Casio CFX-9850G PLUS Troubleshooting guide

Program Library
1Prime Factor Analysis
2Greatest Common Measure
3t-Test Value
4Circle and Tangents
5Rotating a Figure
Before using the Program Library
•Be sure to check how many bytes of unused memory is remain-
ing before attempting to perform any programming.
•This Program Library is divided into two sections: a numeric
calculation section and a graphics section. Programs in the
numeric calculation section produce results only, while graphics
programs use the entire display area for graphing. Also note
that calculations within graphics programs do not use the
multiplication sign (×) wherever it can be dropped (i.e. in front of
open parenthesis).
Chapter
22

412
PROGRAM SHEET
Program for Prime Factor Analysis
Description
Produces prime factors of arbitrary positive integers
For 1 < m< 1010
Prime numbers are produced from the lowest value first. “END” is displayed at the
end of the program.
(Overview)
mis divided by 2 and by all successive odd numbers (d= 3, 5, 7, 9, 11, 13, ....) to
check for divisibility.
Where dis a prime factor, mi= mi–1/dis assumed, and division is repeated until
mi+ 1 <d.
Example [1]
119 = 7 ×17
[2]
440730 = 2 ×3 ×3 ×5 ×59 ×83
[3]
262701 = 3 ×3 ×17 ×17 ×101
Preparation and operation
•Store the program written on the next page.
•Execute the program as shown below.
Step Key operation Display Step Key operation Display
No. 1

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P
Lbl
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Goto
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Lbl
Lbl
Lbl
R
0
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M
:
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"
2
Frac
B
B
A
B
A
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F
M
^
(
A
I
+
÷
^
^
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A
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2
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A
N
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÷
÷
1
⇒
→
×
÷
D
T
→
2
2
→
Goto
B
B
B
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:
A
=
:
:
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A
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C
8
:
–
→
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Frac
4
=
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2
A
⇒
(
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0
Goto
:
=
Goto
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⇒
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1
1
÷
Goto
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⇒
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B
7
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mi
d
mi +1
Goto
3
)
:
9
→
=
Goto
:
B
0
5
:
⇒
:
No. 1
Line Program
File
name
Memory Contents

414
PROGRAM SHEET
Program for
Greatest Common Measure
Description
Euclidean general division is used to determine the greatest common measure for two
interers aand b.
For |a|, |b| < 109, positive values are taken as < 1010
(Overview)
n0=max (|a|, |b|)
n1= min (|a|, |b|)
nk–2
nk= nk–2 – ––– nk–1
nk–1
k= 2, 3....
If nk= 0, then the greatest common measure (c) will be nk–1.
Example [1] [2] [3]
When a= 238 a= 23345 a= 522952
b= 374 b= 9135 b= 3208137866
↓↓ ↓
c= 34 c= 1015 c= 998
Preparation and operation
•Store the program written on the next page.
•Execute the program as shown below.
Step Key operation Display Step Key operation Display
No. 2

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Lbl
Abs
B
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Lbl
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Lbl
M
1
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<
→
2
=
→
3
N
:
→
A
C
:
0
A
:
"
A
⇒
:
( – )
⇒
:
B
F
A
:
Goto
B
(
Goto
C
^
A
"
Abs
2
→
lnt
3
→
Goto
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?
B
:
A
(
:
B
1
T
→
→
:
A
:
:
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→
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2
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÷
Goto
"
B
)
:
B
:
×
"
B
?
–
→
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a, n0
b, n1
nk
B
)
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→C:
No. 2
Line Program
File
name
Memory Contents

416
t =(x – m)
xn–1
n
Program for t-Test Value
Description
The mean (sample mean) and sample standard deviation can be used to obtain a t-test
value.
x: mean of xdata
xσn–1 : sample standard deviation of xdata
n: number of data items
m: hypothetical population standard deviation (normally
represented by µ , but mis used here because of variable
name limitations)
Example To determine whether the population standard deviation for sample data 55, 54, 51,
55, 53, 53, 54, 52, is 53.
Perform a t-test with a level of significance of 5%.
Preparation and operation
•Store the program written on the next page.
•Execute the program as shown below.
No. 3
Step Key operation Display Step Key operation Display
The above operation produces a t-test value of t(53) = 0.7533708035. According to the t-distribution
table in the next page, a level of significance of 5% and a degree of freedom of 7 (n– 1 = 8 – 1 = 7)
produce a two-sided t-test value of approximately 2.365. Since the calculated t-test value is lower
than the table value, the hypothesis that population mean mequals 53 is accepted.
PROGRAM SHEET

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∞
0.2
3.078
1.886
1.638
1.533
1.476
1.440
1.415
1.397
1.383
1.372
1.341
1.325
1.316
1.310
1.306
1.303
1.301
1.299
1.296
1.292
1.289
1.285
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0.1
6.314
2.920
2.353
2.132
2.015
1.943
1.895
1.860
1.833
1.812
1.753
1.725
1.708
1.697
1.690
1.684
1.679
1.676
1.671
1.664
1.658
1.651
1.645
12.706
4.303
3.182
2.776
2.571
2.447
2.365
2.306
2.262
2.228
2.131
2.086
2.060
2.042
2.030
2.021
2.014
2.009
2.000
1.990
1.980
1.970
1.960
63.657
9.925
5.841
4.604
4.032
3.707
3.499
3.355
3.250
3.169
2.947
2.845
2.787
2.750
2.724
2.704
2.690
2.678
2.660
2.639
2.617
2.596
2.576
0.05 0.01
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T
{
5
l-Var
Lbl
(
"
Goto
5
4
List
0
x
T
0
T
5
,
1
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–
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5
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M
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→
xσn– 1
,
_
_
1
1
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÷
5
n
5
)
,
→
5
T
3
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mt
,53,
P (Probability)
Degree
of Freedom
No. 3
Line Program
•t-distribution table
The values in the top row of the table show the probability (two-sided
probability) that the absolute value of tis greater than the table values
for a given degree of freedom.
Memory Contents
File
name
M:aM
T:aT

418
PROGRAM SHEET
Program for Circle and Tangents
Description
Formula for circle:
x2+ y2= r2
Formula for tangent line passing
through point A (x', y'):
y– y' = m(x– x')
*mrepresents the slope of
the tangent line
With this program, slope mand intercept b(= y' – mx') are obtained for lines drawn from
point A (x', y') and are tangent to a circle with a radius of r. The trace function is used to read
out the coordinates at the points of tangency, and factor zoom is used to enlarge the graph.
Example
To determine mand bfor the following values:
r= 1
x' = 3
y' = 2
Notes
•The point plotted for A cannot be moved. Even if it is moved on the graph, the calculation is
performed using the original value.
•An error occurs when r= x'.
•Be sure to always perform a trace operation whenever you select trace and the message
TRACE is on the display.
Preparation and operation
•Store the program written on the next page.
•Execute the program as shown below.
No. 4
Memory Contents
0X
Y
A
(x',y')
r
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T
Prog
"
R
Prog
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Plot
R
(
Lbl
Graph Y=
"
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M
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(–)
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:
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Goto
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)
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→
+
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x2
^
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1
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^
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1
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x2
"
–
R
B
A
Z
Goto
0
(
B
A
Z
Goto
5
→
x2
–
)
A
→
x2
P
)
_
x–1
_
_
:
x2–
Factor
A
F
x2
_
)
→
x–1
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→
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N_
No. 4
Line Program
File
name

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View
Window
35
36
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1
2
1
2
Prog
S
Graph Y=
Goto
Lbl
Graph Y=
Prog
:
Lbl
"
W
1
C
Graph Y=
Graph Y=
"
=
N
3
9
M
"
Goto
3
E
I
(–)
,
I
(–)
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2
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(
W
6
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N
N
6
1
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(
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⇒
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X
I
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D
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C
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(
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–
–
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)
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)
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1
=
+
"
,
1
B
C
(–)
⇒
_
I
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Goto
R
.
9
C
1
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L
,
E
3
"
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No. 4
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File
name
File
name