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valentina_108 [34]

4 years ago

14

Can all one-variable inequalities be simplified to a two-step inequality?

1 answer:

iren2701 [21]4 years ago

6 0

Next , solve as you would solve any two -step inequality. Since 2 is added to , you can get by itself on one side of the inequality by subtracting 2 from both sides. Then, to get by itself on one side of the inequality, you need to multiply both sides by 2.D

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3 years ago

Find the cdf F(x) associated with each of the following probability density functions. Sketch the graphs of f(x) and F(x).

wolverine [178]

Answer:

See steps below

Step-by-step explanation:

a)

$\bf f(x)=3(1-x)^2\;(0$

$\bf F(x)=\int_{0}^{x}f(t)dt=\int_{0}^{x}3(1-t)^2dt=3\int_{0}^{x}(1-t)^2=1-(1-x)^3$

The cdf associated with f is

$\bf \boxed{F(x)=1-(1-x)^3}$ for 0<x<1

<h3>See picture 1 </h3>

The median is a point x such that

F(x) = ½

so, the median is

$\bf 1-(1-x)^3=1/2\rightarrow (1-x)^3=1/2\rightarrow \boxed{x=1-\sqrt[3]{2}}$

The 25th percentile equals the 1st quartile and is a point x such

F(x) = ¼

and the 25th percentile is

$\bf 1-(1-x)^3=1/4\rightarrow (1-x)^3=3/4\rightarrow \boxed{x=1-\sqrt[3]{3/4}}$

b)

$\bf f(x)=\frac{1}{x^2}\;(1$

$\bf F(x)=\int_{1}^{x}f(t)dt=\int_{1}^{x}\frac{dt}{t^2}=1-\frac{1}{x}$

The cdf associated with f is

$\bf \boxed{F(x)=1-\frac{1}{x}}$ for x>1

<h3>See picture 2 </h3>

The median is

$\bf 1-\frac{1}{x}=\frac{1}{2}\rightarrow \boxed{x=2}$

The 25th percentile is

$\bf 1-\frac{1}{x}=\frac{1}{4}\rightarrow \boxed{x=4/3}$

c)

f(x) = 1/3 for 0<x<1 or 2<x<4

$\bf F(x)=\int_{0}^{x}\frac{dt}{3}=\frac{x}{3}\;(0$

$\bf F(x)=\frac{1}{3}+\int_{2}^{x}\frac{dt}{3}=\frac{1}{3}+\frac{x-2}{3}=\frac{x-1}{3}\;(2$

The cdf associated with f is

$\bf F(x)=\frac{x}{3}$ for 0<x<1

$\bf F(x)=\frac{x-1}{3}$ for 2<x<4

<h3>See picture 3 </h3>

The median is

$\bf \frac{x-1}{3}=1/2\rightarrow x=1+3/2\rightarrow \boxed{x=5/2}$

The 25th percentile is

$\bf \frac{x}{3}=1/4\rightarrow \boxed{x=3/4}$

4 0

4 years ago