PLEASE ANSWER + BRAINLIEST!!

Find the value of x

raketka [301]

Answer:

65

Step-by-step explanation:

remember a TRIANGLE always have a 180 degrees total angle,

so 45+70=115

180-115=65

5 0

3 years ago

How many 2/3 cup servings are there in a container that holds 7 cups?

avanturin [10]

Answer:

10.5

Step-by-step explanation:

4 0

3 years ago

Which statements about this system of equations are true? Check all that apply. - x + 6y = 16 8x - 6y = -2 The x-variable will b

frutty [35]

Answer:

The true statements are:

The y-variable will be eliminated when adding the system of equations

There is only one solution to the system of equations is

Step-by-step explanation:

* Lets explain how to solve the problem

- We use the elimination method to solve the system of the

linear equation

- The solution is one of three cases

# Exactly one solution ⇒ the 2 lines which represented the equations

intersect each other at one point

# No solution ⇒ the 2 lines which represented the equations are

parallel to each other

# Infinite solutions ⇒ the two lines are coincide

- In the system of the linear equations of the problem we have two

linear equations -x + 6y = 16 and 8x - 6y = -2

- To solve we must to eliminate one of the two variables

∵ The y's in the two equations have the same coefficients and

different signs

∴ We add the equations to eliminate y

∴ (-x + 8x) + (6y - 6y) = 16 + -2

∴ 7x = 14 ⇒ divide both sides by 7

∴ x = 2

- Substitute the x in any one of the two equations by 2

∴ -2 + 6y = 16 ⇒ add 2 to both sides

∴ 6y = 18 ⇒ divide both sides by 6

∴ y = 3

∴ The solution of the system of the equations is (2 , 3) ⇒ only one

solution

- Lets check the statements to find the true statements

# The x-variable will be eliminated when adding the system of

equations is not true

# The y-variable will be eliminated when adding the system of

equations is true

# The sum of the system of equations is - x + 6y is not true

# There is only one solution to the system of equations is true

6 0

3 years ago

Read 2 more answers

The spinner at the right is divided into eight equal parts. Find the theoretical probability of

finlep [7]

Answer:

$\frac{3}{4}$ probability of landing on a number greater than 2 on the spinner.

Step-by-step explanation:

A theoretical probability is given by the number of desired outcomes divided by the number of total outcomes.

In this question:

8 possible outcomes(all numbers from 1 to 8).

6 desired outcomes(3, 4, 5, 6, 7 and 8, that is, all the numbers greater than 2). So

$p = \frac{6}{8} = \frac{3}{4}$

$\frac{3}{4}$ probability of landing on a number greater than 2 on the spinner.

4 0

3 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

4 years ago