If R = 10x²y² z² and errors in x,y, z are 0.03,0.01, 0.02 respectively at x=3, y=1, Z=2. Calculate the absolute errer and percentage errer in evaluating R..
Answers
Answer:
Correct option is
B
6 %
Given, A=
y
q
z
r
x
p
Take ln and then differentiate ,
A
ΔA
=±(p
x
Δx
+q
y
Δy
+r
z
Δz
)
Thus the maximum percentage error in A:
A
ΔA
×100=p
x
Δx
×100+q
y
Δy
×100+r
z
Δz
×100=(3×1)+(2×0.5)+(1×2)=6 %
Answer:
Percentage error = 12% and absolute error = 43.2
Step-by-step explanation:
Given :- R = 10x²y²z², error in x, y and z are 0.03, 0.01, 0.02 respectively at x=3, y=1, z=2.
To find :- The absolute error and percentage error in R.
Solution:-
R = 10x²y²z²
The given values are x = 3, y = 1 and z = 2.
∴ R = 10( 3)²( 1² )( 2 )²
= 10( 9 )( 4 )
= 360
The given error in x = 0.03 = 3%, y = 0.01 = 1%, z = 0.02 = 2%
ΔR = 2 Δx + 2 Δy + 2 Δz
R x y z
= 2(3) + 2(1) + 2(2)
= 6 + 2 + 4
= 8 + 4
= 12 %
Therefore, the percentage error in R = 12%.
So, relative error = 12/100 = 0.12
Absolute error = relative error * exact value
= 0.12 × 360
= 43.2
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