The draw back would be that many crimes would go unpunished. Some victims could be to scared, some would prefer to forget about the crime and focus on other things in their lives.
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.

Evaluate using a calculator:

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:

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


c.
If:

Then:

Evaluate at t = 20:


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:

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>
Hello!
Here is your answer:
Yes, it's correct, it is Googl.
I hope I could help!
Answer:
b. Increased frequency of birth defects in fish populations in the pond
Explanation: