User Manual

m12 0.
12 & 41 k 1.
8 & 13 k 2.
5 & 2 k 3.
23 & 200 k 4.
15 & 71 k 5.
Ra 5.357506761
Rb –3.120289663
R© 0.503334057
x=10→y’=? 10 @y 24.4880159
y=22→x’=? 22 @x 9.63201409
@≠ –3.432772026
@≠ 9.63201409
xy
12 41
813
52
23 200
15 71
k[]
30
m10
k 1.
0.
40 & 2 k 2.
50 k 3.
]]]
45 & 3 k
45.
]
] 60 k
3.
60.
DATA
30
40
40
50
DATA
30
45
45
45
60
↓
Σx = x
1
+ x
2
+ ··· + x
n
Σx
2
= x
1
2
+ x
2
2
+ ··· + x
n
2
x =
Σx
n
Σxy = x
1
y
1
+ x
2
y
2
+ ··· + x
n
y
n
Σy = y
1
+ y
2
+ ··· + y
n
Σy
2
= y
1
2
+ y
2
2
+ ··· + y
n
2
y =
Σy
n
σy =
Σy
2
– ny
2
n
sy =
Σy
2
– ny
2
n – 1
sx =
Σx
2
– nx
2
n – 1
σx =
Σx
2
– nx
2
n
k&~£pnzw^
¢PZWvrab©
xy≠ƒ«»√
m10 0.
95 k 1.
80 k 2.
k 3.
75 & 3 k 4.
50 k 5.
R~ 75.71428571
Rp 12.37179148
Rz 530.
Rw 41’200.
R£ 13.3630621
L= 178.5714286
64.43210706
m11 0.
2 & 5 k 1.
k 2.
12 & 24 k 3.
21 & 40 & 3 k 4.
15 & 25 k 5.
Ra 1.050261097
Rb 1.826044386
Rr 0.995176343
R£ 8.541216597
R¢ 15.67223812
x=3 → y’=? 3 @y 6.528394256
y=46 → x’=? 46 @x 24.61590706
@ƒ 60 @√)= 0.102012
@» 0.5 ±)= 0.691463
DATA
95
80
80
75
75
75
50
x=
σx=
Rn
7. n=
Σx=
Σx
2
=
sx=
sx
2
=
xy
25
25
12 24
21 40
21 40
21 40
15 25
— ×10+50=
(95–x )
sx
( 95 -K~)
/K£ * 10
+ 50 =
X2 =
N2 =
X3 =
x = 60 → P(t) ?
t = –0.5 → R(t) ?
t = ––––
x – x
σx
Standardization conversion formula
Rumus penukaran pemiawaian
CÙng thŸc biân «∞i chuƒn hÛa
m (2-VLE)
m20
2x + 3y = 4 2 ® 3 ® 4 ®
5x + 6y = 7 5 ® 6 ® 7
x = ? ® [x]
–1.
y = ? ® [y] 2.
det(D) = ? ® [det(D)] –3.
m (3-VLE)
m21
x + y – z = 9 1 ® 1 ® 1 ±® 9 ®
6x + 6y – z = 17 6 ® 6 ® 1 ±® 17 ®
14x – 7y + 2z = 42 14 ® 7 ±® 2 ® 42
x = ? ® [x]
3.238095238
y = ? ® [y] –1.638095238
z = ? ® [z] –7.4
det(D) = ? ® [det(D)] 105.
m (QUAD, CUBIC)
m22
3x
2
+ 4x – 95 = 0 3 ® 4 ®± 95
x1 = ? ®
5.
x2 = ? ® –6.333333333
@® 5.
m23
5x
3
+4x
2
+3x +7=0 5 ® 4 ® 3 ® 7
x1 = ? ®
–1.233600307
i
x2 = ? ® 0.216800153
i
@≠
+
1.043018296
i
x3 = ? ® 0.216800153
i
@≠
–
1.043018296
i






a
1
x + b
1
y = c
1
a
2
x + b
2
y = c
2
a
1
b
1
a
2
b
2
D =
a
1
x + b
1
y + c
1
z = d
1
a
2
x + b
2
y + c
2
z = d
2
a
3
x + b
3
y + c
3
z = d
3
a
1
b
1
c
1
a
2
b
2
c
2
a
3
b
3
c
3
D =