Answer:
<h2>
88859.375 & f(n)= 28000(0.75)×</h2>
Step-by-step explanation:
using the information on the problem a function can can be made
f(n)= 28000(0.75)×
where x is the amount of years
plug in 4 for x in the equation to get
f(4)=8859.375
Recall the Maclaurin expansion for cos(x), valid for all real x :

Then replacing x with √5 x (I'm assuming you mean √5 times x, and not √(5x)) gives

The first 3 terms of the series are

and the general n-th term is as shown in the series.
In case you did mean cos(√(5x)), we would instead end up with

which amounts to replacing the x with √x in the expansion of cos(√5 x) :

Using the Poisson distribution, there is a 0.8335 = 83.35% probability that 2 or fewer will be stolen.
<h3>What is the Poisson distribution?</h3>
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:

The parameters are:
- x is the number of successes
- e = 2.71828 is the Euler number
is the mean in the given interval.
The probability that a rental car will be stolen is 0.0004, hence, for 3500 cars, the mean is:

The probability that 2 or fewer cars will be stolen is:

In which:




Then:

0.8335 = 83.35% probability that 2 or fewer will be stolen.
More can be learned about the Poisson distribution at brainly.com/question/13971530
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