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| revprev | Admin | Jan 18'24 at 2:34 | +23 | ||
| revcurprev | Admin | Jan 18'24 at 2:34 | +23 | ||
| revcurprev | Admin | Jan 18'24 at 2:33 | +23 | ||
| revcurprev | Admin | Jan 18'24 at 0:22 | +26 | m | |
| revcur | Admin | (Created page with "'''Answer: B''' Since <math>S_{0}(t)=1-F_{0}(t)=\left(1-\frac{t}{\omega}\right)^{\frac{1}{4}}</math>, we have <math>\ln \left[S_{0}(t)\right]=\frac{1}{4} \ln \left[\frac{\omega-t}{\omega}\right]</math>. Then <math>\mu_{t}=-\frac{d}{d t} \log S_{0}(t)=\frac{1}{4} \frac{1}{\omega-t}</math>, and <math>\mu_{65}=\frac{1}{180}=\frac{1}{4} \frac{1}{\omega-65} \Rightarrow \omega=110</math>. <math>e_{106}=\sum_{t=1}^{3}{ }_{t} p_{106}</math>, since <math>{ }_{4} p_{106}=0</mat...") | Jan 15'24 at 20:09 | +781 |