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VARVARA [1.3K]
3 years ago
11

Given an exponential function for compounding interest, A(x) = P(.95)x, what is the rate of change?

Mathematics
2 answers:
d1i1m1o1n [39]3 years ago
5 0

The answer should be -5% I hope this helped

statuscvo [17]3 years ago
4 0

Answer:

Option 3

The rate of change is -5%                

Step-by-step explanation:

Given : An exponential function for compounding interest, A(x) = P(.95)^x

To find : What is the rate of change?

Solution :

The general form of an exponential function is:

f(x) = a(1+r)^x

Where, a is the initial amount,

(1+r) is the rate of change,

r is the growth or decay factor

We have given, A(x) = P(.95)^x

Rate of change is

1+r=0.95

r=0.95-1

r=-0.05

Convert to percent,

r=-0.05\times 100=-5\%

Therefore, Option 3 is correct.

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The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per
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Answer:

1.76% probability that in one hour more than 5 clients arrive

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per hour.

This means that \mu = 2

What is the probability that in one hour more than 5 clients arrive

Either 5 or less clients arrive, or more than 5 do. The sum of the probabilities of these events is decimal 1. So

P(X \leq 5) + P(X > 5) = 1

We want P(X > 5). So

P(X > 5) = 1 - P(X \leq 5)

In which

P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353

P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707

P(X = 2) = \frac{e^{-2}*2^{2}}{(2)!} = 0.2707

P(X = 3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1804

P(X = 4) = \frac{e^{-2}*2^{4}}{(4)!} = 0.0902

P(X = 5) = \frac{e^{-2}*2^{5}}{(5)!} = 0.0361

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P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9824 = 0.0176

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