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ANEK [815]
2 years ago
11

A function f gives the number of stray cats in a town t years since the town started

Mathematics
1 answer:
Bingel [31]2 years ago
6 0

Answer:

  f(0) = 243 . . . cats at the start

Step-by-step explanation:

The exponential function tells you the number of cats after t years is ...

  f(t) = 243(1/3)^t

When t=0, the exponential factor is 1, so the value of the function is the multiplying constant.

  f(0) = 243

This means there were 243 stray cats in town when the animal control program started.

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Answer:

a. E(x) = 3.730

b. c = 3.8475

c. 0.4308

Step-by-step explanation:

a.

Given

0 x < 3

F(x) = (x-3)/1.13, 3 < x < 4.13

1 x > 4.13

Calculating E(x)

First, we'll calculate the pdf, f(x).

f(x) is the derivative of F(x)

So, if F(x) = (x-3)/1.13

f(x) = F'(x) = 1/1.13, 3 < x < 4.13

E(x) is the integral of xf(x)

xf(x) = x * 1/1.3 = x/1.3

Integrating x/1.3

E(x) = x²/(2*1.13)

E(x) = x²/2.26 , 3 < x < 4.13

E(x) = (4.13²-3²)/2.16

E(x) = 3.730046296296296

E(x) = 3.730 (approximated)

b.

What is the value c such that P(X < c) = 0.75

First, we'll solve F(c)

F(c) = P(x<c)

F(c) = (c-3)/1.13= 0.75

c - 3 = 1.13 * 0.75

c - 3 = 0.8475

c = 3 + 0.8475

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c.

What is the probability that X falls within 0.28 minutes of its mean?

Here we'll solve for

P(3.73 - 0.28 < X < 3.73 + 0.28)

= F(3.73 + 0.28) - F(3.73 + 0.28)

= 2*0.28/1.3 = 0.430769

= 0.4308 -- Approximated

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