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3 years ago

In the diagram, which angles are vertical angles? Check all that apply.

Mathematics

2 answers:

Igoryamba3 years ago

7 0

Answer: ∠ABE and ∠CBD

∠ABC and ∠EBD

Step-by-step explanation:

The vertical angles are the pair of the opposite angles that are formed when two lines crosses each other .

They are also known as vertically opposite angles.

In the given picture , it can be seen that two lines AD and ECare intersecting each other at vertex B.

Then, the pair of angles are vertical angles are :

∠ABE and ∠CBD

∠ABC and ∠EBD

dybincka [34]3 years ago

5 0

The vertical angles made is ABE and CBD, Thus meaning that your answer is B

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The admission fee at a local zoo is $1.50 for children and $5.00 for adults. On a certain day, 3000 people enter the zoo and $9,

vlabodo [156]

1600 children and 1400 adults attended

<h3>How to determine the number of adults?</h3>

Let the children be x and adult be y.

So, we have the following equations:

x + y = 3000

1.5x + 5y = 9400

Make x the subject in x + y = 3000

x = 3000 - y

Substitute x = 3000 - y in 1.5x + 5y = 9400

1.5(3000 - y) + 5y = 9400

Expand

4500 - 1.5y + 5y = 9400

Evaluate the like terms

3.5y = 4900

Divide both sides by 3.5

y = 1400

Substitute y = 1400 in x = 3000 - y

x = 3000 - 1400

Evaluate

x = 1600

Hence, 1600 children and 1400 adults attended

Read more about system of equations at:

brainly.com/question/14323743

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2 years ago

What four consecutive intergers have a sum of -34

Ivenika [448]

Answer:

-7, -8 , -9 and -10 are such four integers.

Step-by-step explanation:

Let the first integer = k

So, the next three consecutive integers are ( k+1), ((k + 1) + 1) and (((k + 1) + 1) +1)

or, k , (k +1), (k +2) and (k +3) are four consecutive integers.

Now, their sum is -34.

⇒k + (k +1) + (k +2) + (k +3) = -34

or, 4k + 6 = -34

or, 4k = - 34 - 6 = -40

⇒ k = -40/ 4 = -10

Hence, the first integer k = -10

The next integer = k + 1 = -10 + 1 = -9

Similarly next two integers are -8 and - 7.

So -7, -8 , -9 and -10 are such four integers.

7 0

3 years ago

F(1) = -3<br> f(n) = 2 · f(n − 1) +1<br> f(2)=

raketka [301]

Answer:

-5

Step-by-step explanation:

f(2) = 2 * f(1) + 1

f(2) = 2 * -3 + 1

f(2) = -6 + 1

f(2) = -5

6 0

3 years ago

Compare and order from least to greatest 4/5, 3/12 and 5/6?

jonny [76]

The Answer: 3/12, 5/6, 4/5...

7 0

3 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

3 years ago