$$w_a = \frac{0.2312-123.1}{1.52}$$
My professor told me to use Taylor series/the formula for the propagation of the error: $f(x)-f(x_a) = f'(x_a)*(x-x_a)$ but I am not sure know how to apply it here.
I did:
$$w = f(x,y,z) = \frac{x-y}{z}$$
$$|E(x_a)| \le 0.5*10^{-4} \\ |E(y_a)| \le 0.5 * 10^{-1} \\ |E(z_a)| \le 0.5*10^{-2} \\ f_x = 1/z\\ f_y = -1/z \\ f_z = -\frac{x-y}{z^2}\\ |E(w_a)| = f_x(x_a)E(x_a)+f_yE(y_a)+f_z(z_a)E(z_a) \Leftrightarrow \\ |E(w_a)| \le (\frac{1}{1.52})(0.5*10^{-4})+ (-\frac{1}{1.52})(0.5*10^{-1})+ (-\frac{0.2312-123.1}{(1.52)^2})(0.5*10^{-2}) = .233$$
Looks wrong. Help?