Find each percent 40 to 25

.3 3. True or False? For any integer m, 2m(3m + 2) is divisible by 4. Explain to get credit.

Sav [38]

Answer with explanation:

We have to prove that, For any integer m, 2 m×(3 m + 2) is divisible by 4.

We will prove this result with the help of Mathematical Induction.

⇒For Positive Integers

For, m=1

L HS=2×1×(3×1+2)

=2×(5)

=10

It is not divisible by 4.

⇒For Negative Integers

For, m= -1

L HS=2×(-1)×[3×(-1)+2]

=-2×(-3+2)

= (-2)× (-1)

=2

It is not divisible by 4.

False Statement.

4 0

3 years ago

I want to know how to solve this equation

Sphinxa [80]

Answer:

one property of log is that if the log expressions have the same base (in this case, 2), then you can multiply the added logs.

The answer would then be D

3 0

3 years ago

The table shows the solution to the equation|2x-5|-2=3: which is the first step

Anastaziya [24]

|2x - 5| - 2 = 3 |add 2 to both sides

|2x - 5| = 5 ⇔ 2x - 5 = 5 or 2x - 5 = -5 |add 5 to both sides

2x = 10 or 2x = 0 |divide both sides by 2

x = 5 or x = 0

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