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$</span><span class="mn">54.5</span>billionin 2013to<span class="mo">$</span><span class="mn">74.6</span>billion1in 2015. Find an exponential function to model the revenue as a function of years since 2013. What is the continuous percent growth rate, per year, of revenue? Round your answer to three decimal places. </p> <p>A(t)=? </p> <p> </p> <p>The form XXX XXXXXXXXXXX XXXXXXXXX XX: A(t) = X * e^(r * t) </p> <p>XX XXXX XXXX the values of these variables/constants: P, r </p> <p>This XXXXXXXX gives XXXXXXX conditions. XXX <b>XXXXXXX XXXX, t<XXX>0</XXX> = 0</b> (Zero XXXXXXXXXXX to XXXX), and the <b>XXXXXXX revenue X(t<sub>X</sub>) = X(X) = 54.5 billion</b>. </p> <p>XX XXX these initial XXXXXXXXXX to XXXXX XXX XXX XXXXXXXX <b>X</b>: </p> <ol><li>Plug in 0 XXX t : A(X) <ol><li>We use the XXXXXXX form X(t) = X * e^(r * t) </li><li> X(<b>X</b>) = P * e^(r * <b>0</b>) </li><li> X(<b>0</b>) = P * e^X </li><li> A(<b>X</b>) = X * X </li><li> X(<b>X</b>) = X </li></ol></li><li>We XXXX, from earlier, that A(0) = XX.X billion </li><li>We now have two ways to represent A(0) <ol><li>X(X) = X </li><li>A(X) = XX.5 </li></ol></li><li>Thus X = XX.X </li></ol> <p>XXX that XX know what P XX, all XX XXXX XX find XX r. </p> <p>We XXXXX with X(t) =<b> P</b> * e^(r * t). XXXX update it with XXX XXXXX XX X we XXXX found. </p> <p> X(t) = <b>(XX.5)</b> * e^(r * t) </p> <p> </p> <p>XXX XXXXX XX another condition XXXX is given: In 2015, XXXXX XXXX $XX.X billion. XXX t = X (corresponds XX XXXX, 2 years XXXXX XXXX), then X(X) = XX.X XXXXXXX. </p> <p>Lets XXX this XX solve XXX r: </p> <ol><li>Plug in X for t : A(X) <ol><li>We use X(t) = (XX.5) * e^(r * t) </li><li> A(<b>2</b>) = (XX.X) * e^(r * <b>X</b>) </li><li> A(2) = (XX.X) * e^(XX) </li></ol></li><li>We XXXX know XXXX XXXXXXX XXXX A(2) = XX.X billion </li><li>XX XXX know XXX ways XX XXXXXXXXX A(X) <ol><li>A(X) = (XX.5) * e^(XX) </li><li>A(2) = XX.6 XXXXXXX </li></ol></li><li>Thus 74.6 = (54.X) * e^(XX) </li><li>XXXXX XXX r <ol><li>74.6 = (XX.5) * e^(XX) </li><li>/54.X = /XX.5 </li><li>(XX.6/54.5) = e^(2r) </li><li>XX( XX.6/54.X ) = XX( e^(2r) ) </li><li>ln( XX.X/54.5 ) = XX </li><li>/2 = /2 </li><li>XX( XX.X/XX.X )/2 = r </li></ol></li><li>Thus r = ln( XX.6/XX.X )/2 = 0.15696990277 </li></ol> <p>XXXX constant<b> r,</b> XXXXXXXXXX XXX continuous XXXXXX rate XXX XXXX XX XXXXXXX. We convert to percentage, XX.XXXXXXXXX % XXX then XXXXX to XXXXX XXXXXXX XXXXXX: 15.7%. </p> <p>XXXX we have <b>A(t) = 54.X e^(0.157 t)</b>. </p> <p>Please XXX XX XXXX if you XXXX any other XXXXXXXXX 😊 XXXX luck! </p> <ul><li>-Connor Myers </li></XX> <p> </p>">
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