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2 years ago

Can someone actually help and not delete their answer I rlly need your help I’ll make u brainliest or whatever

Mathematics

1 answer:

Brrunno [24]2 years ago

5 0

Answer:

p = -30 + 20

Hope this helps!

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Pls help this is major , ill mark you brainliest x.

fenix001 [56]

Answer:

Step-by-step explanation:

5 0

3 years ago

A cylinder has a base diameter of 16 inches and a height of 18inches. What is the volume in cubic inches, of the nearest tenths

Stolb23 [73]

Answer:

3619.11 cubic inches

Step-by-step explanation:

volume of a cylinder is $V=\pi r^2h$

diameter is 16 in, so radius (r) is 8 in

plug in values: $V=\pi 8^2(18) = 3619.11$ cubic inches

8 0

3 years ago

What is the equation of the line in slope-intercept form?

Sladkaya [172]

Answer:

equation: y = -6x -20

Step-by-step explanation:

pass (0,-20) (-10,40)

y = mx + b

y intercept (b) = -20

slope (m) = (40 - -20) / (-10 - 0) = 60 / -10 = - 6

equation: y = -6x -20

3 0

3 years ago

Solve for x. 1 2 x + 3 2 (x + 1) − 1 4 = 5 A) 5 2 B) 15 4 C) 15 8 D) 17 8

Hitman42 [59]

Answer: x=-13/44 or -.2954

8 0

3 years ago

Read 2 more answers

The random variable X has the following probability density function: fX(x) = ( xe−x , if x > 0 0, otherwise. (a) Find the mo

dusya [7]

Answer:

Follows are the solution to the given points:

Step-by-step explanation:

Given value:

$\to f_X (x) \ \ xe^{-x} \ , \ x>0$

For point a:

Moment generating function of X=?

Using formula:

$\to M(t) =E(e^{tx})= \int^{\infty}_{-\infty} \ e^{tx} f(x) \ dx$

$M(t) = \int^{\infty}_{-\infty} \ e^{tx}xe^{-x} \ dx = \int^{\infty}_{0} \ xe^{(t-1)x} \ dx$

integrating the values by parts:

$u = x \\\\dv = e^{(t-1)x}\\\\dx =dx \\\\v= \frac{e^{(t-1)x}}{t-1}\\\\M(t) =[\frac{e^{(t-1)x}}{t-1}]^{-\infty}_{0} -\int^{\infty}_{0} \frac{e^{(t-1)x}}{t-1} \ dx\\\\$

$= \frac{1}{t-1} (0) - [\frac{e^{(t-1)x}}{(t-1)^2}]^{\infty}_{0}\\\\=\frac{1}{(t-1)^2}(0-1)\\\\=\frac{1}{(t-1)^2}\\\\$

Therefore, the moment value generating by the function is $=\frac{1}{(t-1)^2}$

In point b:

$E(X^n)=?$

Using formula: $E(X^n)= M^{n}_{X}(0)$

form point (a):

$\to M_{X}(t)=\frac{1}{(t-1)^2}$

Differentiating the value with respect of t

$M'_{X}(t)=\frac{-2}{(t-1)^3}$

when $t=0$

$M'_{X}(0)=\frac{-2}{(0-1)^3}= \frac{-2}{(-1)^3}= \frac{-2}{-1}=2\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M'''_{X}(t)=\frac{(-2)(-3)(-4)}{(t-1)^5}\\\\M''_{X}(0)=\frac{(-2)(-3)(-4)}{(0-1)^5}= \frac{24}{(-1)^5}= \frac{24}{-1}=-24\\\\\therefore \\\\M^{K}_{X} (t)=\frac{(-2)(-3)(-4).....(k+1)}{(t-1)^{k+2}}\\\\$

$M^{K}_{X} (0)=\frac{(-2)(-3)(-4).....(k+1)}{(-1)^{k+2}}\\\\\therefore\\\\E(X^n) = \frac{(-2)(-3)(-4).....(n+1)}{(-1)^{n+2}}\\\\$

7 0

3 years ago