Multiply the number by 100 and you have a percent. As a fraction, you already wrote it (although you missed the exponent in the denominator): .
Okay, this might be pretty basic, but I am in the process of approximating the percent error in computing the area of a square. One of the square's side is 19 cm and has a possible error of 0.05 cm.
I got to where dA = +- 1.9cm^{2}
Now, I got dA/A = 1.9/19 which equals, about .0052631579
With that being said, I have gone completely blank and have no idea on how to turn that decimal into a fraction of a percent. Thank you for your help!
If my math is too sloppy, please explain how to make it look better(actually showing numerator on top of denominator, etc)
So each side could be as small as 19-.05= 18.95 cm or as large as 19+ .05= 19.05 cm. In the first case the area is square centimeters. In the second case the area is square centimeters. We can write that approximately as square cm. We could also use a linear approximation from the start by taking square centimeters. The "percent error" (also called "relative error" is .
I got to where dA = +- 1.9cm^{2}
Now, I got dA/A = 1.9/19 which equals, about .0052631579
With that being said, I have gone completely blank and have no idea on how to turn that decimal into a fraction of a percent. Thank you for your help!
If my math is too sloppy, please explain how to make it look better(actually showing numerator on top of denominator, etc)
When you say the area of a square is x*y with the assumption that x=y, You will have two errors to consider for the area. Your question shows only one error dx. To find the error on the area, you have to find first the fractional error on x. this is dx/x. Since this error is repeated twice, then the error on the area is 2*(dx/x) =2*(0.05/19) = 0.1/19 = 0.053 ---> 5.3%
Yap. Just as I said. Take a look at "An introduction to Error Analysis", John Taylor. I taught that book in physics lab some years ago.
If the errors dx and dy (delta x, delta y)are independent, then the error on the product xy would be sqrt((dx)^2 + (dy)^2). this one will gve about 0.7% error on the area. There are seral ways of calculating the errors. The error of the problem as asked is just what I said.