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2 years ago

4X-y=6 3X+2y=21 Commomn core

Mathematics

1 answer:

Alex73 [517]2 years ago

3 0

Assuming you need to find the solution, it's x=3 and y=6
I used elimination

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PLEASE IS DUE IN A FEW MINUTES..

hjlf

Looks like you are screwed good luck:)

3 0

2 years ago

Jeremy randomly selected a round tile and a square tile from the sets shown below. what is the probability that jeremy selected

klio [65]

The probability that Jeremy randomly selected a round tile and a square tile from the sets is given as 1/40 (Option D).

<h3>What is probability?</h3>

Probability in mathematics is the branch of maths that speaks to the likelihood of an event occurring. Probability is usually given or described to occur between 0 and 1.

1 is the likelihood that the event will occur while zero is the certainty that an event will NOT occur.

Taken in decimal form, the likelihood of the above event occurring will very slim, because, 1/40 = 0.025

Learn more about Probability at:

brainly.com/question/24756209

5 0

1 year ago

Plz help me with the answer

UkoKoshka [18]

Answer:

Step-by-step explanation:

Put x = 9 to the equation 3x + 4y = 43

(3)(9) + 4y = 43

27 + 4y = 43 <em>subtract 27 from both sides</em>

4y = 16 <em>divide both sides by 4</em>

y = 4

5 0

2 years ago

Let X1 and X2 be independent random variables with mean μand variance σ².

My name is Ann [436]

Answer:

a) $E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu$

So then we conclude that $\hat \theta_1$ is an unbiased estimator of $\mu$

$E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu$

So then we conclude that $\hat \theta_2$ is an unbiased estimator of $\mu$

b) $Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}$

$Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}$

Step-by-step explanation:

For this case we know that we have two random variables:

$X_1 , X_2$ both with mean $\mu = \mu$ and variance $\sigma^2$

And we define the following estimators:

$\hat \theta_1 = \frac{X_1 + X_2}{2}$

$\hat \theta_2 = \frac{X_1 + 3X_2}{4}$

Part a

In order to see if both estimators are unbiased we need to proof if the expected value of the estimators are equal to the real value of the parameter:

$E(\hat \theta_i) = \mu , i = 1,2$

So let's find the expected values for each estimator:

$E(\hat \theta_1) = E(\frac{X_1 +X_2}{2})$

Using properties of expected value we have this:

$E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu$

So then we conclude that $\hat \theta_1$ is an unbiased estimator of $\mu$

For the second estimator we have:

$E(\hat \theta_2) = E(\frac{X_1 + 3X_2}{4})$

Using properties of expected value we have this:

$E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu$

So then we conclude that $\hat \theta_2$ is an unbiased estimator of $\mu$

Part b

For the variance we need to remember this property: If a is a constant and X a random variable then:

$Var(aX) = a^2 Var(X)$

For the first estimator we have:

$Var(\hat \theta_1) = Var(\frac{X_1 +X_2}{2})$

$Var(\hat \theta_1) =\frac{1}{4} Var(X_1 +X_2)=\frac{1}{4} [Var(X_1) + Var(X_2) + 2 Cov (X_1 , X_2)]$

Since both random variables are independent we know that $Cov(X_1, X_2 ) = 0$ so then we have:

$Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}$

For the second estimator we have:

$Var(\hat \theta_2) = Var(\frac{X_1 +3X_2}{4})$

$Var(\hat \theta_2) =\frac{1}{16} Var(X_1 +3X_2)=\frac{1}{4} [Var(X_1) + Var(3X_2) + 2 Cov (X_1 , 3X_2)]$

Since both random variables are independent we know that $Cov(X_1, X_2 ) = 0$ so then we have:

$Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}$

7 0

2 years ago

data set A is {30, 45, 32, 50, 33, 40, 44, 32}. Data set B is {28, 43, 30, 48, 35, 42, 46, 34}. which statement best compares th

3241004551 [841]

Answer:

the mean in set B is equal to the mean in set A (option C)

Question:

The question is incomplete as the answer choices were not given.Let's consider the following question:

Data set A is {30, 45, 32, 50, 33, 40, 44, 32}. Data set B is {28, 43, 30, 48, 35, 42, 46, 34}. which statement best compares the two data sets?

a) median for set A is equal to the median for set B

b) Range for set A is greater than range for set B

c) The mean in set B is equal to the mean in set A

Step by step explanation:

We can describe a data set using four ways:

Center, spread, shape and unusual features.

Let's consider the center and spread.

Center: This is the median of the distribution.

Spread: This is the variation of the data set. If the range is wide, the spread is larger and If the range is small, the spread is smaller.

Rearranging the data set:

A = {30, 32, 32, 33, 40, 44, 45, 50}

B = {28, 30, 34, 35, 42, 43, 46, 48}

From the data:

The median for set A = (33+40)/2 = 73/2= 36.5

The median for set B = (35+42)/2 = 77/2= 38.5

Range = highest value - lowest value

The data ranges from 30 to 50 (range = 20) for A

The data ranges from 28 to 48 (range = 20) for B

Mean for set A = (30+32+32+33+40+4445+50)/8 = 306/8 = 38.25

Mean for set B = (28+30+34+35+42+43+46+48)/8= 306/8 = 38.25

In both data set the mean is equal to 38.25.

Therefore the statement that best compares the two data sets is the mean in set B is equal to the mean in set A (C)

8 0

3 years ago