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poizon [28]
4 years ago
6

Convert the decimal to a fraction. 0.6 repeating

Mathematics
2 answers:
NikAS [45]4 years ago
6 0

Answer: 2/3

Step-by-step explanation:

liraira [26]4 years ago
4 0
2/3






0.666... is equivalent to 2/3
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Suppose a random variable X has the following probability density function:f(x)=1/x, 1≤x≤C, or f(x)=0 otherwise a) what must the
Sergeeva-Olga [200]

Answer:

a)  C=e^1=e

b) P(X

c) E(X) =\int_{1}^e x \frac{1}{x} dx =x \Big|_1^e \ =e-1

E(X^2) =\int_{1}^e x^2 \frac{1}{x} dx =\frac{x^2}{2} \Big|_1^e \ =\frac{1}{2}(e^2 -1)

Var(X)=E(X^2)-[E(X)]^2= \frac{1}{2}(e^2 -1) -(e-1)^2 = 0.242

Step-by-step explanation:

a) what must the value of C be so that f(x) is a probability density function?

In order to be a probability function we need this condition:

\int_{1}^C \frac{1}{x} dx =1

And solving the left part of the integral we have:

ln(x) \Big|_1^C \ =1

ln(C)-ln(1)=1, so then C=e^1=e

b) find P(X<2)

We can find this probability on this way using the density function:

P(X

c) find E(X) and Var(X)

We can find the expected value on this way:

E(X) =\int_{1}^e x \frac{1}{x} dx =x \Big|_1^e \ =e-1

In order to find the Var(X) we need to find the second moment given by:

E(X^2) =\int_{1}^e x^2 \frac{1}{x} dx =\frac{x^2}{2} \Big|_1^e \ =\frac{1}{2}(e^2 -1)

And now we can use the following definition:

Var(X)=E(X^2)-[E(X)]^2= \frac{1}{2}(e^2 -1) -(e-1)^2 = 0.242

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3 years ago
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Answer:

d

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