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klio [65]
3 years ago
5

I will mark you brainliest for answering

Mathematics
2 answers:
frez [133]3 years ago
8 0

Answer:

28%

Step-by-step explanation:

Finger [1]3 years ago
8 0

Answer:

25% of the questions were correct

Step-by-step explanation:

14+42=56

14/56=0.25

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Kyle took a total of 9 pages of notes during a 3hr class. How many hours of class will it take for Kyle to take 12 pages of ntoe
Vladimir [108]

Answer:

4 hours

Step-by-step explanation:

9÷3=1

1 hour for 3 pages

12÷3= 4

4 hours

5 0
1 year ago
Help Quickly ASAP! look at the screenshot I gave you<br><br> POINT:20
igomit [66]

Answer:

41 i think

Step-by-step explanation:

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8 0
3 years ago
What is the approximate volume of a cylinder with a diameter of 10 meters and a height of 5 meters? Use 3.14 for pi. A. 392.5 m3
EastWind [94]
For cylinder
d=10m
So r=5m
h=5m
volume=pi×r^2×h
=3.14×5×5×5
=3.14×125
=392.5m^3
So option a is correct.
3 0
3 years ago
Read 2 more answers
Find the product of z1 and z2, where z1 = 5(cos 60° + i sin 60°) and z2 = 7(cos 210° + i sin 210°)
Verizon [17]

Answer:

cos210=cos(180+30)=−cos30=−√32 . sin210=sin(180+30)=−sin30=−12 . 3(cos210+isin210) =3(−√32)+3i(−12). −(32)√3−(32)i. My favourite way of seeing that sin30=12 and cos30=√32 is ...

Step-by-step explanation:

But that's what I say personally

3 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
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