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3 years ago

1. Find the slope of a line that passes through the points (-1,-4) and (5, -1).

1 answer:

7 0

The answer is A. 1/2

(-1,-4) are (x1, y1)
(5,-1) are (x2, y2)
So you subtract y2-y1/x2-x1 = -1-(-4)/5-(-1)
-1-(4) = 3
5-(-1) = 6
3/6= 1/2
Slope is 1/2 making A your answer

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I need help on all of them

Aneli [31]

Answer:

Part A

1) -4x+17

2) -45x+32

3) -17x

Part B

4) 4x+28

5) -12x+15

6) -1x-6

Part C

7) 2x+19

8) -4x-4

9) -14x-24

10) 3x-2

Step-by-step explanation:

I did the work in a notebook. I know it's right.

Hope this helps. Have a nice day you amazing bean child.

4 0

3 years ago

Let

Digiron [165]

What I gather from the question is that $X$ has second moment $E(X^2)=81$ and variance $V(X) = 58$ , and you're asked to find the expectation and variance of the random variable $Y=2X+10$ .

From the given second moment and variance, we find the expectation of $X$ :

$V(X) = E(X^2) - E(X)^2 \implies E(X) = \sqrt{E(X^2) - V(X)} = \sqrt{23}$

Expectation is linear, so

$E(Y) = E(2X+10) = 2 E(X) + 10 = \boxed{2\sqrt{23} + 10}$

Using the same variance identity, we have

$V(Y) = V(2X+10) = E((2X+10)^2) - E(2X+10)^2$

and

$E((2X+10)^2) = E(4X^2 + 40X + 100) = 4E(X^2) + 40E(X) + 100 = 424 + 40\sqrt{23}$

so that

$V(Y) = V(2X+10) = (424 + 40\sqrt{23}) - (2\sqrt{23} + 10)^2 = \boxed{232}$

Alternatively, we can use the identity

$V(aX+b) = a^2 V(X) \implies V(2X+10) = 4V(X) = 232$

5 0

2 years ago

Subtract 428,731- 175 ,842

bearhunter [10]

Your answer is 252,889

7 0

3 years ago

3. Solve the inequality –2(z + 5) + 20 > 6.

12345 [234]

$\boxed{-2\left(z+5\right)+20>6\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:z$

Solution:

Given inequality is:

$-2(z+5)+20>6$

We have to solve the given inequality

$-2\left(z+5\right)+20>6\\\\\mathrm{Subtract\:}20\mathrm{\:from\:both\:sides}\\\\-2\left(z+5\right)+20-20>6-20\\\\\mathrm{Simplify}\\\\-2\left(z+5\right)>-14$