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2 years ago

. can someone help pls

Mathematics

1 answer:

disa [49]2 years ago

6 0

Answer:

The answer is 3,906.25

Step-by-step explanation:

All you have to do is divide 15,625 by 4! :)

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Write in slope intercept form an equation of the line that passes through the given points

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slope intercept form equation, y = mx + b

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2 years ago

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2 years ago

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the length of a rectangle is 3 feet more than twice the width. if the width is increased by four feet the perimeter of the new r

densk [106]

Answer:

The area of the original rectangle = 189 sq ft.

Step-by-step explanation:

Let, the width of the rectangle = m feet

So, the length of the rectangle = (2 m + 3) ft

Now, new width w' = (m + 4)

Perimeter of new rectangle (P') = 68 ft

Perimeter of a rectangle = 2 (LENGTH + WIDTH)

⇒68 ft = 2( L + W')

or, 68 ft = 2 [( 2m + 3) + (m + 4)]

or, 2 (3m + 7) = 68

or, 3m + 7 = 68/2 = 34

⇒ 3m = 34 - 7 = 27

⇒ m = 27/3 = 9

or, m = 9 ft

Hence, the original width of the rectangle = 9 ft

Original Length of the rectangle = 2m + 3 = 2(9) + 3 = 21 ft

Now, Area of the Rectangle = LENGTH X WIDTH

= 21 ft x 9 ft = 189 sq ft

Hence,the area of the original rectangle = 189 sq ft.

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3 years ago

Judy has $10 to buy watermelons and grapes. Judy plans to buy two watermelons that cost $6 and as many pounds of grapes as she c

lawyer [7]

Judy can buy at least 5 pounds of grapes.

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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