All categories

2 years ago

PLEASE HELP

Mathematics

1 answer:

shtirl [24]2 years ago

8 0

Step-by-step explanation:

1*-2=-2, -2*-2=4, -2*4=-8, -8*-32=256

You might be interested in

Mike has a credit rating of 720. Tyler has a credit rating of 560. Mike and Tyler apply for identical loans from Park Bank. Mike

Eva8 [605]

Answer:

6.2%

Step-by-step explanation:

Credit rating is an evaluation of the credit risk of a borrower, that how often a person is going to repay their debt, by credit rating it predicts the ability of the debtor to payback.

Mike has credit rating = 720

Tyler has credit rating = 560

Both are approved for loan. Mike's Credit score is higher, which means he is a much safer debtor as compared to Tyler. Mike will be able to pay back much easily than Tyler. Therefore Mike interest rate is 3.2%

Interest rate of Tyler is higher as he is not that trusted and has low Credit rating. Tyler is approved for a loan that charged 3 percentage points higher because of his inferior credit rating so it interest on the loan will be

Interest = 3.2%+3% = 6.2%

4 0

2 years ago

Read 2 more answers

Select all ratios equivalent to 20:15

Radda [10]

Answer:

b and c

Step-by-step explanation:

6 0

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