PLZ HELP I NEED TO PASS MATH LOL, What is the value of the expression (−10)(−7)? Enter your answer in the box.

A bar graph in which the bars are arranged in frequency order is

Kruka [31]

It is the Pareto Graph. It is a bar graph in which bars are arranged in frequency order. Pareto graph is the combination of bar and line graph that displays the data in a descending order. It has an 80/20 rule.

4 0

3 years ago

Denver has a current temperature of -5°, while Barcelona has a current temperature of 54°. What is the difference in temperature

irga5000 [103]

Answer:

59°

Step-by-step explanation:

Difference=54-(-5)

=54+5=59

7 0

2 years ago

PLS HELP ASAP I DONT HAVE TIME AND IT ALSO DETECT IF ITS RIGHT OR WRONG!

IgorLugansk [536]

It’s C. Rational number

7 0

3 years ago

How to solve X + y = 2<br> X - y = 4

SpyIntel [72]

So y+x=2
x-y=4
one way is subsituteion where yousubisutte one of the number for another exg

x-y=4
add y to both sides
x=y+4
subsitute y+4 for x in x+y=2
y+4+y=2
2y+4=2
subtract 4 from both sides
2y=-2
divide both sides by 2
y=-1
subsiutte
x+y=2
x-1=2
x=3

or elimination

x+y=2
x-y=4
add the two equations and cancel the y terms

x+x+y-y=2+4
2x=6
divide both sides by 2
x==3
subsitute
x+y=2
3+y=2
y=-1

so x=3
y=-1

5 0

3 years ago

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