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REY [17]
3 years ago
8

I need help with this question!!

Mathematics
1 answer:
kkurt [141]3 years ago
3 0

Answer:

24

Step-by-step explanation:

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What is the length of the hypotenuse of the triangle? triangle A B C. Side A C is 7 centimeters and side B C is 3 centimeters. H
Maksim231197 [3]

Answer: √58 cm or 7.62 cm

Step-by-step explanation:

The Pythagorean Theorem is a²+b²=c². Since we are given the lengths of a and b, we can plug them into this formula to find the length of the hypotenuse.

7²+3²=c²

49+9=c²

58=c²

c=√58 or 7.62 cm

6 0
3 years ago
The volume of a cube is related to the area of s face by the formula v= a^3/2. What is the volume of a cube whose face has an ar
Sophie [7]

Answer:

  1000 m³

Step-by-step explanation:

Put the number into the formula and do the arithmetic.

  v = a^(3/2) = (100 m²)^(3/2) = (√100)³ m³ = 1000 m³

The volume is 1000 cubic meters.

4 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 nth 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
Determine whether the given value is a solution of the equation.
serious [3.7K]

Answer:

84 is a solution.

Step-by-step explanation:

To find out if 84 is a solution or not, we plug in the value and simplify.

  • (84)/14 = 6
  • 6 = 6

Therefore, 84 is a solution.

6 0
3 years ago
G(b) = 5b- 9 h(b) = (b - 1)^2 Evaluate. (hog)(-6) =
laiz [17]

Given:

The two functions are:

g(b)=5b-9

h(b)=(b-1)^2

To find:

The value of (h\circ g)(-6).

Solution:

We have,

g(b)=5b-9

h(b)=(b-1)^2

We know that,

(h\circ g)(b)=h(g(b))

(h\circ g)(b)=h(5b-9)

(h\circ g)(b)=[(5b-9)-1]^2

(h\circ g)(b)=[5b-10]^2

Putting b=-6, we get

(h\circ g)(-6)=[5(-6)-10]^2

(h\circ g)(-6)=[-39-10]^2

(h\circ g)(-6)=[-49]^2

(h\circ g)(-6)=2401

Therefore, the value of (h\circ g)(-6) is 2401.

7 0
3 years ago
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