Consider the baggage check-in process of a small airline. Check-in data indicate that from 9 a.m. to 10 a.m., 220 passengers che
1 answer:
You might be interested in
Welll the y-intercept is negative 2 so on the y axis the first point is negative 2. 1/4 is the rise over run so you would go up from your first point one unit and over to the left four units or vise versa down one unit and over to the right four units, or the image
The maximum error in the calculated surface area is 24.19cm² and the relative error is 0.0132.
Given that the circumference of a sphere is 76cm and error is 0.5cm.
The formula of the surface area of a sphere is A=4πr².
Differentiate both sides with respect to r and get
dA÷dr=2×4πr
dA÷dr=8πr
dA=8πr×dr
The circumference of a sphere is C=2πr.
From above the find the value of r is
r=C÷(2π)
By using the error in circumference relation to error in radius by:
Differentiate both sides with respect to r as
dr÷dr=dC÷(2πdr)
1=dC÷(2πdr)
dr=dC÷(2π)
The maximum error in surface area is simplified as:
Substitute the value of dr in dA as
dA=8πr×(dC÷(2π))
Cancel π from both numerator and denominator and simplify it
dA=4rdC
Substitute the value of r=C÷(2π) in above and get
dA=4dC×(C÷2π)
dA=(2CdC)÷π
Here, C=76cm and dC=0.5cm.
Substitute this in above as
dA=(2×76×0.5)÷π
dA=76÷π
dA=24.19cm².
Find relative error as the relative error is between the value of the Area and the maximum error, therefore:
As above its found that r=C÷(2π) and r=dC÷(2π).
Substitute this in the above
Hence, the maximum error in the calculated surface area with the circumference of a sphere was measured to be 76 cm with a possible error of 0.5 cm is 24.19cm² and the relative error is 0.0132.
Learn about relative error from here brainly.com/question/13106593
#SPJ4