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aleksklad [387]

2 years ago

15

a sum of money is divided among ethan June way and Raj in the ratio of 5:6:9 after eating gave $50 to his mother the ratio becom

es 3:4:6 find the amount of money ethan has after giving $50 to his mother

1 answer:

Natasha_Volkova [10]2 years ago

5 0

The answer you are looking for is 35

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Find the solution(s) to (x- 3) = 49. Check all that apply.

natima [27]

Answer:

B.x= -4 and E. x= 10 are the answers

4 0

2 years ago

Minnie has 4 different stamps. She needs 3 to mail a large letter. How many ways can she pick the 3 that will be used?

timurjin [86]

Number of ways of picking a smaller number out of a bigger number is called combination.
Number of way of picking 3 stamps out of 4 is 4C3 (4 combination 3) = 4! / (3! x (4 - 3)!) = 4! / (3! x 1!) = (4 x 3 x 2 x 1) / (3 x 2 x 1 x 1) = 24 / 6 = 4 ways.

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

Which of the following are solutions to the quadratic equation? Check all that

LenKa [72]

Answer:

Step-by-step explanation:

2x2+5x−10=x2+4

Step 1: Subtract x^2+4 from both sides.

2x2+5x−10−(x2+4)=x2+4−(x2+4)

x2+5x−14=0

For this equation: a=1, b=5, c=-14

1x2+5x+−14=0

Step 2: Use quadratic formula with a=1, b=5, c=-14.

x=

−b±√b2−4ac

2a

x=

−(5)±√(5)2−4(1)(−14)

2(1)

x=

−5±√81

2

x=2 or x=−7

Answer:

x=2 or x=−7

5 0

3 years ago

Read 2 more answers

What's 454.895 rounded to the nearest cent?????? Please help me!! I just want to make sure I got the answer right.

Nastasia [14]

454.90, the 5 makes the 9 go up which in turn makes the 8 also go up.

3 0

2 years ago