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3 years ago

How did institutional changes between the 1960s and 1980s change state legislatures abilities to develop public policy?

1 answer:

4 0

The Institutional changes between the 1960s and 1980s shifted the public policy from emphasizing public interest to upholding and satisfying the regulations. Among the greatly affected was the educational system. The structures of the different learning institutions were changed to accommodate dynamic market models and work around the expected outcomes of learning.

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1. The rate at which people enter a movie theater on a given day is modeled by the function S defined by S(t) = 80 -12 cos 6 The

Arlecino [84]

Hi there!

To find the total amount of people that have ENTERED by t = 20, we must take the integral of the appropriate function.

$\text{Amount that entered} = \int\limits^{20}_{10} {S(t)} \, dt \\\\ = \int\limits^{20}_{10} {80 - 12cos(\frac{t}{5})} \, dt$

Evaluate using a calculator:

$= 899.97 \approx \boxed{900\text{ people}}$

To solve, we can find the total amount of people that have entered of the interval and subtract the total amount of people that have left from this value.

In other terms:
$\text{Amount of people} = \int\limits^{20}_{10} {S(t)} \, dt - \int\limits^{20}_{10} {R(t)} \, dt$

We can evaluate using a calculator (math-9 on T1-84):

$\text{\# of people} = \int\limits^{20}_{10} {80-12cos(\frac{t}{5})} \, dt - \int\limits^{20}_{10} {12e^{\frac{t}{10}}+20} \, dt$

$= 899.97 - 760.49 = 139.47 \approx \boxed{139 \text{ people}}$

If:
$P(t) = \int\limits^t_{10} {S(t) - R(t)} \, dt$

Then:

$\frac{dP}{dt} = P'(t)= \frac{d}{dt}\int\limits^t_{10} {S(t) - R(t)} \, dt = S(t) - R(t)$

Evaluate at t = 20:

$S(20) = 80 - 12cos(\frac{20}{5}) = 87.844\\\\R(20) = 12e^{\frac{20}{10}} + 20 = 108.669$

$S(20) - R(20) = 87.844 - 108.669 = -20.823$

This means that at t = 20, there is a <u>NET DECREASE</u> of people at the movie theater of around 20.823 (21) people per hour.

To find the maximum, we must use the first-derivative test.

Set S(t) - R(t) equal to 0:

$80 - 12cos(\frac{t}{5}) - 12e^{\frac{t}{10}} - 20 = 0\\\\60 - 12(cos(\frac{t}{5}) + e^{\frac{t}{10}})= 0$

Graph the function with a graphing calculator and set the function equal to y = 0:

According to the graph, the graph of the first derivative changes from POSITIVE to NEGATIVE at t ≈ 17.78 hours, so there is a MAXIMUM at this value.

<u>Thus, at t = 17.78 hours, the amount of people at the movie theater is a MAXIMUM.</u>

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