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Marysya12 [62]
4 years ago
11

Convert to a decimal using long division: 5/11 AndHow did you decide when you have calculated enough decimal places?

Mathematics
1 answer:
podryga [215]4 years ago
3 0
Hope this helps. I wrote it out to make sure it was right too.

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Which of the following expressions is equivalent to −56−13?Select all that apply.
Dmitrij [34]
So -56 -13 = -69
So the answer would be 69?
You didn’t show the options.
5 0
3 years ago
Can i get some help and a good explenation
denis-greek [22]

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The correct answer is $2,50

Step-by-step explanation:

The unit rate is what we use to determine the price of a single unit, when something is sold in a group, as in this case 5 ounces.

4 0
3 years ago
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A concrete pillar has the shape of a cylinder. It has a radius of 6 meters and a height of 4 meters. If concrete costs $82 per c
Gnesinka [82]

82

Step-by-step explanation:

6 0
3 years ago
Suppose Burger King has run a major advertising campaign in the hopes of increasing monthly sales. To investigate the effectiven
Alexxandr [17]

Answer:

The right answer is:

a.H0: μd = 0; H1: μd > 0

Step-by-step explanation:

The claim that want to be tested is that the sales were significantly increased after the commercial, indicanting that the advertisement campaign was effective.

This claim is usually expressed in the alternative hypothesis as it has to have enough evidence to prove that it is true.

Then, the alternative hypothesis H1 should state that the difference (sales after - sales before) is higher than 0.

The null hypothesis would state that the difference is not significantly different from 0, or, in other words, that the sales are the same before and after and that the variation is due to pure chance.

Then, the null hypothesis H0 would state that the difference is equal to 0.

The right answer is:

a.H0: μd = 0; H1: μd > 0

4 0
4 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
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