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3 years ago

Is the expression 2(5-4) numerical or algebraic Please help asap

Mathematics

2 answers:

Alexxandr [17]3 years ago

7 0

Answer:

The expression is numerical.

Step-by-step explanation:

It is numerical because it contains numbers and operations. Algebraic expressions contain variables.

nadya68 [22]3 years ago

4 0

Answer:

Numerical

Step-by-step explanation:

This is a numerical expression because there are no variables; there are only numbers.

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I don’t understand this assignment!!

Rashid [163]

Answer:

Equation of the Line: y = 1.25x + 5

Equation in Standard Form: 5x - 4y = -20

Step-by-step explanation:

y = mx + b

$\frac{17.5 - 22.5}{10 - 14}$ = $\frac{-5}{-4}$ = $\frac{5}{4}$ or 1.25

y = 1.25x + 5

Ax + By = C (A must be positive, and no fractions or decimals)

y = 1.25x + 5

-1.25x + y = 5

times -4

5x - 4y = -20

7 0

3 years ago

Let X ~ N(0, 1) and Y = eX. Y is called a log-normal random variable.

Cloud [144]

If $F_Y(y)$ is the cumulative distribution function for $Y$ , then

$F_Y(y)=P(Y\le y)=P(e^X\le y)=P(X\le\ln y)=F_X(\ln y)$

Then the probability density function for $Y$ is $f_Y(y)={F_Y}'(y)$ :

$f_Y(y)=\dfrac{\mathrm d}{\mathrm dy}F_X(\ln y)=\dfrac1yf_X(\ln y)=\begin{cases}\frac1{y\sqrt{2\pi}}e^{-\frac12(\ln y)^2}&\text{for }y>0\\0&\text{otherwise}\end{cases}$

The $n$ th moment of $Y$ is

$E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy$

Let $u=\ln y$ , so that $\mathrm du=\frac{\mathrm dy}y$ and $y^n=e^{nu}$ :

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du$

Complete the square in the exponent:

$nu-\dfrac12u^2=-\dfrac12(u^2-2nu+n^2-n^2)=\dfrac12n^2-\dfrac12(u-n)^2$

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du$

But $\frac1{\sqrt{2\pi}}e^{-\frac12(u-n)^2}$ is exactly the PDF of a normal distribution with mean $n$ and variance 1; in other words, the 0th moment of a random variable $U\sim N(n,1)$ :