3 years ago

Help please, geometry question.

Mathematics

1 answer:

Inessa05 [86]3 years ago

5 0

I can help, whats the question

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Let X1, . . . ,Xn be an i.i.d. random sample from a N(0, 1) population. Define Y1 = 1 n n X i=1 Xi , Y2 = 1 n n X i=1 |Xi| . Cal

nalin [4]

For each $1\le i\le n$ , $E[X_i]=0$ , so that

$\displaystyle E[Y_1]=E\left[\frac1n\sum_{i=1}^nX_i\right]=\frac1n\sum_{i=1}^nE[X_i]=0$

Meanwhile,

$\displaystyle E[Y_2]=\frac1n\sum_{i=1}^nE[|X_i|]$

Each of the $X_i$ have PDF

$f_{X_i}(x)=\dfrac1{\sqrt{2\pi}}e^{-x^2/2}$

for $x\in\Bbb R$ . From this we get

$E[|X_i|]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty|x|e^{-x^2/2}\,\mathrm dx=\sqrt{\frac2\pi}\int_0^\infty xe^{-x^2/2}\,\mathrm dx=\sqrt{\frac2\pi}$

$\implies E[Y_2]=n\sqrt{\dfrac2\pi}$

8 0

2 years ago

Is 90 1/10th of 900?

mestny [16]

Yes, 90 is 1/10 of 900

Because

90 = 1/10 x 900
90 = 90

5 0

2 years ago

Solve for x: 3|x-3| 2=14<br> a. No solution<br> b. x=-1,x=8.3<br> c. x=0,x=7<br> d. x=-1,x=7

VMariaS [17]

I am looking on the answers, and there is only one case, when a or b or c or d pass: 3|x-3| + 2 = 14. So I assume, that before two is plus. Then:

3|x-3|+2=14 |minus 2
3|x-3|=12 |divide 3
|x-3|=4

From absolute value definition you've got two ways:

x-3=4 or x-3=-4
x=7 or x=-1

And answer d) passes

3 0

3 years ago

IIIIII NEEEEEED HEEEELPPPPPP ANYYYY ANSWERRRSS APPRECIATEDDDDD

Marta_Voda [28]

Answer:

B. Infinitely many solutions

6 0

3 years ago

Different shapes are drawn on cards and then the cards are placed in a bag. The number of cards for each shape is shown in the t

Goshia [24]

70% or 91/130.

There are 91 circle cards and 130 total.

6 0

3 years ago

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