Mathematics: HELP!! REAL answers only please.

HELP!! REAL answers only please. - id7409390, sdonachy7

Ответ:

emilypzamora11

01.02.2022

Step-by-step explanation:

So the angle of 40 is a vertical angle with the one opposite of it, meaning that the triangle on the left would have two values of 40 and 41, to find the value of the 3rd angle of the left triangle, you can subtract 40 and 41 from 180 which is the total number of degrees in a triangle to get 99, then you can find the degrees of the triangle on the right's bottom angle by subtracting 99 from 180, meaning the bottom left angle would be 81, then you can do the same to find the other angle by subtracting 114 from 180 to get 66. Lastly you can add 81 and 66 to get 147, then subtracting 147 from 180 you would get x=33

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Ответ:

alishabhappy1

03.01.2022

Mathematics

Answers:

- mean: 2.667
- variance: 0.889

Explanation:

To get the mean and variance of x, we need to verify first whether...
- x is discrete or continuous random variable
- f is probability mass or probability density function

because if we cannot verify the 2 statements above, we can't compute the mean and the variance.

Since 0 < x < 4, x is a continuous random variable because x can be any positive number less than, which includes a non-integer.

Note that if the random variable is continuous and $0 \leq f(x) \leq 1$ for any values of x in the domain of f, then f is a probability density function (PDF).

Note that

$0 \ \textless \ x \ \textless \ 4
\\ \Leftrightarrow 0.125(0) \ \textless \ 0.125x \ \textless \ 0.125(4)
\\ \Leftrightarrow 0 \ \textless \ 0.125x \ \textless \ 0.5
\\ \Leftrightarrow 0 \ \textless \ f(x) \ \textless \ 0.5 
\\ \Rightarrow 0 \ \textless \ f(x) \ \textless \ 1 (\text{Because }0 \ \textless \ f(x) \ \textless \ 0.5 \ \textless \ 1)$

Hence, for any x in the domain of f, 0 < f(x) < 1. Moreover, since x is a continuous random variable, thus f is a PDF.

First, we use the following notations for mean and variance:

E[x] = mean of x
Var[x] = variance of x

Since f is a probability density function, we can use the following formulas for the mean and the variance of x:

$\boxed{\text{mean of }x = E[x] = \int_{-\infty}^{\infty}{xf(x)}dx}$

$\boxed{\text{Variance of }x = \text{Var}[x] = E[x^2] - (E[x])^2}}$

To compute for the mean of x,

$\text{mean of }x = E[x]
\\ = \int_{-\infty}^{\infty}{xf(x)}dx}
\\ = \int_{-\infty}^{\infty}{x(0.125x)}dx}
\\ \boxed{\text{mean of }x = \int_{-\infty}^{\infty}{0.125x^2}dx}}$

The integral seems complicated because of the infinity sign. But because the domain of f is the set of positive numbers less than 4, that is,

$\text{domain of }f = \left \{x : 0 \ \textless \ x \ \textless \ 4 \right \}$

the bounds of the integral for the mean can be changed from $-\infty \ \textless \ x \ \textless \ \infty$ to $0 \ \textless \ x \ \textless \ 4$ so that

$\boxed{\text{mean of }x = \int_{-\infty}^{\infty}{0.125x^2}dx = \int_{0}^{4}{0.125x^2}dx}$

Hence, the mean is computed as

$\text{mean of }x = \int_{0}^{4}{0.125x^2}dx
\\ = \left[ \frac{0.125x^3}{3} \right]_{0}^{4}
\\
\\ = \left[ \frac{0.125(4)^3}{3} \right] - \left[ \frac{0.125(0)^3}{3} \right] 
\\
\\ \boxed{\text{mean of }x = \frac{8}{3} \approx 2.667}$

Since the formula for variance is computed as

$\text{Variance of }x = \text{Var}[x] = E[x^2] - (E[x])^2$

we must first compute for E[x^2]

for which

$E[x^2] = \int_{-\infty}^{\infty}{x^2 f(x)}dx
\\
\\ = \int_{-\infty}^{\infty}{x^2(0.125x)}dx
\\
\\ \boxed{E[x^2] = \int_{-\infty}^{\infty}{0.125x^3}dx}$

Similar to the computation of integral of the mean, we take note that

$\text{domain of }f = \left \{x : 0 \ \textless \ x \ \textless \ 4 \right \}$

so that we can change the bounds of the integral, that is,

$\boxed{E[x^2] = \int_{-\infty}^{\infty}{0.125x^3}dx = \int_{0}^{4}{0.125x^3}dx}$

Hence,

$E[x^2] = \int_{0}^{4}{0.125x^3}dx
\\
\\ = \int_{0}^{4}{0.125x^3}dx
\\
\\= \left[ \frac{0.125x^4}{4} \right]_{0}^{4} 
\\ 
\\ = \left[ \frac{0.125(4)^4}{4} \right] - \left[ \frac{0.125(0)^4}{4} \right] 
\\ 
\\ \boxed{E[x^2] = 8}$

Because $E[x] = \frac{8}{3}$ ,

$\text{Variance of }x 
\\
\\ = E[x^2] - (E[x])^2
\\
\\ = 8 - \left( \frac{8}{3} \right)^2
\\
\\ \boxed{\text{Variance of }x = \frac{8}{9} \approx 0.889}$

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HELP!! REAL answers only please.

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