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3 years ago

Plz help..............refresh the page before u answer

Mathematics

1 answer:

Dmitriy789 [7]3 years ago

7 0

If you factor the denominator of the left multiplicand...

[(a-4)(a+3)]/[a(a-4)] you can see that the (a-4)s cancel out leaving:

(a+3)/a

now you have

(a+3)/a * 2a^3/((a+3)(a-1)) so the (a+3)s cancel out leaving:

1/a * 2a^3/(a-1) which is:

2a^3/(a(a-1)) so now a^1 cancel leaving

2a^2/(a-1)

You might be interested in

When 27 times x squared times z all over negative 3 times x squared times z to the sixth power is completely simplified, the exp

marysya [2.9K]

Answer:

-5 or 5 depending on how you write the simplified form

Step-by-step explanation:

$\dfrac{27x^2z}{3x^2z^6}=9x^{2-2}z^{1-6}\\\\\boxed{=9z^{-5}=\dfrac{9}{z^5}}$

The applicable rules of exponents are ...

(a^b)(a^c) = a^(b+c)

1/a^b = a^-b

Some folks like their simplified form to contain only positive exponents. Other folks avoid unnecessary fractions. Choose the form that satisfies your grader.

4 0

2 years ago

An airliner travels 45 miles in 6 minutes. what is the speed in miles per hour?

Romashka-Z-Leto [24]

6 minutes/60minutes = 0.1 hrs
45miles/0.1hour = 450 miles per hour or mph

7 0

2 years ago

Read 2 more answers

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

2 years ago

Select the statement that best justifies the conclusion based on the given information.

artcher [175]

Hope this helps but I think it is A.)addition property of equality.

6 0

3 years ago

Read 2 more answers

What is the inverse of the function shown( in the form of a graph)

NemiM [27]

Answer:

$y = - 1x + b$

Step-by-step explanation:

The inverse of a slope is the exact opposite slope of the original. I'll show a function below where you can easily solve future problems if you need to.

$- \frac{1}{m} x$

m is the slope, so for the equation above, you simply insert the original slope, being 1, in the equation.

$- \frac{1}{1} x \\ - 1x$

So the inverse slope would be -1.

6 0

2 years ago