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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 XXXX for XXXXXXXXXXX XXXXXXXXX XX: A(t) = X * e^(r * t) </p> <p>XX XXXX find XXX values of XXXXX XXXXXXXXX/constants: X, r </p> <p>XXXX question XXXXX XXXXXXX XXXXXXXXXX. The <b>XXXXXXX XXXX, t<sub>0</XXX> = 0</b> (Zero XXXXXXXXXXX XX XXXX), XXX XXX <b>XXXXXXX revenue A(t<XXX>0</sub>) = A(0) = 54.X billion</b>. </p> <p>XX XXX these XXXXXXX conditions to solve XXX XXX XXXXXXXX <b>X</b>: </p> <ol><li>Plug in 0 for t : A(X) <ol><li>We XXX XXX XXXXXXX XXXX X(t) = X * e^(r * t) </li><li> A(<b>X</b>) = X * e^(r * <b>X</b>) </li><li> X(<b>0</b>) = X * e^0 </li><li> A(<b>0</b>) = X * X </li><li> A(<b>0</b>) = X </li></ol></li><li>XX know, from earlier, XXXX A(0) = XX.X billion </li><li>We now XXXX XXX ways XX XXXXXXXXX A(0) <ol><li>A(X) = P </li><li>A(X) = 54.5 </li></ol></li><li>Thus P = 54.5 </li></ol> <p>XXX XXXX we know what X is, all XX XXXX XX find XX r. </p> <p>We start XXXX X(t) =<b> P</b> * e^(r * t). XXXX XXXXXX it XXXX XXX value XX X we XXXX XXXXX. </p> <p> A(t) = <b>(XX.5)</b> * e^(r * t) </p> <p> </p> <p>Now XXXXX XX XXXXXXX condition that is XXXXX: XX XXXX, XXXXX made $XX.X XXXXXXX. XXX t = 2 (XXXXXXXXXXX XX 2015, 2 XXXXX XXXXX 2013), then A(X) = 74.6 XXXXXXX. </p> <p>Lets use this XX solve XXX r: </p> <ol><li>XXXX in X for t : X(2) <ol><li>We XXX X(t) = (54.X) * e^(r * t) </li><li> X(<b>X</b>) = (XX.X) * e^(r * <b>2</b>) </li><li> X(2) = (XX.5) * e^(XX) </li></ol></li><li>XX XXXX know XXXX earlier XXXX X(2) = XX.X billion </li><li>XX XXX know XXX XXXX XX XXXXXXXXX X(2) <ol><li>X(2) = (54.5) * e^(2r) </li><li>X(2) = XX.6 XXXXXXX </li></ol></li><li>XXXX 74.X = (XX.5) * e^(2r) </li><li>XXXXX for r <ol><li>74.6 = (XX.X) * e^(2r) </li><li>/54.X = /XX.5 </li><li>(XX.X/54.5) = e^(2r) </li><li>ln( 74.6/XX.X ) = ln( e^(2r) ) </li><li>ln( 74.6/54.X ) = 2r </li><li>/2 = /X </li><li>ln( 74.6/54.X )/X = r </li></ol></li><li>Thus r = ln( XX.6/XX.5 )/X = X.15696990277 </li></ol> <p>XXXX constant<b> r,</b> represents XXX continuous XXXXXX rate XXX XXXX of XXXXXXX. We XXXXXXX XX XXXXXXXXXX, 15.XXXXXXXXX % and then round to XXXXX XXXXXXX places: 15.X%. </p> <p>XXXX we have <b>A(t) = XX.X e^(0.157 t)</b>. </p> <p>Please XXX XX know XX you have any other XXXXXXXXX 😊 XXXX XXXX! </p> <XX><li>-Connor XXXXX </li></XX> <p> </p>">
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