All categories

3 years ago

Are these factors equivalent or nonequivalent?

Mathematics

1 answer:

dusya [7]3 years ago

6 0

I'm pretty sure the above fractions are equivalent because if you reduce 5s/15t then it equals s/3t.

You might be interested in

Which expressions are equivalent to 4d+6+2d4d+6+2d ? Choose all answers that apply: Choose all answers that apply: (Choice A) A

Sliva [168]

Answer:

Options A and C are correct choices.

Step-by-step explanation:

We have been given an expression $4d+6+2d$ and we are asked to choose all the equivalent expressions from given choices.

Let us simplify our given expression.

$4d+6+2d=(4d+2d)+6$

$4d+6+2d=6d+6$

Let us see which of our given choices matches with our answer.

A. $2(3d+3)$

Upon using distributive property $a(b+c)=a*b+a*c$ we will get,

$2*3d+2*3$

$6d+6$

Therefore, option A is a correct choice.

B. $6(d+6)$

Upon using distributive property $a(b+c)=a*b+a*c$ we will get,

$6*d+6*6$

$6d+36$

Since this expression is not equivalent to our answer, therefore, option B is not a correct choice.

C. $(3d+3)+(3d+3)$

Upon removing parenthesis we will get,

$3d+3+3d+3$

Upon combining like terms we will get,

$(3d+3d)+(3+3)$

$6d+6$

Therefore, option C is the correct choice.

7 0

3 years ago

If f(x) = 6x – 4, what is f(x) when x = 8?Which represents where f(x) = g(x)?

Alexxx [7]

F(x) = 44 as for having 8 as x

7 0

3 years ago

Read 2 more answers

Which equation is not true 12/10=6/5 3/1=30/10 6/2=2/3 8/6=16/12

blsea [12.9K]

Answer:

third one is not true means wrong

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

$E[U^0]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du=1$

so we end up with

$E[Y^n]=e^{\frac12n^2}$

3 0

3 years ago

the art club receives $10 plus $2 for every sculpture they sell for a fundraiser. the expression 2x+10 represents the amount the

Len [333]

2x + 10...the common number between these 2 terms is 2...so factor the 2 out

2(x + 5) <==

8 0

3 years ago