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Pani-rosa [81]
3 years ago
7

Please help, will mark brainliest.

Mathematics
1 answer:
dsp733 years ago
3 0

9514 1404 393

Answer:

  a) x = {-1, 3, 4}

  b) (0.472, 13.128)

  c) (3.528, -1.128)

  d) x < 0.472 U x > 3.528

  e) 0.472 < x < 3.528

Step-by-step explanation:

This is a cubic (odd degree) function with a positive leading coefficient, so it will be increasing until the first turning point, and after the last turning point. It will be decreasing between the turning points.

a) A graph shows the zeros to be x = -1, x = 3, x = 4.

b) A graph shows the local maximum to be approximately (0.472, 13.128). The x-coordinate of this point is exactly 2-√(7/3).

c) The local minimum is about (3.528, -1.128). Its x-coordinate is exactly 2+√(7/3).

d) As stated above, the increasing intervals are (-∞, 0.428) ∪ (3.528, ∞).

e) The decreasing interval is (0.428, 3.528).

_____

<em>Additional comments</em>

The sum of odd-degree term coefficients is the same as the sum of coefficients of even-degree terms, so you know one of the roots is -1. Factoring that out gives the quadratic x^2 -7x +12 = (x -3)(x -4), so the other two roots are 3 and 4.

The derivative is 3x^2 -12x +5 = 3(x -2)^2 -7, so its roots (turning points of f(x)) are 2±√(7/3).

I find a graphing calculator can show me the roots and turning points easily.

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Let X1 and X2 be independent random variables with mean μand variance σ².
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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

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Step-by-step explanation:

For this case we know that we have two random variables:

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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

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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)

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Var(\hat \theta_1) = Var(\frac{X_1 +X_2}{2})

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