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Mathematics: Post your doubts here!

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The ages, x years, of 150 cars are summarised by Σx = 645 and Σ x^2= 8287.5. Find Σ(x − x~)^2, where
x~ denotes the mean of x.

ANY one plz
I just wanted to make sure the answer is correct.

So
they want
Σ(x − x~)^2
so for me this was very hard, since we dont have individual numbers, so i went to the formula book and found the first equation helpful.
If you check it ( i cant write it down) you will see the numerator is similar to the wanted question, so basically I made it the subject of formula, and i got
Σx^2 - n(x~)^2 as the formula

Now it is easy
we have Σx^2 value
and n = 150
and mean is Σx/n
 
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681
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153
applepie1996
We have to find the point p where the perpendicular distance to the two planes is same.

first we will for an equation the way the paper gave.
therefore
for plane m
|x+2y-2z-1|/sqrt((1^2) + (2^2) + (-2^2) = |x+2y-2z-1|/3 [P.S, sqrt is square root]
for plane n
|2x-2y+z-7|/sqrt((2^2)+(-2^2)+(1^2) = |2x-2y+z-7|/3


since you are finding a point where both distance is same , therefore.
|x+2y-2z-1|/3 = |2x-2y+z-7|/3 (3 can be cancelled both sides)
=
|x+2y-2z-1| = |2x-2y+z-7|

now we will sub the line values in the equation.
the x component (1+2t) [P.S, t is lamda ]
the y comp. (1+t)
the z comp. (-1+2t)

therefore.
| (1+2t) +2 ( 1+t) -2(-1+2t) -1 | = |2(1+2t) - 2(1+t) + (-1+2t) - 7 |
simplifying it you get
|4| = |-8+4t|
now to solve modulus, we do squaring method
16 = 64-64t+16t^2
simplify
2 = 8 - 8t +2t^2
2t^2 - 8t + 6 = 0
solve it and get
t = 3 or t= 1
when t=3 the position of point is [P.S, u do this by sub t value in line equation)
(7, 4, 5) => OA
when t = 1 the position of point is
(3, 2, 1) => OB

BA (or AB, same thing) = OA - OB
= (4 , 2, 4)

now find the mod
sqrt(4^2 + 2^2 + 4^2)
= 6

I dint do every step in detail, (very long) if u dont get any step, just ask. [Pray for me please.]
 
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applepie1996
We have to find the point p where the perpendicular distance to the two planes is same.

first we will for an equation the way the paper gave.
therefore
for plane m
|x+2y-2z-1|/sqrt((1^2) + (2^2) + (-2^2) = |x+2y-2z-1|/3 [P.S, sqrt is square root]
for plane n
|2x-2y+z-7|/sqrt((2^2)+(-2^2)+(1^2) = |2x-2y+z-7|/3


since you are finding a point where both distance is same , therefore.
|x+2y-2z-1|/3 = |2x-2y+z-7|/3 (3 can be cancelled both sides)
=
|x+2y-2z-1| = |2x-2y+z-7|

now we will sub the line values in the equation.
the x component (1+2t) [P.S, t is lamda ]
the y comp. (1+t)
the z comp. (-1+2t)

therefore.
| (1+2t) +2 ( 1+t) -2(-1+2t) -1 | = |2(1+2t) - 2(1+t) + (-1+2t) - 7 |
simplifying it you get
|4| = |-8+4t|
now to solve modulus, we do squaring method
16 = 64-64t+16t^2
simplify
2 = 8 - 8t +2t^2
2t^2 - 8t + 6 = 0
solve it and get
t = 3 or t= 1
when t=3 the position of point is [P.S, u do this by sub t value in line equation)
(7, 4, 5) => OA
when t = 1 the position of point is
(3, 2, 1) => OB

BA (or AB, same thing) = OA - OB
= (4 , 2, 4)

now find the mod
sqrt(4^2 + 2^2 + 4^2)
= 6

I dint do every step in detail, (very long) if u dont get any step, just ask. [Pray for me please.]
i will
and pray for me too pleez im not well :(
 
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457
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I just wanted to make sure the answer is correct.

So
they want
Σ(x − x~)^2
so for me this was very hard, since we dont have individual numbers, so i went to the formula book and found the first equation helpful.
If you check it ( i cant write it down) you will see the numerator is similar to the wanted question, so basically I made it the subject of formula, and i got
Σx^2 - n(x~)^2 as the formula

Now it is easy
we have Σx^2 value
and n = 150
and mean is Σx/n
which book??
 
Messages
1,601
Reaction score
553
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123
applepie1996
We have to find the point p where the perpendicular distance to the two planes is same.

first we will for an equation the way the paper gave.
therefore
for plane m
|x+2y-2z-1|/sqrt((1^2) + (2^2) + (-2^2) = |x+2y-2z-1|/3 [P.S, sqrt is square root]
for plane n
|2x-2y+z-7|/sqrt((2^2)+(-2^2)+(1^2) = |2x-2y+z-7|/3


since you are finding a point where both distance is same , therefore.
|x+2y-2z-1|/3 = |2x-2y+z-7|/3 (3 can be cancelled both sides)
=
|x+2y-2z-1| = |2x-2y+z-7|

now we will sub the line values in the equation.
the x component (1+2t) [P.S, t is lamda ]
the y comp. (1+t)
the z comp. (-1+2t)

therefore.
| (1+2t) +2 ( 1+t) -2(-1+2t) -1 | = |2(1+2t) - 2(1+t) + (-1+2t) - 7 |
simplifying it you get
|4| = |-8+4t|
now to solve modulus, we do squaring method
16 = 64-64t+16t^2
simplify
2 = 8 - 8t +2t^2
2t^2 - 8t + 6 = 0
solve it and get
t = 3 or t= 1
when t=3 the position of point is [P.S, u do this by sub t value in line equation)
(7, 4, 5) => OA
when t = 1 the position of point is
(3, 2, 1) => OB

BA (or AB, same thing) = OA - OB
= (4 , 2, 4)

now find the mod
sqrt(4^2 + 2^2 + 4^2)
= 6

I dint do every step in detail, (very long) if u dont get any step, just ask. [Pray for me please.]
JazakAllah :)
InshaAllah i will pray for you :)
and May Allah fulfill all your wishes :)
 
Messages
457
Reaction score
317
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73
I just wanted to make sure the answer is correct.

So
they want
Σ(x − x~)^2
so for me this was very hard, since we dont have individual numbers, so i went to the formula book and found the first equation helpful.
If you check it ( i cant write it down) you will see the numerator is similar to the wanted question, so basically I made it the subject of formula, and i got
Σx^2 - n(x~)^2 as the formula

Now it is easy
we have Σx^2 value
and n = 150
and mean is Σx/n
I made it the subject of formula, and i got
Σx^2 - n(x~)^2 as the formula
sorry but how?
 
Messages
681
Reaction score
1,731
Points
153
I made it the subject of formula, and i got
Σx^2 - n(x~)^2 as the formula
sorry but how?
Yes, that is correct, exactly like i told you..
now sub the value of Σx^2 = 8287.5
and n = 150
and the mean is 645/150 = 4.3
then take its square and multiply with 150
subtract from 8287.5
 
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681
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I don't have any website link but I have them in hard copy. Do you want me to scan them ??
Well, if it is okay for you, and wont bother you then yes, very much wanted..!! Jazaki Allah khairan!
 
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I don't have any website link but I have them in hard copy. Do you want me to scan them ??
seriously i need p3 integration notes too !! i haven't practiced it for so long now.. its all gone to some recycle bin in my head
 
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