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3 years ago

Micah and Lucy are in the school play.

Mathematics

2 answers:

Georgia [21]3 years ago

7 0

I think it’s the 2nd option

den301095 [7]3 years ago

4 0

Answer:

52,56

Step-by-step explanation:

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What is a fraction of 42 is 7

kramer

$\frac{7}{42}\ of\ 42\ is\ equal\ 7\\\\check:\frac{7}{42}\cdot42=\frac{7}{42}\cdot\frac{42}{1}=\frac{7}{1}=7$

3 0

3 years ago

-9(m+3) +14=-49 please show work

yKpoI14uk [10]

Answer:

m=4

Step-by-step explanation:

−9(m+3)+14=−49

Step 1: Simplify both sides of the equation.

−9(m+3)+14=−49

(−9)(m)+(−9)(3)+14=−49(Distribute)

−9m+−27+14=−49

(−9m)+(−27+14)=−49(Combine Like Terms)

−9m+−13=−49

−9m−13=−49

Step 2: Add 13 to both sides.

−9m−13+13=−49+13

−9m=−36

Step 3: Divide both sides by -9.

−9m

−9

−36

−9

m=4

8 0

3 years ago

AB=A, B, equals Round your answer to the nearest hundredth.

suter [353]

Step-by-step explanation:

$\sin(20 \degree) = \frac{3}{AB} \\ AB = \frac{3}{ \sin(20 \degree) } \\ AB = \frac{3}{0.342} \\ AB = 8.77$

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

Will spent $3.75 for 3 pounds of granola. What is his unit rate in dollars per pound?

Drupady [299]

The unit rate is $1.25 per pound

6 0

3 years ago