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revAdmin (Created page with "'''Solution: B''' <math display="block"> 260=\frac{100}{(1-d / 4)^8} e^{\int_2^5 1 /(t+1) d t}=\frac{100}{(1-d / 4)^8} e^{\left.\ln (t+1)\right|_2 ^5}=\frac{100}{(1-d / 4)^8} e^{\ln (6)-\ln (3)}=\frac{100}{(1-d / 4)^8} e^{\ln (6 / 3)}=\frac{100}{(1-d / 4)^8} 2 \text {. } </math> So <math>1-d / 4=(200 / 260)^{.125}</math> so <math>d=4 *\left(1-(200 / 260)^{.125}\right)=0.129</math>. '''References''' {{cite web |url=https://web2.uwindsor.ca/math/hlynka/392oldtests.h...")Nov 26'23 at 17:49+632