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4 years ago

What 6,589 to the nearest thousand

2 answers:

6 0

6,589 to the nearest thousand would be <u>7,000</u> since you have high numbers as 5,8, and 9 that allows you to round up your answer to 7,000.

BlackZzzverrR [31]4 years ago

6 0

6,589 to the nearest thousand is 7,000. Look to the number right of the hundreds place and if its below 4 you round down and if its above 5 you round up.

<h2><em>-Rhear</em></h2>

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Which statement is true?

SVEN [57.7K]

Answer:

$y=log_{1} x$ is not a logarithmic function because the base is equal to 1.

Step-by-step explanation:

A logarithmic function is of the form: $y=log_{b}x$

For a logarithmic function to exist, we must fulfill certain conditions. The conditions are

i. The base, $b$ , is greater than 0. ${b>0)$

ii. The base, $b$ , is not equal to 1. $(b\ne 1)$

iii. $x$ must be greater than 0. $(x>0)$

Now, if we observe all the choices given, we conclude that the third choice is correct as the base there is 1. If base is equal to 1, then it violates the definition of logarithmic functions. So, the correct choice is:

$y=log_{1} x$ is not a logarithmic function because the base is equal to 1.

6 0

3 years ago

Read 2 more answers

Multiply or divide as indicated. Y^-9•y^-8•y^10

BabaBlast [244]

Answer:

1 / y^7

Step-by-step explanation:

Y^-9•y^-8•y^10

= y^10 / y^9•y^8

= y^10 / y^17

= 1 / y^7

8 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

3(9-8x-4x)+8(3x+4)=11<br> how do I work this out???

Arte-miy333 [17]

Remember PEMDAS is the order of operations. parenthesis,exponents,multiply,divide,add,subtract. so -8x-4x is -12x. now simplified the problem is 3(9-12x) + 8(3x+4)=11 . now I would distribute the numbers before the parentheses and it becomes 27-36x + 24x+32 =11 . now combine like terms. 59-12x =11 . subtract 59 on both sides. -12x=-48. divide -12 on both sides. x=4. :-)

7 0

3 years ago

1 point<br> Solve: 0.75 (x + 200) = 150.5 (2 – 28)<br> type your answer...

adelina 88 [10]

Answer:

$\huge \: x = - 5417.33$