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SVETLANKA909090 [29]
3 years ago
15

Please help solve all of these-

Mathematics
1 answer:
lana [24]3 years ago
7 0

Answer:

3−y

2. 8

4.​​2​bh

Step-by-step explanation:

5(2x+y)=15

2 Divide both sides by 55.

2x+y=\frac{15}{5}2x+y=

​5

​

​15

​​

3 Simplify  \frac{15}{5}

​5

​

​15

​​   to  33.

2x+y=32x+y=3

4 Subtract yy from both sides.

2x=3-y2x=3−y

5 Divide both sides by 22.

x=\frac{3-y}{2}x=

​2

​

1. ​3−y

​​ 2. 8

2.Add 2y2y to both sides.

x=-8+2yx=−8+2y

2 Regroup terms.

x=2y-8x=2y−8

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Alejandro rounded the number 76.518 to 76.5. Jiro rounded the same number to 76.52. Who is correct? Explain your reasoning
sergeinik [125]
Alejandro is correct because you do not need to add another number to 76.518, as Jiro did, but you need to round it to the nearest tenth.

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4 0
3 years ago
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So can someone show there work in this one?
mihalych1998 [28]

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4 0
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Please help me it’s due today! The one on the top is just helping explain the bottom questions
Vinvika [58]

Answer:

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Step-by-step explanation:

3 0
3 years ago
Circles in the Coordinate Plane<br> Acellus<br> Complete the equation of this circle:<br> А<br> 1
s344n2d4d5 [400]

Answer:

(x -+2)² + (y - 4)² = 36

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

Here (h, k ) = (- 2, 4) and r = 6 , then

(x - (- 2))² + (y - 4)² = 6² , that is

(x + 2)² + (y - 4)² = 36 ← equation of circle

5 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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