All categories

3 years ago

Which is the graph of the system of equations? x + 2y = 2 4x - y = 4

1 answer:

4 0

Solve for x for x: x+2y=2
x+2y +-2y=2+-2y (add -2y on both sides.)
x=-2y+2
Substitute : -2y+2 for x in
4x-y=4:
4(-2y+2)-y=4
-9y+8=4 Simplify both sides of the equation.
-9y+8+-8=4+-8 Add -8 to both sides.
-9y=-4
-9/-9=-4-9 Divide both sides by -9
y=4/9
Substitute 4/9 for y in
x=-2y+2
x=-2(4/9)+2
x=10/9
y=4/9,x=10/9

You might be interested in

All I need is the answer I have legit no idea what I'm doing.

rusak2 [61]

Answer:

yes, you are right, that hexagon follows all of the guidelines

Step-by-step explanation:

wait no, plot points at the correct ones, then draw line straight down, then a reflection of the initial line. Draw points for the corresponding sides

5 0

3 years ago

Find the area of each sector.

sergey [27]

Answer:

D. $\frac{525\pi}{8}$ m²

Step-by-step explanation:

Area of the sector = ½*r²∅

Where,

Radius (r) = 15 m

Angle in radians (∅) = $\frac{7\pi}{12}$

Plug in the values into the equation

Area of sector = $\frac{1}{2} * 15^2 * \frac{7\pi}{12}$

Area of sector = $\frac{1}{2} * 225 * \frac{7\pi}{12}$

Area of sector = $\frac{1*225*7\pi}{2*12}$

Area of sector = $\frac{1,575\pi}{24}$

Simplify

Area of sector = $\frac{525\pi}{8}$

5 0

2 years ago

In an RPG game, each player can choose among 5 basic classes: Mage, Warrior, Archers, Priests, Monks. Each basic class can advan

grigory [225]

Answer:

In RPGs a Character Class is a designation that determines a player's abilities and fighting style (and depending on the game possibly even their origin, education, and home area) often in the form of a job or archetype. A character class is defined by the abilities that it lends to a character — as such, two different characters with the same class are theoretically interchangeable, in that they have the same "power set" and can play the same role in gameplay because of their similar abilities. However, character class systems can come with varying levels of customization — ranging from characters of a given class being literally identical to having so much variety that character class is no longer even a good indicator of that character's abilities. While most common in fantasy Role-Playing Games, they have recently began to appear in other genres, such as trading card games and MOBAs.

Step-by-step explanation:

7 0

1 year ago

On average how fast did he drive in hours per mile

Sindrei [870]

Answer:

38 miles per hour

Step-by-step explanation:

<em>Simple just divide 342/9 to get 38</em>

4 0

2 years ago

Let X1,X2......X7 denote a random sample from a population having mean μ and variance σ. Consider the following estimators of μ:

Viefleur [7K]

Answer:

a) In order to check if an estimator is unbiased we need to check this condition:

$E(\theta) = \mu$

And we can find the expected value of each estimator like this:

$E(\theta_1 ) = \frac{1}{7} E(X_1 +X_2 +... +X_7) = \frac{1}{7} [E(X_1) +E(X_2) +....+E(X_7)]= \frac{1}{7} 7\mu= \mu$

So then we conclude that $\theta_1$ is unbiased.

For the second estimator we have this:

$E(\theta_2) = \frac{1}{2} [2E(X_1) -E(X_3) +E(X_5)]=\frac{1}{2} [2\mu -\mu +\mu] = \frac{1}{2} [2\mu]= \mu$

And then we conclude that $\theta_2$ is unbiaed too.

b) For this case first we need to find the variance of each estimator:

$Var(\theta_1) = \frac{1}{49} (Var(X_1) +...+Var(X_7))= \frac{1}{49} (7\sigma^2) = \frac{\sigma^2}{7}$

And for the second estimator we have this:

$Var(\theta_2) = \frac{1}{4} (4\sigma^2 -\sigma^2 +\sigma^2)= \frac{1}{4} (4\sigma^2)= \sigma^2$

And the relative efficiency is given by:

$RE= \frac{Var(\theta_1)}{Var(\theta_2)}=\frac{\frac{\sigma^2}{7}}{\sigma^2}= \frac{1}{7}$

Step-by-step explanation:

For this case we assume that we have a random sample given by: $X_1, X_2,....,X_7$ and each $X_i \sim N (\mu, \sigma)$

Part a

In order to check if an estimator is unbiased we need to check this condition:

$E(\theta) = \mu$

And we can find the expected value of each estimator like this:

$E(\theta_1 ) = \frac{1}{7} E(X_1 +X_2 +... +X_7) = \frac{1}{7} [E(X_1) +E(X_2) +....+E(X_7)]= \frac{1}{7} 7\mu= \mu$

So then we conclude that $\theta_1$ is unbiased.

For the second estimator we have this:

$E(\theta_2) = \frac{1}{2} [2E(X_1) -E(X_3) +E(X_5)]=\frac{1}{2} [2\mu -\mu +\mu] = \frac{1}{2} [2\mu]= \mu$

And then we conclude that $\theta_2$ is unbiaed too.

Part b

For this case first we need to find the variance of each estimator:

$Var(\theta_1) = \frac{1}{49} (Var(X_1) +...+Var(X_7))= \frac{1}{49} (7\sigma^2) = \frac{\sigma^2}{7}$

And for the second estimator we have this:

$Var(\theta_2) = \frac{1}{4} (4\sigma^2 -\sigma^2 +\sigma^2)= \frac{1}{4} (4\sigma^2)= \sigma^2$

And the relative efficiency is given by:

$RE= \frac{Var(\theta_1)}{Var(\theta_2)}=\frac{\frac{\sigma^2}{7}}{\sigma^2}= \frac{1}{7}$

5 0

3 years ago