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Tomtit [17]
3 years ago
8

If sales of a new kind of hydrogen car really take off in test markets where it is introduced, what macroeconomic

Advanced Placement (AP)
1 answer:
Veseljchak [2.6K]3 years ago
4 0

Answer:

  • large-scale investment in hydrogen car infrastructure  
  • increased funding for research and development into advancing hydrogen car technology

Explanation:

If test markets show that hydrogen cars can be successful, the market will try to take advantage of that and that would include investing heavily into the production of hydrogen cars.

There will also be an increase in the funding for research in hydrogen cars because investors would like to find a way to have an edge over competitors seeing as hydrogen cars are about to increase in production.

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In the context of the passage, "coarse" (sentence 1) is best interpreted as
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Explanation:

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Is it important for you that your career is aligned with your values? Why or why not? Give an example that demonstrates your ans
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3 years ago
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!

a.

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}}

b.

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}}

c.

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.

d.

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>

8 0
2 years ago
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