A textbook of Computer Based Numerical and Statiscal Techniques part 4 ppt
A textbook of Computer Based Numerical and Statiscal Techniques part 4 ppt
... 32. In a ∆ABC, b = 9.5 cm, c = 8.5 cm and A = 45
o
, find allowable errors in b, c, and
A such that the area of ∆ABC may be determined nearest to a square centimeter.
Sol. Let area of the ∆ABC be ... standards:
IEEE 7 54 is for binary arithmetic, and IEEE 8 54 covers decimal arithmetic as well. The only IEEE
7 54, adopted almost universally by computer manufacturers. U...
A textbook of Computer Based Numerical and Statiscal Techniques part 1 ppt
... This page
intentionally left
blank
This page
intentionally left
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A TEXTBOOK OF
COMPUTER BASED NUMERICAL
AND
STATISTICAL TECHNIQUES
Anju Khandelwal
M.Sc., Ph.D.
Department of Mathematics
SRMS ... Based Numerical and Statistical Techniques is primarily written
according to the unified syllabus of Mathematics for B. Tech. II year and M.C .A. I year students
of...
A textbook of Computer Based Numerical and Statiscal Techniques part 6 pptx
... negative.
Thus f(0.5) is negative and f(1) is positive. Then the root lies between 0.5 and 1.
40
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
Second approximation: The second approximation to the root ... (approximation) a and b (where (a > b)) such that
f (a) . f(b) < 0.
Step 2: Evaluate the mid point x
1
of a and b given by x
1
=
1
2
(a + b) and...
A textbook of Computer Based Numerical and Statiscal Techniques part 10 ppt
... x = 2 [Ans. 1.756]
(c) tan x = x [Ans. 4. 49 34]
84
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
x
2
= 1 –
10
120
−
−−
(–1) = 1 – 0. 047 62 = 0.95238
Hence, f(x
2
) = 0.138 24
Now ... f(x
5
) = – 0.00002
ALGEBRAIC AND TRANSCENDENTAL EQUATION
77
Since, imaginary roots occur in conjugate pairs roots are 1.915 ± 1.908i upto 3 places of
decimal. Assuming other...
A textbook of Computer Based Numerical and Statiscal Techniques part 19 pptx
... marks obtained by 49 2 candidates in a certain examination:
040 4 045 4550505555606065
. 21 043 547 43279
Marks
No of Candidates
−−−−−−
Find out
(a) No. of candidates, if they secure more than 48 but ... the argument half way
between the arguments at q and r is
+
B
A,
24
where A is the arithmetic mean of q and r and B is arithmetic
mean of 3q – 2p – s and 3r – 2s...
A textbook of Computer Based Numerical and Statiscal Techniques part 20 pptx
... following table:
11
44 5 5 6
22
172.2903 162.888 153.7 245 145 .3375 137. 648 3
Rate
Annuity Value
178
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
Here, we have h = 20, a = 99
∴ u =
70 99
1 .45
20
−
=−
Now ... of candidates who obtained marks between certain limits
are as follows:
Marks
No of candidates
−−−−−019203 940 5960798099
. 41 62 65 50 17
Find no. of can...
A textbook of Computer Based Numerical and Statiscal Techniques part 27 ppt
... result.
Example 7. The following are the mean temperatures (°F) on three days, 30 days apart round the pds.
of summer and winter. Estimate the app. dates and values of max. and min. temperature.
INTERPOLATION ... of 20th Jan. and
its value can be obtained similarly.
[f(x)]
min
= [f(x)]
1.1 84
= 63. 647 °F approximately.
Example 8. The mode of a certain frequency Curve y =...
A textbook of Computer Based Numerical and Statiscal Techniques part 30 pptx
... dxa xdxa xdxa xdx
1111
3
5/2 2 4
012
0000
x dxa xdxa xdxa xdx=++
∫∫∫∫
or Simplifying above equations, we get
12
0
012
0
12
2
233
2
2 345
2
345 7
aa
a
a
aa
aaa
++=
++ =
++=
2 84
COMPUTER BASED NUMERICAL ... both
dy
dx
and the curvature
2
2
dy
dx
are the same for the pair of cubics that join at each
point. The spline have possess the given properties.
278
COMPUTER BASED...
A textbook of Computer Based Numerical and Statiscal Techniques part 35 pptx
...
x
+−+
Tabulating from
4x =
to
4. 4x =
we obtain
(4. 3) 1.0 142 56y =
and
(4. 4) 1.018701y =
. Ans.
Example 5. Solve the equation
21,yxy
′
=+
given that
0,y =
at
0,x =
by the use of Taylor ... fields of Engineering and Science, we come across physical and natural phenomena which,
when represented by mathematical models, happen to be differential equations. For examp...
A textbook of Computer Based Numerical and Statiscal Techniques part 40 pptx
... = 54. 8
376
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
largest coefficient of y. We continue this process till last equation. This procedure is known as
partial pivoting. In general, ...
4. 4x =
and
0
0
z =
for
1
y
and we obtain
()
1
1
51 8.8 0 4. 22
10
y
=−−=
Similarly, we obtain
()
1
1
61 4. 4 2 4. 22 4. 816
10
z
=−−×=
Now for second approximation, we o...
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