How do i write a linear equation in slope intercept form for 16x - 4y = 2

Mathematics

1 answer:

marusya05 [52]3 years ago

8 0

Slope intercept form: y=mx+b
16x-2=4y
y=4x-1/2<<<< answer

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The sum of the digits of a two-digit number is 5. If the number is multiplied by 3, the result is 42. Write and solve a system o

Elenna [48]

A two digit number has a tens digit and a ones digit.
Let's say x = tens digit and y = ones digit

"The sum of the digits is 5"
x + y = 5

The next phrase is "the number multiplied by 3 is 42" but we need to represent the number using the digits. So they need to be multiplied first by their place value and added together. [Example: 34 = 3(10) + 4(1)]

The number is: 10x + y
3(10x + y) = 42

The system of equations: (two equations for two unknowns)
x + y = 5
30x + 3y = 42

Then you can use substitution or elimination to combine and solve.
I'll use elimination, multiply the entire top equation by -3 and add the equations together. y will cancel out

-3x - 3y = -15
30x + 3y = 42
------------------
27x + 0 = 27
x = 1

then plug x = 1 into either equation to find y

1 + y = 5
y = 4

remember the x and y represent digits so the number xy is 14

6 0

3 years ago

PLEASE HELP ASAP!!

Angelina_Jolie [31]

Answer:

Step-by-step explanation:

EGB and EHD are the same angles.

4 0

2 years ago

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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 $U\sim N(n,1)$ :