Q1: Detecting and removing errors from the data given 012345678910 0 50 100 150 200 250 f(x) = 1.19 x³ − 11.09 x² + 27.8 x R² = 0.99 Plot with error My DataCurve Error = (1-0.9895) x 100% = 1.05%
012345678910 0 50 100 150 200 250 f(x) = x³ − 8.05 x² + 13.24 x + 18.89 R² = 1 Plot with error removed My DataCurve Error = (1-1) x 100% = 0%
Q2: Plot fori=32+50sin(314.2t)+20sin(628.4t−π 2) The expected response was a sinusoidal wave centered at 32. However, due two two different frequency components with different magnitudes, the current results in the above waveform. The double frequency component lags by 180owhich is the reason why the frequency components do not superimpose each other in every half a cycle. Q3: High-Pass filter magnitude and phase response The transfer function of an analog high-pass filter is: sRC 1+sRC=s s+1 RC
Q4: The z-score is defined as:z=x−~x σ In this case,σ=0.9, x=13.5 (a)~x< 12.0A,z=13.5−12 0.9=1.67 The p-value for z < 1.67 is: 0.4525 (b)~xbetween 12.5A and 13.0A z1=13.5−12.5 0.9=1.11 z2=13.5−13 0.9=0.56 P = 0.3643 - 0.2123 = 0.1520 (c)~x> 14 z=13.5−14 0.9=−0.56
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