₦airaland Forum

Karmanaut:
Continuation of question 1.
Pardon my handwriting.

idris4r83:
Abeg help me solve this

Karmanaut:
tetralogyfallot:

How to solve:
Create a table for finding the mean:
One column for Amount, another for frequency, and another for the cumulative frequency.
In the places where there are unknowns write the cumulative frequency in terms of unknowns, ie in algebraic form.
For example The second cumulative frequency will be 18 + x., the Third 43 + x.
Then since we have the value of the median and the total frequency we can find an unknown which will enable us find the other unknown.
Use the formula for finding the median of a grouped data.
Estimated Median = L + {w * [ (n/2) − cfb] }/ fm
Where
L is the lower class boundary of the group containing the median
n is the total number of data
cfb is the cumulative frequency of the groups before the median group
fm is the frequency of the median group
w is the group width.

After setting up the table you'll see that the median class is 41-60, because it's cumulative frequency contains half of the total frequency.


So our L will be 41,
N = 87, from the question
cfb will be 18+x as it is the cumulative frequency of the class before the median class.
fm, the frequency of the median class is 25.
And we were given the median to be 48.
The class width is obviously 20.


So you substitute your values in and solve for x.
To find y:
Since the total frequency is 87, when you've gotten x, summing the frequencies should give you 87.
So you have 18 + x + 25 + y +15 = 87.
When you get x substitute it then get y.

I hope you can take it from here?

If you solve it correctly you should get:
x=16.75; y = 12.25.

Would have written the solution and uploaded it but Almighty PHCN has struck and I'm too lazy to turn on the generator.
If you don't understand then maybe I'll write it and upload when there's power.

PS: Two people uploaded the same question, are you two the same person or is this an assignment?
Cheers.

tetralogyfallot:
Thanx bro, its an assignment and the other username is my friend.

Karmanaut:
Do you understand?

tetralogyfallot:
I don't get it bro, i will prefer you to upload it. Thanx in anticipation.

Karmanaut:
You're in luck.
There's light now.

Karmanaut:
Same answer.
Will do the rest later, perhaps this night.

ayokunlei:
thanks alot

Karmanaut:
Same answer.
Will do the rest later, perhaps this night.

agentofchange1:
HEY buddy, still expecting , guess u'v been busy lately .

mojeed4:
If the population of a country double in 50 years, in how many years will it treble, under the assumption that the rate of increase is proportional to the number of inhabitants?


please help with this question... given as an assignment.,. course code - differential equation II.

Karmanaut:
Answer: 79 years.
Solution is attached.
Hope you understand.
PS you should listen more in class and ask questions whenever you don't understand a concept.

mojeed4:
cooool! I swear you be bad ass...
Never really understood the part where the solution is x(t) = ce^kt... I mean the whole concept of that particular part generally.

I guess mainly because I skip a Lotta maths classes though.. and always late to class whenever I attend one.

MaxGraviton:
Senior men and Women in the house. Pls I need help with this. . . Thanks in Advance.

bolkay47:
1/2sqrt(x).Arcos[2/sqrt(x)]+k

MaxGraviton:
Senior men and Women in the house. Pls I need help with this. . . Thanks in Advance.

Karmanaut:
Geez, I forgot.
No light so I have to type the answer.
2)2xy" + 3y' -y= 0
Recurrence relation is an = an-1/(2n+2r+1)(n+r)
The roots of the indicial equation are -1/2 and 0.
The first series is y1(x) = a0( 1 + x/3 + x²/630+....)
which is the same as the series of sinh√(2x) * 1/√(2x)
Therefore y1(×) = (a0*sinh√(2x))/ √(2×)
The second series is
y2= b0(1+x/30 + ) which is the same as cosh(2×)/√×
y2= (b0 * cosh√(2×)/√×)

You can rewrite it in terms of e.
By using the trigonometric identities:
sinh(x) = 1/2(e^x -e^-x)
cosh (x) = 1/2(e^x +e[/sup]-×[/sup])
Define P1 and P2 as the linear combination of the two solutions with P1= a0/2√2 + b0/√2 and P2 = b0/√2 - a0/2
y(x) = (P1 * e^√(2×))/√× - (P2 * e^-√(2×)) /√(2×)
Which simplifies to
e^-√(2×) (2P1 e^2√(2×) + P2 * √2) / 2√×

agentofchange1:

$√[x/(4-x)] dx = $ √x/√(4-x) dx

put t² = 4 - x

=> x= 4-t²

dx= -2t dt

thus, we get

-2$ √(4-t² ) dt

set t=2sin@ , @= arcsin(t/2)

dt=2cos@d@

=> -2$ √(4-4sin² @) . 2cos@d@

=> -8$ cos² @ d@

=>- 4$(1+cos2@) d@

(since cos² @ = 0.5(1+cos2@)

= -4 [ @ + 0.5sin2@)]+C

= -4[ arcsin(t/2) + sin[arcsin(t/2) *cos[arcsin(t/2)]] +C

now simplify further n replace. t

guess that's it, hope u get?

agentofchange1:
ok boss..
thou art great... God bless and increase thee.
as usual av solved it (them ) conversely .
i presume thou art a graduate of Matimatizz...

Karmanaut:
Far from it. I'm just an internet troll.

MaxGraviton:
Thanks a lot. But pls I'm finding it dificult to understand it. Do you mind sending a shot of the solution?

agentofchange1:
ok... leme... c....

MaxGraviton:
Thanks a lot cheif. Expecting...

bolkay47:
1/2sqrt(x).Arcos[2/sqrt(x)]+k

agentofchange1:
sorry was tired yesterday..slept off

here

MaxGraviton:
Pls Chief, does it have to be y=2Sin @ can't it be just y=sin @.
Like...What's the reason for your choice of 2Sin @ ? Thanks .

masperano:
What informed my choice is the denominator, I chose y=2sin@ to get rid of the "4" using the trigonometry identity. Sin^2@+Cos^2@=1. if the denominator was say 6-y^2 , i would have gone for y=sqrt(6)sin@ , so that when i square it i can get rid of 6 at the denominator using the aforementioned identity. The take home message is if you are integrating a somewhat "non standard function(in the context of integration)" one has to cleverly look for ways to reduce the integrand to some thing you can integrate straight up(as observed in the problem, i reduced it to the form of sin^2@).

PS: There are several solutions to that problem. A different solution has been given by agentofchange. Now it is always very good for one to always do what i call "sanity check" by differentiating your answer to see if you can recover the question(the integrand) this is because differentiation is the opposite of integration(you should be able to recover what you integrated by differentiating). I have however used a soft ware known as mathematica to integrate this same question to demonstrate to you that you can get several representation of the solution. I usually use Mathematica or Maple to do my messy integrations and differentiations or solve my differential equations or PDE when i know i am likely to make a mistake.

MaxGraviton:
Thanks a lot cheif.

But one question cheif, Why does it have to be U=2sin@

Why can't it just be U=signatu @ or U=cos @ .
Coz from the look of things up there in the solution, those Trig notatations were overcrowding the solution. . . SO does it have to be like that?