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2 years ago

Convert 9/25 to a percent

Mathematics

2 answers:

Arte-miy333 [17]2 years ago

4 0

Answer:

36%

Step-by-step explanation:

Divide 9 and 25 as decimals.

Mademuasel [1]2 years ago

3 0

Answer:

36%

Step-by-step explanation:

If you know that 25 times 4 equals a dollar, then a dollar equals 100.

4 times 25 = 100

Then you have to do the same to the 9.

9 times 4 is 36.

36/100 is 36%

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In a parallel circuit, the supply voltage is 120 V, 60 Hz. The resistance is 30 Ω and the inductive reactance is 40 Ω. What is t

kvasek [131]

Given : supply voltage = 120V;60 Hz

Inductive Reactance $(X_{L}) = 40$ Ω

The inductive reactance is given by the formula, $X_{L}=Lw$ , where L is inductance and $w$ is given by formula, $w = 2\pi(frequency)$

Apply this, we can say that, $L = \frac{X_{L}}{2\pi (frequency)} = \frac{40}{2(3.14)(60)} = 0.106157 H = 106.157 mH$

8 0

2 years ago

Which one is it ????

Inessa05 [86]

Answer:

465371295673

Step-by-step explanation:

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6 0

3 years ago

Two straight lines VZ and YW intersect at X

Alexeev081 [22]

Answer:

∠WVX = 35°

Step-by-step explanation:

Look at the rough draw.

As you can see,

VW ║ YZ

So ∠XWV is congruent to ∠XYZ through alternate interior angle.

We can solve ∠WVX by considering the two other angles, 88° and 57°.

Total angle of a triangle is 180°.

So,

? + 88 + 57 = 180

? + 145 = 180

? = 180 - 145

? = 35°

5 0

3 years ago

Decide if the following biconditional statement

garik1379 [7]

true because if you add up 4 90°you get 360

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