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3 years ago

How do you put the equation 2x+y=2 into slope intercept form

Mathematics

2 answers:

Dahasolnce [82]3 years ago

5 0

Answer:

Step-by-step explanation:

you put the equation in slope and intercept form by making y the subject of the formula

i.e

y = -2x + 2

-2 is the slope and 2 is the intercept

HACTEHA [7]3 years ago

3 0

Answer:

Step-by-step explanation:

2x + y = 2

Making y as subject

y = - 2x + 2

And the equation is y = mx + c

Where m is slope = -2

Y intercept = 2

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How to multiply fractions and mixed numbers together?

Basile [38]

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5 0

3 years ago

What Is the slope intersept of this equation Y-3=4(x+2)

lilavasa [31]

Answer:

Step-by-step explanation:

Step 1) distribute 4 to (x+2). 4(x+2)= 4x+8

Step 2) add 3 to both sides. y-3+3= 4x+8+3

Step 3) simplify by solving for y. y=4x+11

*The slope is the coeficcient (or the number infront of x), so the slope is 4

7 0

4 years ago

Read 2 more answers

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

-4+2(12)<br> ————<br> -3(-3)-(-1)

dsp73

Answer:

20/10 is 2

final answer is 2

8 0

3 years ago

Please explain how to help me learn please

devlian [24]

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