La corrupcion en mexico

Solo se ocuparon 8 datos ya que fueron los unicos marcados por el chispeador en la cinta.
N | Distancias (y) en (m) | Tiempos (x) |
1 | .005 | 1/50 = 0.02 |
2 | .014 | 2/50 = 0.04 |
3 | .026 | 3/50 = 0.06 |
4 | .042 | 4/50 = 0.08 |
5 | .061 | 5/50 = 0.1 |
6 | .085 | 6/50 = 0.12 |
7 | .112 | 7/50 = 0.14 |
8 | .144 | 8/50 = 0.16 |
9 | .179 | 9/50 = 0.18 |
10| .218 | 10/50 = 0.2 |
11 | .2671 | 11/50= 0.22 |
∑ | 1.1351 | 1.32 |







(xi ti) | (xi ti2) | ti2 | ti3 | ti4 |
0.0001 | 0.000002 | 0.0002 | 0.000008 | 0.00000016 |
0.00056 | 0.0000224 | 0.0016 | 0.000064 | 0.00000256 |
.00156 | 0.0000936 | 0.0036 | 0.000216 | 0.00001296 |
0.00336 | 0.0002688 | 0.0064 | 0.000512 | 0.00004096 |
0.0061 | 0.00061 | 0.01 | 0.001 |0.0001 |
0.0102 | 0.0001224 | 0.0144 | 0.001728 | 0.00020736 |
0.01568 | 0.0021952 | 0.0196 | 0.002744 | 0.00038416 |
0.02304 | 0.0036864 | 0.0256 | 0.004096 | 0.00065536 |
0.03222 | 0.0057996 | 0.0324 | 0.005832 | 0.00104976 |
0.0436 | .00872 | 0.04 | 0.008 | 0.0016 |
0.0586762 | .01292 | 0.0484 | 0.010648 | 0.00234256 |
0.1950962 | .15079 | .2022 | 0.034848 | 0.00639528 |sistema de ecuaciones
L1=i=1nxi=1.32
L2=i=1nxi ti=.1950962

L3i=1nxi ti2=0.2022.

a11 a0+ a12 a1+ a13 a2= l1
a12 a0+ a22 a1+ a23 a2= l2
a13 a0+ a32 a1+ a33 a2= l3
Por lo tanto tenemos lo siguiente
a11=N ; a12=i=1nti ; a13=i=1nti2
a21=a12; a22=i=1nti2; a23=i=1nti3
a31=a13; a32=a23 ; a33=i=1nti4
Sustituyendo los datos en el sistema deecuaciones y resolviéndolo nos queda:
11 a0 + 1.32 a1 + 0.2022 a2= 1.1351
1.32 a0 + 0.2022 a1 + 0.034848 a2=0.1950962
0.2022 a0 + 0.034848 a1 + 0.0639528 a2=0.15079
Realizado por el método de sustitución tenemos que:
8a0 + 0.72a1 + 0.0816a2 = 0.545
0.72a0 + 0.0816a1 + 0.010368a2 = 0.0675
0.0816a0+ 0.010368a1+0.00140352a2 = 0.00138805
Despejando a0
a0=0.545-0.72a1-0.0816a2 8
Sustituyendo el valor de a0 en las otras ecuaciones:
0.720.545-0.72a1-0.0816a2 8+0.0816a1 + 0.010368a2 = 0.0675
0.08160.545-0.72a1-0.0816a2 8+0.010368a1+ 0.00140352a2 = 0.00138805

Realizando las operaciones para obtener un sistema de dos ecuaciones:
Primera ecuación
0.3924-0.5184a1-0.058752a28 +0.0816a1 + 0.010368a2 =0.0675

0.04905 – 0.0648a1 –0.007344a2+0.0816a1 + 0.010368a2 = 0.0675

0.0168a1 + 0.003024a2 = 0.01845..............................1ª ecuación
Segunda ecuación
0.044472-0.058752a1-0.00665856a28+0.010368a1+ 0.00140352a2 = 0.00138805

0.005559–0.007344 a1 –0.00083232 a2 +0.010368a1+ 0.0014035a2 = 0.00138805

0.003024a1 +0.00057118 a2 =0.00417095 ...................2ª ecuaciónQuedando el sistema de ecuaciones de la siguiente manera:
0.0168a1 + 0.003024a2 = 0.018450.003024a1 +0.00057118 a2 =0.00417095

Resolviendo este sistema encontramos a1 y a2
a1=0.01845-0.003024a20.0168
Sustituyendo
0.0030240.01845-0.003024a20.0168+0.00057118a2 =0.00417095
0.000055792 – 0.000009144a2 +0.00057118a2 =0.00417095
0.000562036a2 = 0.004226742
a2 = 0.004226742 0.000562036
a2=0.752041Sustituyendo en a1:
a1=0.01845-0.003024(0.752041)0.0168
a1=0.01845+0.0022740.0168
a1=0.0207240.0168
a1=0.123357

Sustituyendo nuestros valores de a1 y a2 para encontrar el valor de a0 lo haremos en el despeje que hicimos para a0:

a0=0.545-0.72a1-0.0816a2 8

a0=0.545-0.720.123357-0.0816(0.752041)8
a0=0.545-0.088817-0.0613668
a0=0.3948178
a0=0.04935

a0=0.04935 a1=0.123357a=0.752041

Realizando la transformación con la formula yi=xi/ti

| distancias en (m) Xi | / | tiemposti | transformacion a “yi” | yiti | (ti)2 |
| 0.005 | / | 0.02 | 0.25 | 0.005 | 0.0004 |
| 0.015 | / | 0.04 | 0.375 | 0.015 | 0.0016 |
| 0.027 | / | 0.06 | 0.45 | 0.027 | 0.0036 |
| 0.047 | / | 0.08 | 0.5875 | 0.047 | 0.0064 |
| 0.069 | / | 0.1 | 0.69 | 0.069 | 0.01 |
|...