All categories

3 years ago

What is m A.)120° B.)75° C.)80° D.)85°

Mathematics

1 answer:

vlada-n [284]3 years ago

7 0

Answer:

120

Step-by-step explanation:

linear angle (straight line) = 180

180-60 = 120

You might be interested in

Help can someone help me find the value of ClX Y and Z I would appreciate it luvvs ;) ♡

Stolb23 [73]

$\\ \sf\longmapsto 9x+46=35$

$\\ \sf\longmapsto 9x=35-46$

$\\ \sf\longmapsto 9x=-11$

$\\ \sf\longmapsto x=\dfrac{-11}{9}$

6 0

3 years ago

In all, 1000 students took a statistics exam worth 150 points. The first quartile (or Q1) for all 1000 scores is 90 points. Abou

Brums [2.3K]

Answer:

The number of students who scored more than 90 points is 750.

Step-by-step explanation:

Quartiles are statistical measures that the divide the data into four groups.

The first quartile (Q₁) indicates that 25% of the observation are less than or equal to Q₁.

The second quartile (Q₂) indicates that 50% of the observation are less than or equal to Q₂.

The third quartile (Q₃) indicates that 75% of the observation are less than or equal to Q₃.

It is provided that the first quartile is at 90 points.

That is, P (X ≤ 90) = 0.25.

The probability that a student scores more than 90 points is:

P (X > 90) = 1 - P (X ≤ 90)

= 1 - 0.25

= 0.75

The number of students who scored more than 90 points is: 1000 × 0.75 = 750.

4 0

3 years ago

Read 2 more answers

How do you solve this math problem-2(x+3y)=18

ss7ja [257]

Greetings!

Solve for <em>x</em>.
$-2(x+3y)=18$
Distribute the Parenthesis. <em>How?</em> Multiply all of the terms inside the parenthesis by the number outside the parenthesis.
$(-2)*x+(-2)*3y=18$
$-2x-6y=18$
Add 6y to both sides.
$(-2x-6y)+6y=(18)+6y$
$-2x=18+6y$
Divide both sides by -2.
$(-2x)/-2=(18+6y)/-2$
The Answer Is:
$x=-9-3y$
$x=-3y-9$

Hope this helps.
-Benjamin

7 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 $n$ 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:

$nu-\dfrac12u^2=-\dfrac12(u^2-2nu+n^2-n^2)=\dfrac12n^2-\dfrac12(u-n)^2$

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

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

Find an equation of the line passing through the points (5, 21) and (-5, -29)

gogolik [260]

Trata 5-29+-5=21 porque ya as tenido este

4 0

3 years ago