Question

attempt 1: 10 attempts remaining. the following model represents the total spent on education funding, in millions of dollars, in a state from 2005 to 2015, where the time (t = 0) represents the year 2005: (e(t)=27ln(2692 - 281t)) find the rate at which the total spent on education funding in that state was changing in the year 2013. (round your answer to the nearest integer.) the total spent on education funding was by million dollars per year in the year 2013. submit answer

Explanation:

Step1: Determine the value of t for 2013

Since \(t = 0\) represents 2005, for 2013, \(t=2013 - 2005=8\).

Step2: Differentiate the function \(E(t)\)

We use the chain - rule. If \(y = 27\ln(u)\) where \(u = 2692-281t\), then \(\frac{dy}{du}=\frac{27}{u}\) and \(\frac{du}{dt}=- 281\). By the chain - rule \(\frac{dE}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}\). So \(E^\prime(t)=\frac{27\times(-281)}{2692 - 281t}=\frac{-7587}{2692 - 281t}\).

Step3: Evaluate \(E^\prime(t)\) at \(t = 8\)

Substitute \(t = 8\) into \(E^\prime(t)\): \(E^\prime(8)=\frac{-7587}{2692-281\times8}=\frac{-7587}{2692 - 2248}=\frac{-7587}{444}\approx - 17\).

Answer:

The total spent on education funding was decreasing by 17 million dollars per year in the year 2013.