All even numbers greater than 2 are composite

How many 1/3 1b. portions can you get out of a 6 2/3 bag of nuts?help me pls

Crank

Answer:

20

Step-by-step explanation:

Divide 6 2/3 by 1/3

7 0

3 years ago

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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

Subtract (2x3 + 9x − 8) − (4x2 − 15x + 7).

STatiana [176]

2x^3 + 9x - 8 - (4x^2 - 15x + 7)....distribute thru the parenthesis
2x^3 + 9x - 8 - 4x^2 + 15x - 7....combine like terms
2x^3 - 4x^2 + 24x - 15 <==

7 0

3 years ago

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At Saturday night’s football game there were 18 less fans than half the fans at Friday night’s game. There were “x” fans at Frid

Valentin [98]

X divided by 2 take away 18 hope it helps.

7 0