If r and y are floating-point numbers, then the evaluation of f(x,y) = may be very inaccurate due to cancellation. For example, with base b = 10, precision k = 4 idealized chopping arithmetic, ifl 12.34 and y = 0.9555, then the following results are obtained: A(-) = fl(152.2756) = 152.2 or 0.1522x 103. fl(y*) =A(0.91298025) = 0.9129 A(2,2 + y2)-A(152.2 + 0.9129) = A(153.1 129) = 153.1 A(Vz? + y*) =A(V153.1) =A( 12.37335 ) = 12.37 f (v/r2 + y2 + x-A( 12.37-12.34) = A(0.03) = 0.03 fl(f(z, y)-A(yN12 + y2 + -A(0.9555/0.03) = A(31.85) = 31.85 However, the correct value of f(x, y) is 25.868066..., so the floating-point approximation has a large relative error (around 23%). Note that f(x,y) can be rewritten as g(x, y) (a) Using base b = 10, precision k = 4, idealized chopping arithmetic, =-12.34 and y = 0.9555, evaluate A(g(x,y)) and determine the relative error (b) For each of the specified data in the table below, place an X in the appropriate box to indicate which of the formulas f(z,y) or g(x,y) is more accurate in precision k = 4 floating-point arithmetic, or if they are both accurate. Put exactly one X in each row of the table. (No justification for your answers is required. It is NOT necessary to do any floating-point computation to answer this question.) data f(x, y) more accurateg(, y) more accurate both accurate 0.1234. y = 12.34 = 12.34, y =-0.9123 T =-0.1234. y =-0.005678