Write an equation of the line that passes through (-2,5) and (2,-1)

Solve x/3 = x+2/2. X=

Leni [432]

Answer:

I will assume that the term "x+2/2" is meant to be "(x + 2)/2)." Otherwise the equation would read (x/3) = x + 1

Step-by-step explanation:

(x/3) = (x + 2)/2

x = 3*(x+2)/2 [Multiply both sides by 3]

x = (3x + 6)/2

x = (3/2)x + 6/2

x - (3/2)x = 3

-0.5x = 3

x = -6

====================

Check:

Does (x/3) = (x + 2)/2 for x = -6?

(-6/3) = (-6 + 2)/2

-2 = -4/2

-2 = -2 YES

6 0

2 years ago

The difference between the observed value of the dependent variable and the value predicted using the estimated regression equat

Elenna [48]

Answer:

For this case we define the dependent variable as Y and the independent variable X. We assume that we have n observations and that means the following pairs:

$(x_1, y_1) ,....,(x_n,y_n)$

For this case we assume that we want to find a linear regression model given by:

$\hat y = \hat m x +\hat b$

Where:

$\hat m$ represent the estimated slope for the model

$\hat b$ represent the estimated intercept for the model

And for any estimation of the dependent variable $\hat y_i , i=1,...,n$ is given by this model.

The difference between the observed value of the dependnet variable and the value predicted using the estimated regression equation is known as residual, and the residual is given by this formula:

$e_i = y_i -\hat y_i , i=1,...,n$

So the best option for this case is:

d. residual

Step-by-step explanation:

For this case we define the dependent variable as Y and the independent variable X. We assume that we have n observations and that means the following pairs:

$(x_1, y_1) ,....,(x_n,y_n)$

For this case we assume that we want to find a linear regression model given by:

$\hat y = \hat m x +\hat b$

Where:

$\hat m$ represent the estimated slope for the model

$\hat b$ represent the estimated intercept for the model

And for any estimation of the dependent variable $\hat y_i , i=1,...,n$ is given by this model.

The difference between the observed value of the dependnet variable and the value predicted using the estimated regression equation is known as residual, and the residual is given by this formula:

$e_i = y_i -\hat y_i , i=1,...,n$

So the best option for this case is:

d. residual

7 0

3 years ago

2.5% of a population are infected with a certain disease. There is a test for the disease, however the test is not completely ac

Aleks04 [339]

Answer:

$P(disease/positivetest) = 0.36116$

Step-by-step explanation:

This is a conditional probability exercise.

Let's name the events :

I : ''A person is infected''

NI : ''A person is not infected''

PT : ''The test is positive''

NT : ''The test is negative''

The conditional probability equation is :

Given two events A and B :

P(A/B) = P(A ∩ B) / P(B)

$P(B) >0$

P(A/B) is the probability of the event A given that the event B happened

P(A ∩ B) is the probability of the event (A ∩ B)

(A ∩ B) is the event where A and B happened at the same time

In the exercise :

$P(I)=0.025$

$P(NI)= 1-P(I)=1-0.025=0.975\\P(NI)=0.975$

$P(PT/I)=0.904\\P(PT/NI)=0.041$

We are looking for P(I/PT) :

P(I/PT)=P(I∩ PT)/ P(PT)

$P(PT/I)=0.904$

P(PT/I)=P(PT∩ I)/P(I)

0.904=P(PT∩ I)/0.025

P(PT∩ I)=0.904 x 0.025

P(PT∩ I) = 0.0226

P(PT/NI)=0.041

P(PT/NI)=P(PT∩ NI)/P(NI)

0.041=P(PT∩ NI)/0.975

P(PT∩ NI) = 0.041 x 0.975

P(PT∩ NI) = 0.039975

P(PT) = P(PT∩ I)+P(PT∩ NI)

P(PT)= 0.0226 + 0.039975

P(PT) = 0.062575

P(I/PT) = P(PT∩I)/P(PT)

$P(I/PT)=\frac{0.0226}{0.062575} \\P(I/PT)=0.36116$

8 0

3 years ago

An___ graph consists of distinct, isolated points

Bess [88]

Answer:

I believe the answer is a Discrete graph.

Step-by-step explanation:

Hope my answer has helped you and if not then I am sorry.

4 0

3 years ago

Graph the system of linear inequalities <br><br> y>x-3<br> y<-x+2

Musya8 [376]

Answer:

it'll be a graph like this. the blue line is y<-x+2 and you would shade everything below this line because it's less than. the green is y>x-3 so you would shade everything above this line because it is greater than.

5 0

3 years ago