Eyohimself:

A certain characteristic in a large population has a distribution that is symmetric about the mean m. If 68 percent of the distribution lies within one standard deviation d of the mean, what percent of the distribution is less than m + d ?
A. 16%B. 32%C. 48%D. 84%E. 92%

The explanation for this is much easier with a diagram.

Let G(z) = P(x is less than z).
Since the distribution is symmetric about the mean m, then G(m) = P(x < m) = 50%.

The statement "68 percent of the distribution lies within one standard deviation d of the mean" means that:
G(m+d) - G(m-d) = 68% = P(m-d<x<m+d)

So, we are looking for G(m+d) which is also equal to P(x < m+d)

But P(x < m+d) = P(x < m) + P(m < x < m+d)
= P(x < m) + 0.5*[P(m-d < x < m+d)] due to symmetric nature of distribution about the mean m
= G(m) + 0.5*(G(m+d) - G(m-d)) just substituting
= 50% + 0.5(68%)
= 50% + 34%
= 84% (Answer = D)

_{NOTE: there are several other ways of doing this.}

Nice maths tread

Eyohimself:

He must sell it for :


0.11*170270 + 0.055*170270 + 170270 = $198 364.55 ( $198 365--to the nearest $)

( Assuming they are making the gains on the cost price).

Eyohimself,

The agent must sell for $199,395(nearest $). 
Only Ann's gain is made on the initial cost price.
To the Agent the current cost price is 170270 + (0.11*170270) = $188,999.70
So, selling price is 188,999.70 + (0.055*188,99.70) = $199,394.68 ($199,395)

ADint:

Eyohimself,

The agent must sell for $199,395(nearest $).
Only Ann's gain is made on the initial cost price.
To the Agent the current cost price is 170270 + (0.11*170270) = $188,999.70
So, selling price is 188,999.70 + (0.055*188,99.70) = $199,394.68 ($199,395)

Thanks for noting that. I made my calculations based on the assumption that they both had

gains on the initial cost price which I think is quite possible. Your contribution is welcomed.

You based your solution on the assumption that the Agent made the gain on Ann's selling price which

from common logic, is more probable. Thanks for the insight.

I thought the other way round.

hi every one,

i'm happy to see o much going on on this thread.boy i ' have a lot of catching up to do.
eyohimself how is your prep.i'm sure you are in top shape.
i'm brewing some stuff.
Richyblacky
you are such a blessing on this thread.

integrating sec x

sec x=

sec x (sec x +tan x)/(sec x +tan x)=sec² x+sec x tan x/(sec x +tan x)

integerate:

ln(sec x +tan x)

There are 2 rooms. One has 3 light switches and the second one has three light bulbs. U can only enter the one with the light bulbs once. How do u deteremine which of the three light switches corresponds to three of the light bulbs in the second room

jay bee:

There are 2 rooms. One has 3 light switches and the second one has three light bulbs. You can only enter the one with the light bulbs once. How do u deteremine which of the three light switches corresponds to three of the light bulbs in the second room

1. Go into room with switches

2. Do the following:
i. Put the first switch ON, and leave it ON
ii. Put the second switch ON, leave it ON for a while, and then switch it OFF
iii. Ignore the third switch

3. Go into room with bulbs:
i. The bulb that is ON is controlled by the first switch
ii. The one that is OFF, but WARM is controlled by the second switch
iii. The one that is OFF, but COLD is controlled by the third switch.

NB: Some rather obvious assumptions were made.

This forum and questions are too hot for me ,am logging out.

Eyohimself,

Howz life? How did the GRE go?

I love these guys.I which i also could be a mathematics guru as these guys up here.It's Cooooooooooooool!

Maths geniuses can someone convert this Euler equation

x^2y'' (x) - 2xy' (x) + 2y (x) = 4x^3

to a constant coefficeint problem and show that the solution is given by

y (x) = Ax + Bx^2 + 2x^3.

Can someone solve this 2nd order differencial equation

y'' = 2y^3 + 8y where y =2 , y' = - 8  when x=π/4

Hey..u guys r gr8
On d above question
A pair should be black n black or green and green ryt?
If there are 5 black and 5 green socks (not pairs)
It implies that we have 10 socks which does not form a complete pair.
But if we take away 2(a black and a green), we are left with 8.with that we can form a complete pair
If we take away 3..there is no way the remainder can form complete pair.
NOTE
If u take away any number of socks,it shouldnt be considered as part of the complete pair.
So on that note
I Feel the ans is 2 being the minimum..
Please if u hav anytin contrary please let me understand..tanx

On d above question
A pair should be black n black or green and green ryt?
If there are 5 black and 5 green socks (not pairs)
It implies that we have 10 socks which does not form a complete pair.
But if we take away 2(a black and a green), we are left with 8.with that we can form a complete pair
If we take away 3..there is no way the remainder can form complete pair.
NOTE
If u take away any number of socks,it shouldnt be considered as part of the complete pair.
So on that note
I Feel the ans is 2 being the minimum..
Please if u hav anytin contrary please let me understand..tan

Mke "t" the subject of d formula in dz Newton's 2nd law of motion : s=ut+1/2.at^2

Pls help me 2 solve this questn..prove that cosec2(x) -cot(x)cosec(x)=1

sportyblinks: Mke "t" the subject of d formula in dz Newton's 2nd law of motion : s=ut+1/2.at^2

d eqn wil be change when multiply through by 2,.then u wil hv at^2+2ut-2s=0 and compared it wit general equatn of quadratic ax^2+bx+c=0 4rm there ur a=a, b=2u, c= -2s.then solve it by use of formula d ans wil be t= -2u+- squareroot of 4u^2 +8as all divided by 2a.,

edusegzy:
6. Jefferson National Expansion Memorial /the Gateway Arch

Hello sir! !!!!!!!!

kindly solve this, Rationalize

√3
/ cube rut 5 + cube rut7

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still on

Y I no fit Sabi maths

candylips:
Maths geniuses can someone convert this Euler equation

x^2y'' (x) - 2xy' (x) + 2y (x) = 4x^3

to a constant coefficeint problem and show that the solution is given by

y (x) = Ax + Bx^2 + 2x^3.

let y = Ax^m
y' = mAx^[m-1]
y'' = m(m-1)Ax(m-2)

substitute all and solve

for the particular integral use Ax^3 + bx^2 + cx + d

thankyouJesus:


let y = Ax^m
y' = mAx^[m-1]
y'' = m(m-1)Ax(m-2)

substitute all and solve

for the particular integral use Ax^3 + bx^2 + cx + d

Hmm! Boss! You dey scare me o

thankyouJesus:

let y = Ax^m y' = mAx^[m-1] y'' = m(m-1)Ax(m-2)
substitute all and solve
for the particular integral use Ax^3 + bx^2 + cx + d

how's school?

logadims:
how's school?

Great, thank you.

mathefaro:
Hmm! Boss! You dey scare me o

My boss

edusegzy:
hello everyone,
i have been going through various threads on nairaland and sincerely i am impressed by how young nigerian are helping each other.Seun you know you have acheived a fit that's a dream for many.keep it up bro.
By a way of introduction i recently graduated from O.A.U ile ife chemical engineering with a first class(i said that humbly) and iam currently serving in an oil company in ikoyi.
i humbly want to contribute my own quota on this site and since mathematics is my strenght i decided to do something along that line
if anyone has any maths question,assignment , from secondary school level to university,gmat,gre and sat maths questions, Kindly bring it on .

All the maths whiz in the house come on let's make this thread exciting

Nice to meet u here,i graduated from OAU too i am preparing for gre too bt i av issues with maths, can i av ur num so i can chat u up privately....my mail is wandeololade@ gmail .com...thanks

U guys are doing great here. Keep it up.