All Answers 358090 Answers

23.000x10/100=2.3

x percent of something it is mean= 

\(x*smth\over 100\)

or 

smth*0.01x

Here is an example. Try to understand the meaning and the definition of each word and try to understand how it is calculated. Good Luck.

Statistical file:
{12, 23, 87, 46, 76, 35, 31, 14, 32, 18.34, 67, 89, 80}


Average (mean): μ=46.9492307692
Absolute deviation: 328.507692308
Mean deviation: 25.2698224852
Minimum: 12
Maximum: 89
Variance: 771.64323787
Standard deviation σ=27.778467162
Corrected sample standard deviation s=28.9127453042
Coefficient of variation cV=0.615830011067
Signal-to-noise ratio SNR=1.62382472765
Median: 35
Quartile Q1: 20.67
Quartile Q2: 35
Quartile Q3: 78
1st decile: 16.17 (Too few data to calculate deciles)
2nd decile: 20.67
3rd decile: 27
4th decile: 31.5
5th decile: 35
6th decile: 56.5
7th decile: 71.5
8th decile: 78
9th decile: 83.5
Interquartile range IQR: 57.33
Lower fence: -65.325
Upper fence: 163.995
Set of outliers: {} - empty set - no outliers found
Interdecile range IDR: 67.33
Mode: {76, 67, 80, 87, 89, 46, 35, 18.34, 14, 23, 31, 32, 12} - multimodal
Geometric mean: 38.104298633
Harmonic mean: 30.239738882
Sum: 610.34
Sum of absolute values: 610.34
Average absolute deviation: 25.2698224852
Range: 77

:-)

19 not 21

you have 3 unowns i think it is imposible to solve or you forget smth

Thank you Melody ! 

I alredy cheked it and it is complete answer :P

Yet another approach is simply to use Pythagoras to find the length of each side, then use the cosine rule to find the angles:

.

Nice, like it a lot 

Thanks, guest......here's the same problem without using any substitutions :

x^1/2 - 3x^1/3 = 3x^1/6 - 9   .... note....we can  re-write this  as

x^3/6 - 3x^2/6 - 3x^1/6 + 9 = 0     factor directly

(x^2/6 - 3) ( x^1/6 -  3)  = 0

So either

x  ^2/6 - 3  = 0  →  x^1/3 - 3  = 0  → x^1/3  = 3   cube each side   x = 27

Or

x^1/6 - 3 = 0 → x^1/6  = 3     raise each side to the 6th power  x = 3^6 = 3^3 * 3^3  = 27^2  = 729

Solve for x:
sqrt(x)-3 x^(1/3) = 3 x^(1/6)-9

Subtract 3 x^(1/6)-9 from both sides:
9-3 x^(1/6)-3 x^(1/3)+sqrt(x) = 0

Simplify and substitute y = x^(1/6):
9-3 x^(1/6)-3 x^(1/3)+sqrt(x)  =  9-3 x^(1/6)-3 (x^(1/6))^2+(x^(1/6))^3  =  y^3-3 y^2-3 y+9  =  0:
y^3-3 y^2-3 y+9 = 0

The left hand side factors into a product with two terms:
(y-3) (y^2-3) = 0

Split into two equations:
y-3 = 0 or y^2-3 = 0

Add 3 to both sides:
y = 3 or y^2-3 = 0

Substitute back for y = x^(1/6):
x^(1/6) = 3 or y^2-3 = 0

Raise both sides to the power of six:
x = 729 or y^2-3 = 0

Add 3 to both sides:
x = 729 or y^2 = 3

Take the square root of both sides:
x = 729 or y = sqrt(3) or y = -sqrt(3)

Substitute back for y = x^(1/6):
x = 729 or x^(1/6) = sqrt(3) or y = -sqrt(3)

Raise both sides to the power of six:
x = 729 or x = 27 or y = -sqrt(3)

Substitute back for y = x^(1/6):
x = 729 or x = 27 or x^(1/6) = -sqrt(3)

Raise both sides to the power of six:
Answer: | x = 729                   or             x = 27

I honestly don't know! How "mach" is your horrid grammar?

???

It's pirsonal why I shouldn't tell you

Ur welcome! What's ur name guest?

Thanks!

Show off... Nice job!

Easy!

2*1÷4

2÷4

.5=1/2

2*1÷4

2*.25

.5=1/2

2*1/4=1/2

Maybe to u... I just thought, "dumbass" :)

If you are trying to solve a right-angle triangle with Hypotenuse being 10 and one other side being 8, then you can use Pythogaras's theorem to do that as follows:

10^=8^2 + c^2

100=64 + c^2

c^2=100 - 64

c^2=36 take the square root of both sides,

c=6-being the 3rd side.

That means 2x1/4, right?

Thanks, Coldplay....although......it was little obvious, wasn't it???

LOL! Funny CPhil!!!

 Curt(3)+Curt(3) 3rdrt(3)+3rdrt(3) 3+3=6 6thrt(3) 

Thanks, heureka......here's another method that is vey similar to its 2D counterpart....

c (1,1,1) a (1,-4,2) b (-5,2,7)

Let vector ca =  < 0, -5, 1>

Let vector cb =  < -6, 1, 6 >

Find the length of ca =  sqrt26]

Find the length of cb =sqrt [73]

Find the dot product of both vectors = [0*-6 + -5*1 + 1*6]   = [1]

And the measure of angle acb can be found as follows :

cos[acb] = dot product of both vectors / [ product of their lengths]   .......so we have

cos[acb]  = 1 / sqrt[73*26]

cos^-1 [ 1/sqrt[73*26]  = acb ≈ 88.684736791499°

Similarly 

Let vector ba = < 6,-6,-5>     and its length = sqrt(97)

Let vector bc = < 6, -1, -6>     and it's length = sqrt(73)

And the dot probuct of both vectors = [36 + 6 + 30]  =[72]

So

cos(abc)  = 72/ sqrt(97*73)

cos^-1 [ 72/ sqrt(97*73)]  = abc  ≈ 31.17077106171°

Lastly

Let vector ab  = < -6, 6, 5>     and its length = sqrt(97)

Let vector ac = <0, 5, -1>     and it's length  = sqrt(26)

And the dot product of the two vectors = [0 + 30 - 5 ]  = [25]

So

cos(bac)  = 25 / sqrt(97*26) 

cos^-1 [ 25 / sqrt(97*26) ] = bac ≈ 60.144492146791°

Just as heureka found......

Here's the easy way: length*width

now the hard way: take off both triangles and put them together into a square and put it together with the rectangle to get it as length times width