What is the answer ?

wel

Answer:

Option B

Step-by-step explanation:

When two intersecting lines cut each other they form four angles. these two pair of angles are called opposite angles or vertical angles. The vertical angles formed by two intersecting segments are equal.

Thus option B is correct: The vertical Angles Formed By Two Intersecting Segments Are Congruent !

8 0

4 years ago

Read 2 more answers

Yes or no? Please help

rusak2 [61]

Answer:

yes

Step-by-step explanation:

6 0

3 years ago

When using substitution to solve this system of equations, what is the result of the first step x=6y+3

myrzilka [38]

First step: you have to plug in 6y+3 into the equation and solve
Example: 6y+3+2y = 5

3 0

3 years ago

A building has n floors numbered 1,2,...,n, plus a ground floor g. at the ground floor, m people get on the elevator together, a

fomenos

Let $X_i$ be the random variable indicating whether the elevator does not stop at floor $i$ , with

$X_i=\begin{cases}1&\text{if the elevator does not stop at floor }i\\0&\text{otherwise}\end{cases}$

Let $Y$ be the random variable representing the number of floors at which the elevator does not stop. Then

$Y=X_1+X_2+\cdots+X_{n-1}+X_n$

We want to find $\mathrm{Var}(Y)$ . By definition,

$\mathrm{Var}(Y)=\mathbb E[(Y-\mathbb E[Y])^2]=\mathbb E[Y^2]-\mathbb E[Y]^2$

As stated in the question, there is a $\dfrac1n$ probability that any one person will get off at floor $n$ (here, $n$ refers to any of the $n$ total floors, not just the top floor). Then the probability that a person will not get off at floor $n$ is $1-\dfrac1n$ . There are $m$ people in the elevator, so the probability that not a single one gets off at floor $n$ is $\left(1-\dfrac1n\right)^m$ .

So,

$\mathbb P(X_i=x)\begin{cases}\left(1-\dfrac1n\right)^m&\text{for }x=1\\\\1-\left(1-\dfrac1n\right)^m&\text{for }x=0\end{cases}$

which means

$\mathbb E[Y]=\mathbb E\left[\displaystyle\sum_{i=1}^nX_i\right]=\displaystyle\sum_{i=1}^n\mathbb E[X_i]=\sum_{i=1}^n\left(1\cdot\left(1-\dfrac1n\right)^m+0\cdot\left(1-\left(1-\dfrac1n\right)^m\right)$
$\implies\mathbb E[Y]=n\left(1-\dfrac1n\right)^m$

and

$\mathbb E[Y^2]=\mathbb E\left[\left(\displaystyle\sum_{i=1}^n{X_i}\right)^2\right]=\mathbb E\left[\displaystyle\sum_{i=1}^n{X_i}^2+2\sum_{1\le i$

Computing $\mathbb E[{X_i}^2]$ is trivial since it's the same as $\mathbb E[X_i]$ . (Do you see why?)

Next, we want to find the expected value of the following random variable, when $i\neq j$ :

$X_iX_j=\begin{cases}1&\text{if }X_i=1\text{ and }X_j=1\\0&\text{otherwise}\end{cases}$

If $X_iX_j=0$ , we don't care; when we compute $\mathbb E[X_iX_j]$ , the contributing terms will vanish. We only want to see what happens when both floors are not visited.

$\mathbb P(X_iX_j=1)=\left(1-\dfrac2n\right)^m$
$\implies\mathbb E[X_iX_j]=\left(1-\dfrac2n\right)^m$
$\implies2\displaystyle\sum_{1\le i$

where we multiply by $n(n-1)$ because that's how many ways there are of choosing indices $i,j$ for $X_iX_j$ such that $1\le i$ .

So,

$\mathrm{Var}[Y]=n\left(1-\dfrac1n\right)^m+2n(n-1)\left(1-\dfrac2n\right)^m-n^2\left(1-\dfrac1n\right)^{2m}$

4 0

3 years ago

A square has an area that is less than 100 m. What is a reasonable range for the graph of the

Tpy6a [65]

Answer:

A

Step-by-step explanation:

0<y<100

4 0

3 years ago

Read 2 more answers