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3 years ago

How many centimeters equals 54 inches

Mathematics

1 answer:

cestrela7 [59]3 years ago

5 0

Answer:

137.16

Step-by-step explanation:

1 inch =2.54 cm

Please give brainliest.

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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

Assume that IQ scores are normally distributed with a mean of 100 and a standard deviation of 16. If 100 people are randomly sel

laiz [17]

Answer:

0.03 is the probability that for the sample mean IQ score is greater than 103.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 100

Standard Deviation, σ = 16

Sample size, n = 100

We are given that the distribution of IQ score is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling =

$\dfrac{\sigma}{\sqrt{n}} = \dfrac{16}{\sqrt{100}} = 1.6$

P( mean IQ score is greater than 103)

P(x > 103)

$P( x > 103) = P( z > \displaystyle\frac{103 - 100}{1.6}) = P(z > 1.875)$

$= 1 - P(z \leq 1.875)$

Calculation the value from standard normal z table, we have,

$P(x > 103) = 1 - 0.970 =0.03= 3\%$

0.03 is the probability that for the sample mean IQ score is greater than 103.

7 0

3 years ago

Please help so I can pass

lora16 [44]

Your answer would be F.

4 0

3 years ago

Ann is growing two different plants for a science project. Plant A grew 4/12 inch the first week and 2/12 inch the second week.

Galina-37 [17]

Start the number line from 0 and put divisions at 1/12. Then mark at 4/12 for plant A and indicate this was its growth in the first week. Then, make another mark up to 6/12 and indicate this was the plant's growth in the second week.
For plant B, make one mark at 7/12 and indicate that this is plant B's growth in the first week.

After two weeks, plant A has a height of 6/12 inches and plant B has a height of 7/12 inches, so B is taller.

8 0

3 years ago

Subtract the following rational expressions

lesya692 [45]

$\dfrac{2k}{3k-7} - \dfrac{3}{8k} = \dfrac{8k(2k)}{8k(3k-7)} - \dfrac{3(3k-7)}{8k(3k-7)} = \dfrac{8k(2k)-3(3k-7) }{8k(3k-7)} = \dfrac{16k^2 - 9k + 21}{24k^2-56}$

5 0

3 years ago