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4 years ago

The world record in the long jump is 29 feet and 4 1/2 inches. how many inches short of 10 yards is the record long jump?

1 answer:

3 0

1 yard=36 inches
36×10=360 inches in 10 yards

29 feet and 4 1/2 inches = 352 1/2 inches
360-352.5=7.5 or 7 1/2

the record is 7 1/2 inches short of 10 yards

hope this helped

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The answer would be 24$

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3 years ago

This figure is made up of a quadrilateral and a semicircle.

Arturiano [62]

Answer:

The correct option is B. The area of the figure is 40.4 units².

Step-by-step explanation:

The line AB divides the figure in two parts one is a rectangle and another is semicircle.

The distance formula is

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The length of AB is

$AB=\sqrt{(2+3)^2+(4-2)^2}=\sqrt{25+4}=\sqrt{29}$

The length of AD is

$AD=\sqrt{(-1+3)^2+(-3-2)^2}=\sqrt{4+25}=\sqrt{29}$

Since AB=AD, therefore ABCD is a square. The area of the of square is

$A_1=a\times a=\sqrt{29}\times \sqrt{29} =29$

The area of square is 29 units².

The area of a semicircle is

$A_2=\frac{\pi}{2}r^2$

Since AB is the diameter of the semicircle, therefore the radius of the semicircle is

$r=\frac{d}{2}=\frac{\sqrt{29}}{2}$

The area of the semicircle is

$A_2=\frac{3.14}{2}(\frac{\sqrt{29}}{2})^2=40.3825$

The area of the figure is

$A=A_1+A_2=29+11.2825=40.3825\approx 40.4$

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3 years ago

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3 years ago

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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:

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$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du$

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

$E[Y^n]=e^{\frac12n^2}$

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