Revision as of 03:08, 20 January 2024 by Admin (Created page with "'''Answer: C''' <math>P^{g}=P^{n}+P^{e} \quad</math> where <math>P^{e}</math> is the expense loading <math>P^{n}=1,000,000 \frac{A_{50}}{\ddot{a}_{50}}=1,000,000\left(\frac{0.18931}{17.0245}\right)=11,119.86</math> <math>P^{e}=P^{g}-P^{n}=11,800-11,120=680</math> {{soacopyright|2024}}")
Exercise
ABy Admin
Jan 20'24
Answer
Answer: C
[math]P^{g}=P^{n}+P^{e} \quad[/math] where [math]P^{e}[/math] is the expense loading
[math]P^{n}=1,000,000 \frac{A_{50}}{\ddot{a}_{50}}=1,000,000\left(\frac{0.18931}{17.0245}\right)=11,119.86[/math]
[math]P^{e}=P^{g}-P^{n}=11,800-11,120=680[/math]
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