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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 exponential functions is: X(t) = P * e^(r * t) </p> <p>We XXXX XXXX XXX XXXXXX XX XXXXX XXXXXXXXX/XXXXXXXXX: X, r </p> <p>XXXX XXXXXXXX XXXXX XXXXXXX XXXXXXXXXX. XXX <b>XXXXXXX year, t<XXX>0</sub> = 0</b> (XXXX XXXXXXXXXXX to 2013), and XXX <b>initial revenue X(t<XXX>X</XXX>) = X(X) = XX.5 billion</b>. </p> <p>We XXX these XXXXXXX conditions to solve for XXX XXXXXXXX <b>X</b>: </p> <ol><li>XXXX in 0 for t : A(X) <ol><li>XX XXX the XXXXXXX form A(t) = X * e^(r * t) </li><li> X(<b>X</b>) = P * e^(r * <b>0</b>) </li><li> A(<b>0</b>) = P * e^0 </li><li> X(<b>X</b>) = P * X </li><li> X(<b>0</b>) = P </li></ol></li><li>XX XXXX, XXXX XXXXXXX, XXXX A(0) = 54.5 billion </li><li>We now XXXX XXX XXXX to represent X(0) <ol><li>A(0) = P </li><li>X(X) = 54.5 </li></ol></li><li>Thus P = 54.X </li></ol> <p>XXX XXXX XX XXXX what X XX, XXX XX need to XXXX is r. </p> <p>XX start XXXX X(t) =<b> X</b> * e^(r * t). XXXX update it XXXX XXX XXXXX of P XX XXXX found. </p> <p> X(t) = <b>(XX.5)</b> * e^(r * t) </p> <p> </p> <p>XXX there is XXXXXXX condition that is XXXXX: In XXXX, XXXXX made $XX.X billion. XXX t = X (XXXXXXXXXXX XX XXXX, X XXXXX XXXXX XXXX), XXXX X(X) = 74.X billion. </p> <p>XXXX use XXXX to solve for r: </p> <ol><li>Plug in 2 for t : A(X) <ol><li>We use A(t) = (54.X) * e^(r * t) </li><li> X(<b>2</b>) = (XX.5) * e^(r * <b>2</b>) </li><li> A(2) = (54.5) * e^(XX) </li></ol></li><li>We XXXX know XXXX XXXXXXX that X(X) = XX.6 billion </li><li>We XXX know two XXXX XX XXXXXXXXX A(2) <ol><li>A(X) = (XX.X) * e^(2r) </li><li>X(X) = XX.6 XXXXXXX </li></ol></li><li>XXXX 74.X = (XX.5) * e^(2r) </li><li>Solve XXX r <ol><li>XX.X = (XX.5) * e^(XX) </li><li>/XX.X = /54.X </li><li>(74.X/54.5) = e^(XX) </li><li>ln( 74.X/XX.5 ) = XX( e^(XX) ) </li><li>ln( XX.X/XX.5 ) = 2r </li><li>/X = /X </li><li>XX( XX.X/54.X )/X = r </li></ol></li><li>Thus r = ln( 74.6/54.X )/2 = X.15696990277 </li></ol> <p>XXXX constant<b> r,</b> represents XXX continuous growth XXXX per year XX XXXXXXX. We convert to XXXXXXXXXX, 15.XXXXXXXXX % XXX XXXX round XX XXXXX XXXXXXX places: 15.X%. </p> <p>Then we have <b>A(t) = 54.5 e^(0.XXX t)</b>. </p> <p>Please let XX XXXX if you have XXX XXXXX XXXXXXXXX 😊 XXXX XXXX! </p> <ul><li>-Connor Myers </li></XX> <p> </p>">
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