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Montano1993 [528]

3 years ago

12

What is 31/4 in simplest form

2 answers:

KATRIN_1 [288]3 years ago

5 0

In fraction form 31/4 is simplified. In decimal form 7.75 would be your answer.

Svetllana [295]3 years ago

4 0

31/4 because 31/4 cannot be divided by anythhing but itself. 31/4 is a prime number.

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3 Fred is playing tennis. For every 2 of his serves that land in, 3 serves land out. If he hit 30 serves in, how many serves lan

sineoko [7]

Answer:

d

Step-by-step explanation:

2:3 = 30:45

6 0

3 years ago

Read 2 more answers

Ja'Kiree has a bunch of quarters and nickels in her piggy bank totaling $8.75. If Ja'Kiree has 26 quarters, how many nickels doe

evablogger [386]

Start off by stating your facts. You have a maximum of $8.75. We know that each quarter is 25 cents so we multiply 25x26 equaling 650. Yet we need a decimal so it would be $6.50. Each nickel is worth 5 cents so then you would minus $8.75 and $6.50 getting us to $2.25. Now divide it by 5 cents and you should get 45 nickels. You can also do this by stating that 10 nickels is 50 cents and go until you get 45

6 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

Find the sum of the geometric series for which a = 160, r = 0.5, and n = 6.

harina [27]

$\\ \sf\longmapsto S_n=\dfrac{a(1-r^n)}{1-r}$

$\\ \sf\longmapsto S_n=\dfrac{160(1-0.5^6)}{1-0.5}$

$\\ \sf\longmapsto S_n=\dfrac{160(1-0.015625)}{0.5}$

$\\ \sf\longmapsto S_n=\dfrac{160(0.984378)}{0.5}$

$\\ \sf\longmapsto S_n=80(0.984378)$

$\\ \sf\longmapsto S_n=78.75$

6 0

3 years ago

Read 2 more answers

Can someone help me

Zolol [24]

Answer:

the first one....

6=w+2.

6 0

3 years ago