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This document contains several math problems involving scientific notation. It asks the student to: 1) Write numbers in scientific notation 2) Perform calculations with numbers in scientific notation such as addition, subtraction, multiplication, and division. 3) Calculate errors from approximations and truncations of decimal numbers.

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dharamshi chopda
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157 views2 pages

SC

This document contains several math problems involving scientific notation. It asks the student to: 1) Write numbers in scientific notation 2) Perform calculations with numbers in scientific notation such as addition, subtraction, multiplication, and division. 3) Calculate errors from approximations and truncations of decimal numbers.

Uploaded by

dharamshi chopda
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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Q.

1 Write the following numbers in scientific notation:


(1.)0.00000834
0.834×10 ^-6 expo form
0.834E−06 Std form
(2.) 25.45879
0.2546×10^2 expo form
0.2546 E02 Std form
(3.) 3400000
0.3400000×10 ^7 expo form
0.3400 E07 Std form
(4.) 33
0.33×10 2 expo form
0.33 E02 Std form
(5.)2,300,000,000
0.2300×10 ^10 expo form
0.2300 E10 Std form

Q-2 Let x = 3.141592653589793 is the value of the constant ratio π correct to 15 decimal places
and x ∗ = 3.14159265 be an approximation of x. Compute the following quantities:
approximate value of x a=3.14159265
(a.)The error
error = |x−¿ x4|
error= |3.1415926589793−3.14159265 |
error= 0.000000035979
(b.) The absolute error
Absolute error =|x−x |
Ae=|3.1415926589793 −3.1415926 |
Ae = 0.00000000359779
(c.) The relative error
Relative error = absolute error/x
Re=0.00000000358979/3.1415926589793
Re=0.000000001142666
Q-3 Approximate the following decimal numbers to three digits by using rounding and chopping
(truncation) rules:
(1.)x1 = 1.34579
X = 1.34579 r.o→x=1.346 T→x=1.345
(2.) x2 = 1.34679
X = 1.34679 r.o→x=1.347 T→x=1.346
(3). x3 = 1.34479
X = 1.34479 r.o→x=1.345 T→x=1.3411
(4.) x4 = 3.34379
X = 3.34379 r.o→x=3.344 T→x=3.343
(5.) x5 = 2.34579
X = 2.34579 r.o→x=2.346 T→x=2.345

Q-4 Example on decimal value given in scientific notation:


a)3.25 x 10^ 3+ 2.63 x 10 ^-1
3.25×10 ^3 + 0.000263×10^-3
0.250263× 10^3
0.3250E04
b) 3.25 x 10^ 3- 2.63 x 10 ^-1
3.25 × 10^3 −¿ 0.000263 × 10^3
3.249737 × 10^3
0.3250 or 0.32497E04
Q-5 (i)Add 0.5467E2 to 0.7253E2.
0.5467 E02 + 0.7253 E02
1.272 E02
0.1237 E02
(ii) Subtract 0.7253E2 from 0.5467E5.
0.7253 E02 – 0.5467 E05
0.1786 E02

Q-6 If y = 𝑥 find the error in y for (i) x = 3 and (ii) x = 5 given the error in x as 0.05
using differential calculation.
error x1 = |xa - x|
= |1.7321 – 0.05|
Error x1 = 1.6821
Error x2 = |xb- x|
=|2.2361 – 0.05|
Error x2 = 2.1861
Q-7 (i) Let x = 0.005998. Find the relative error if x is truncated to 3 decimal digits.
x= 0.005098 xa = 0.005
relative error = |x-xa| |x
relative error = |0.005998 – 0.055| 0.005998
relative error = 0.166
(ii) Let x = 0.00458529. Find the absolute error if x is rounded off to 3 decimal digits.
Absolute error = |x - xa|
Absolute error = 0.0004
Q-8 (i) Divide 0.8888E5 by 0.2000E3
0.8888 / 0.2000 × E05 -E03
4.444 E02
0.4444 E03
(ii) Multiply 0.1234E5 by 0.1111E3
0.1234 ×0.1111 × E05 + E03
0.13709 E08
0.1317 E07
(iii) Subtract 0.7200E – 99 from 0.7734E – 99
0.7734 E-99 –0.7200 E-99
0.0534 E-99
0.5340 E-98
(iv) Add 0.5554E99 to 0.7463E99
0.5554 + 07463 E99
0.5554 E99
0.1302 E100 overflow

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