1, S2, S3 , S4 & S5 ) then asked to form a group of three (probably, study group ) with no repetition allowed.(no repetition of same student(s) in the group or the sample space ) . already noted that S1 is the class rep. & S5 is the assistant class rep. , Now if a group is picked at random , whats the probability (or chance )of picking a group which has its last member as either S1 or S5 ? ( i. e either the class rep. or the assistant )

NB: In combination, we give attention to order of arrangements

total possible arrangement without repetition = 5C3
=10 ways

Now, possible sample space

S={ (s1s2s4),(s1s2s5),(s1s2s6),(s1s4s5),(s1s5s6), (s2s4s5), (s6s4s5),(s6s4s1),(s2s6s5) . . . }

the possible groups continues to about ten of them , i just stopped there cuz, I've gotten what i needed , any more will result to repetition of already exiting group which goes against the given condition

Probability of having either s1 or s5 as last member of the group , could be (s1s2s5) or (s1s4s5) or (s2s4s5) or (s6s4s1) or (s2s6s5) or (s6s4s5)

= 1/10 + 1/10 + 1/10 +1/10 +1/10 + 1/10 =6/10 =3/5 [/b]
Now you guys be the judge

am off this question, happy solving .



Good night!!!!

agentofchange1:



hmmm,

i normally don't like arguing much, but let me try to explain this for the last time & if u guys think otherwise, that's Ok . so be it.

to explain this,

[b]lets say we have 5 students in a class ( named S1, S2, S3 , S4 & S5 ) then asked to form a group of three (probably, study group ) with no repetition allowed.(no repetition of same student(s) in the group or the sample space ) . already noted that S1 is the class rep. & S5 is the assistant class rep. , Now if a group is picked at random , whats the probability (or chance )of picking a group which has its last member as either S1 or S5 ? ( i. e either the class rep. or the assistant )

NB: In combination, we give attention to order of arrangements

total possible arrangement without repetition = 5C3
=10 ways

Now, possible sample space

S={(s1s2s4) ,(),(s1s2s5),(s1s2s6),(s1s4s5),(s1s5s6), (s2s4s5), (s6s4s5),(s6s4s1),(s2s6s5) . . . }

the possible groups continues to about ten of them , i just stopped there cuz, I've gotten what i needed , any more will result to repetition of already exiting group which goes against the given condition

Probability of having either s1 or s5 as last member of the group , could be (s1s2s5) or (s1s4s5) or (s2s4s5) or (s6s4s1) or (s2s6s5) or (s6s4s5)

= 1/10 + 1/10 + 1/10 +1/10 +1/10 + 1/10 =6/10 =3/5 [/b]
Now you guys be the judge

am off this question, happy solving .



Good night!!!!

Good morning bro. It's not that I find joy in arguing with you over this o as you be my oga patapata but I can't help but to point this out;
You said possible sample space

S={(s1s2s4) ,(s1s2s4),(s1s2s5),(s1s2s6),(s1s4s5),(s1s5s6), (s2s4s5), (s6s4s5),(s6s4s1),(s2s6s5) . . . }

Don't you think it's possible that the sample space could also possibly be

S={(s4s1s2) ,(s5s1s2),(s6s1s2),(s5s1s4),(s6s1s5), (s5s2s4), (s5s6s4),(s1s6s4),(s5s2s6), (s5s6s2) . . . } and in which your theory doesn't hold in this case.
Moreover, if you check your sample space, it's already up to 10 and you agreed that we could still continue, so many contradictions.
You see, my boss, whenever they use "no repetition" in this type of question, it simply means the same number can't show up twice in the same group like (s1s1s2). So, if you don't mind, you can also look this up.

Good morning to y'all
#Peace

Seamione:
Cool, that's what I also did

Perhaps u should have added it up. So for instance 24+12=36 and then divided by 60 which will give you the required 3/5(a bit tricky).

masperano:


Perhaps u should have added it up. So for instance 24+12=36 and then divided by 60 which will give you the required 3/5(a bit tricky).

Yeah, I think that what the text did. Actually, I also considered that but then I had to follow the rule of independent events.

To determine the probability of event X or event Y happening, when the two events are not mutually exclusive:
1.) Find the sum of the separate probabilities.
2.) From this sum subtract the probability that both events will occur.
P(X or Y) = P(X) + P(Y)-P(X and Y)

This will give us 12/60 + 24/60 - 12/60 = 2/5

Seamione:
Good morning bro. It's not that I find joy in arguing with you over this o as you be my oga patapata but I can't help but to point this out;
You said possible sample space

S={(s1s2s4) ,(s1s2s4),(s1s2s5),(s1s2s6),(s1s4s5),(s1s5s6), (s2s4s5), (s6s4s5),(s6s4s1),(s2s6s5) . . . }

Don't you think it's possible that the sample space could also possibly be

S={(s4s1s2) ,(s5s1s2),(s6s1s2),(s5s1s4),(s6s1s5), (s5s2s4), (s5s6s4),(s1s6s4),(s5s2s6), (s5s6s2) . . . } and in which your theory doesn't hold in this case.
Moreover, if you check your sample space, it's already up to 10 and you agreed that we could still continue, so many contradictions.
You see, my boss, whenever they use "no repetition" in this type of question, it simply means the same number can't show up twice in the same group like (s1s1s2). So, if you don't mind, you can also look this up.

Good morning to y'all
#Peace

sorry check well, dia nine, i just made small mistake in writing d first group twice. den plus one other 1 i didn't include that will make it 10.

ur still repeating same sample space , no matter how far u go
hmm. i don't think i can explain this any further/longer maybe i should 'nt av used combinatorial approach for this question, i believe u will understand better some day, i really love ur type, who challenges/questions the 'lemmas' . I should have used the method u used ('wrongly ') , but i won't do that now as that will prompt another episode of arguments .m not really feeling strong. ..


nice solving.... keep the fire burning.

#shalom.

integrate log(cosx + secx) showing full workings………
tags: jackpot , agentofchange1 , laplacian and others……

personal59:


I can use laplace to some extent cos I apply it in chemistry very well (am a chemistry student)
But wouldn't expect it sha cos he never teach us that

wow. What's the application to Chemistry?

masperano:


Perhaps u should have added it up. So for instance 24+12=36 and then divided by 60 which will give you the required 3/5(a bit tricky).

not everything you see in a textbook is correct. The high time you realize that, the better.

Required ko, requested ni.


The answer to that question is 2/5. I can bet my lunch on that.


btw, do you know that ''tricky'' means intended to deceive?

Seamione:
Yeah, I think that what the text did. Actually, I also considered that but then I had to follow the rule of independent events.

To determine the probability of event X or event Y happening, when the two events are not mutually exclusive:
1.) Find the sum of the separate probabilities.
2.) From this sum subtract the probability that both events will occur.
P(X or Y) = P(X) + P(Y)-P(X and Y)

This will give us 12/60 + 24/60 - 12/60 = 2/5

Yea after meticulously looking at it from all angles i think you are not incorrect. Could be an erratum in the book.

jackpot:
not everything you see in a textbook is correct. The high time you realize that, the better.

Required ko, requested ni.


The answer to that question is 2/5. I can bet my lunch on that.


btw, do you know that ''tricky'' means intended to deceive?

*Smiles* see who is talking because of the small small sec school/undergrad maths wey u dey solve na im dey shark you abi?

yes sir/ma. I do know it means throwing curved balls.

PS: when replying me next time be more civil

Porthos:
integrate log(cosx + secx) showing full workings………

……

This does not have an anti-derivative; There is no closed form expression for the indefinite integral.
The primitives of ln(cos(x)) cannot be expressed as a finite combination of usual functions.
The analytic expression is complicated and includes a special function (polylogarithm).

integral log(cos(x)+sec(x)) dx = 1/2 (-i Li_2(-e^(2 i x))+i (Li_2((-3+2 sqrt(2)) e^(2 i x))+Li_2(-(3+2 sqrt(2)) e^(2 i x)))+i x^2+2 x log(1+e^(2 i x))-2 log(1+(3-2 sqrt(2)) e^(2 i x)) (x+sin^(-1)(sqrt(2)))-2 log(1+(3+2 sqrt(2)) e^(2 i x)) (x-sin^(-1)(sqrt(2)))-4 i sin^(-1)(sqrt(2)) tan^(-1)((tan(x))/sqrt(2))+2 x log(cos(x)+sec(x)))+constant



Where Li_n(x) is the Polylogarithmic integral function which is given by the integral of dt/ln (t) from 0 to x for all positive real numbers where x=! 1.
the integral for x > 1 has to be interpreted as a Cauchy Principal value

The polylog in this case is specifically the 'dilogarithm' because of the power of n is 2 in the denominator of the sum.
cos (x) = (e^ix + e^-ix)/2

jackpot:
wow. What's the application to Chemistry?

We use it very well under quantum chemistry

Can anyone help confirm this answers? A committee of 3 is to be chosen from 5 men and 4 women.find the probability that there will be one woman in the committee. Is the answer 10/21 or 1/21

personal59:


We use it very well under quantum chemistry

Written your exam yet?
How was it?

lopzy:
Can anyone help confirm this answers? A committee of 3 is to be chosen from 5 men and 4 women.find the probability that there will be one woman in the committee. Is the answer 10/21 or 1/21

5men
4women

Total = 9persons
forming a-3-man committee
Total ways of selection =9C3 =84 ways

now for the given condition, if one woman had to be in the committee, its means to other two gat to be men

thus,

probability that there will be one woman in the committee

5C2*4C1 /9C3= 10x4/84 = 10/21

that's it dear, hope you get?

agentofchange1:


sorry check well, dia nine, i just made small mistake in writing d first group twice. den plus one other 1 i didn't include that will make it 10.

ur still repeating same sample space , no matter how far u go
hmm. i don't think i can explain this any further/longer maybe i should 'nt av used combinatorial approach for this question, i believe u will understand better some day, i really love ur type, who challenges/questions the 'lemmas' . I should have used the method u used ('wrongly ') , but i won't do that now as that will prompt another episode of arguments .m not really feeling strong. ..


nice solving.... keep the fire burning.

#shalom.

I don't know how to say this without sounding arrogant, but I strongly believe I perfectly understand what I need to in this context. It is you rather, who are refusing to change your perspective about the question, which is all good though.

Take a look at the question again If you don't mind

Three digit numbers re formed from the digits 1,2,4,5,6. If repetition is not allowed and a number is picked at random, find the probability that it's a multiple of 5 or an odd number.

The term "repetition is not allowed" here simply means when forming a 3-digit number (just one), no digit should appear more than once. So 124, 412, 214, 142, 421 and 241 are all in conformity with the rule guiding the formation of the numbers (i. 3-digit number from the digits 1,2,4,5,6 ii. No repetition of digits)
and are also totally different numbers such cannot be regarded as repeated numbers. Therefore, they all will make the sample space.
At this point if you still feel what about 3 of us here on this thread has said on this question is wrong, then I give up. Hoping that you change your perspective about it one day.

Mind you, it's not a matter of challenging/questioning your knowledge on this, but about not misleading lots of young ones who are gaining one or two things from this forum.

Happy weekend to y'all
#Peace

Seamione:
I don't know how to say this without sounding arrogant, but I strongly believe I perfectly understand what I need to in this context. It is you rather, who are refusing to change your perspective about the question, which is all good though.

Take a look at the question again If you don't mind
The term "repetition is not allowed" here simply means when forming a 3-digit number (just one), no digit should appear more than once. So 124, 412, 214, 142, 421 and 241 are all in conformity with the rule guiding the formation of the numbers (i. 3-digit number from the digits 1,2,4,5,6 ii. No repetition of digits)
and are also totally different numbers such cannot be regarded as repeated numbers. Therefore, they all will make the sample space.
At this point if you still feel what about 3 of us here on this thread has said on this question is wrong, then I give up. Hoping that you change your perspective about it one day.

Mind you, it's not a matter of challenging/questioning your knowledge on this, but about not misleading lots of young ones who are gaining one or two things from this forum.

Happy weekend to y'all
#Peace

yea bro, i get, its ok now lets forget the whole thing... thanks anyways....

we are all learning.

agentofchange1:



yea bro, i get, its ok now lets forget the whole thing... thanks anyways....

we are all learning.

You're welcome big boss

::::: Some useful mathematical results ::
(α+в+¢)²= α²+в²+¢²+2(αв+в¢+¢α) 1. (α+в)²= α²+2αв+в² 2. (α+в)²= (α-в)²+4αв b 3. (α-в)²= α²-2αв+в² 4. (α-в)²= (α+в)²-4αв 5. α² + в²= (α+в)² - 2αв. 6. α² + в²= (α-в)² + 2αв. 7. α²-в² =(α + в)(α - в) 8. 2(α² + в²) = (α+ в)² + (α - в)² 9. 4αв = (α + в)² -(α-в)² 10. αв ={(α+в)/2}²-{(α-в)/2}² 11. (α + в + ¢)² = α² + в² + ¢² + 2(αв + в¢ + ¢α) 12. (α + в)³ = α³ + 3α²в + 3αв² + в³ 13. (α + в)³ = α³ + в³ + 3αв(α + в) 14. (α-в)³=α³-3α²в+3αв²-в³ 15. α³ + в³ = (α + в) (α² -αв + в²) 16. α³ + в³ = (α+ в)³ -3αв(α+ в) 17. α³ -в³ = (α -в) (α² + αв + в²) 18. α³ -в³ = (α-в)³ + 3αв(α-в)
ѕιη0° =0 ѕιη30° = 1/2 ѕιη45° = 1/√2 ѕιη60° = √3/2 ѕιη90° = 1 ¢σѕ ιѕ σρρσѕιтє σƒ ѕιη тαη0° = 0 тαη30° = 1/√3 тαη45° = 1 тαη60° = √3 тαη90° = ∞ ¢σт ιѕ σρρσѕιтє σƒ тαη ѕє¢0° = 1 ѕє¢30° = 2/√3 ѕє¢45° = √2 ѕє¢60° = 2 ѕє¢90° = ∞ ¢σѕє¢ ιѕ σρρσѕιтє σƒ ѕє¢
2ѕιηα¢σѕв=ѕιη(α+в)+ѕιη(α-в) 2¢σѕαѕιηв=ѕιη(α+в)-ѕιη(α-в) 2¢σѕα¢σѕв=¢σѕ(α+в)+¢σѕ(α-в) 2ѕιηαѕιηв=¢σѕ(α-в)-¢σѕ(α+в)

ѕιη(α+в)=ѕιηα ¢σѕв+ ¢σѕα ѕιηв. » ¢σѕ(α+в)=¢σѕα ¢σѕв - ѕιηα ѕιηв. » ѕιη(α-в)=ѕιηα¢σѕв-¢σѕαѕιηв. » ¢σѕ(α-в)=¢σѕα¢σѕв+ѕιηαѕιηв. » тαη(α+в)= (тαηα + тαηв)/ (1−тαηαтαηв) » тαη(α−в)= (тαηα − тαηв) / (1+ тαηαтαηв) » ¢σт(α+в)= (¢σтα¢σтв −1) / (¢σтα + ¢σтв) » ¢σт(α−в)= (¢σтα¢σтв + 1) / (¢σтв− ¢σтα) » ѕιη(α+в)=ѕιηα ¢σѕв+ ¢σѕα ѕιηв. » ¢σѕ(α+в)=¢σѕα ¢σѕв +ѕιηα ѕιηв. » ѕιη(α-в)=ѕιηα¢σѕв-¢σѕαѕιηв. » ¢σѕ(α-в)=¢σѕα¢σѕв+ѕιηαѕιηв. » тαη(α+в)= (тαηα + тαηв)/ (1−тαηαтαηв) » тαη(α−в)= (тαηα − тαηв) / (1+ тαηαтαηв) » ¢σт(α+в)= (¢σтα¢σтв −1) / (¢σтα + ¢σтв) » ¢σт(α−в)= (¢σтα¢σтв + 1) / (¢σтв− ¢σтα)
α/ѕιηα = в/ѕιηв = ¢/ѕιη¢ = 2я » α = в ¢σѕ¢ + ¢ ¢σѕв » в = α ¢σѕ¢ + ¢ ¢σѕα » ¢ = α ¢σѕв + в ¢σѕα » ¢σѕα = (в² + ¢²− α²) / 2в¢ » ¢σѕв = (¢² + α²− в²) / 2¢α » ¢σѕ¢ = (α² + в²− ¢²) / 2¢α » Δ = αв¢/4я » ѕιηΘ = 0 тнєη,Θ = ηΠ » ѕιηΘ = 1 тнєη,Θ = (4η + 1)Π/2 » ѕιηΘ =−1 тнєη,Θ = (4η− 1)Π/2 » ѕιηΘ = ѕιηα тнєη,Θ = ηΠ (−1)^ηα

1. ѕιη2α = 2ѕιηα¢σѕα 2. ¢σѕ2α = ¢σѕ²α − ѕιη²α 3. ¢σѕ2α = 2¢σѕ²α − 1 4. ¢σѕ2α = 1 − ѕιη²α 5. 2ѕιη²α = 1 − ¢σѕ2α 6. 1 + ѕιη2α = (ѕιηα + ¢σѕα)² 7. 1 − ѕιη2α = (ѕιηα − ¢σѕα)² 8. тαη2α = 2тαηα / (1 − тαη²α) 9. ѕιη2α = 2тαηα / (1 + тαη²α) 10. ¢σѕ2α = (1 − тαη²α) / (1 + тαη²α) 11. 4ѕιη³α = 3ѕιηα − ѕιη3α 12. 4¢σѕ³α = 3¢σѕα + ¢σѕ3α

» ѕιη²Θ+¢σѕ²Θ=1 » ѕє¢²Θ-тαη²Θ=1 » ¢σѕє¢²Θ-¢σт²Θ=1 » ѕιηΘ=1/¢σѕє¢Θ » ¢σѕє¢Θ=1/ѕιηΘ » ¢σѕΘ=1/ѕє¢Θ » ѕє¢Θ=1/¢σѕΘ » тαηΘ=1/¢σтΘ » ¢σтΘ=1/тαηΘ » тαηΘ=ѕιηΘ/¢σѕΘ

#shalom.

hey guys try this asap..

integrate dx/√(1+sinx)

agentofchange1:
hey guys try this asap..

integrate dx/√(1+sinx)

Substituted so much I felt like a coach.

.

Karmanaut:
Substituted so much I felt like a coach.

aahh boss mi. integral of 1/(1-w^2) = arctan(w) ?

agentofchange1:

aahh boss mi. integral of 1/(1-w^2) = arctan(w) ?

Yes. If you differentiate arctan (x) you get (1/(1-x^2)) Do you need the proof?

Karmanaut:
Yes. If you differentiate arctan (x) you get (1/(1-x^2)) Do you need the proof?

yes sir

agentofchange1:
yes sir

A trivial proof is attached.

Karmanaut:

A trivial proof is attached.

hmm. i see...

Karmanaut:
A trivial proof is attached.

k the integral of 1/(1+x^2) =?

agentofchange1:


k the integral of 1/(1+x^2) =?

hi add me to the group with this number 08147276723

agentofchange1:


k the integral of 1/(1+x^2) =?

arctan (x)
It follows the same procedure.
Start from y =arctan (x)
Therefore x = tan (y)(Call this equation 1)
Differentiating x wrt y gives sec^2 (y)
Also from trigonometry sec^2 (x) =1 + tan^2 (x)
Therefore dx/dy = 1 + tan^2 (y) (call this equation 2)
Recall that x =tan (y) from equation 1
Substitute it into equation 2
which guves:
dx/dy=(1+x^2)
Dy/dx Which is required of us is the inverse of this. Turning the equation upside down gives:
dy/dx = 1/(1+x^2)
Q. E. D
I worked backwards from the answer since it is a standard. Do Not Try This At Home.

bolkay47:
hi add me to the group with this number 08147276723

done