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mrs_skeptik [129]
3 years ago
11

(-10x^2+9)+(8x^2-5x+7) (−10x 2 +9)+(8x 2 −5x+7)

Mathematics
1 answer:
zavuch27 [327]3 years ago
7 0

Answer:

-2x^2-5x+16

Step-by-step explanation:

-10x^2-8x^2-5x+9+7=-2x^2-5x+16

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Integrate x<br><img src="https://tex.z-dn.net/?f=x4" id="TexFormula1" title="x4" alt="x4" align="absmiddle" class="latex-formula
Dmitrij [34]

Answer:

The result of the integration is \frac{x^5}{5}

Step-by-step explanation:

Integration of a power of x:

The integration of a power of x is:

\int x^{n} dx = \frac{x^{n+1}}{n+1}

In this question:

n = 4. So

\int x^{4} dx = \frac{x^{4+1}}{4+1} = \frac{x^5}{5}

The result of the integration is \frac{x^5}{5}

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2 years ago
To convert 2.5 hours into minutes, which ratio could you multiply by? Remember that 1 hour = 60 minutes.
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3 years ago
Lake Michigan is the fifth largest lake
nikklg [1K]
Yes with a <span>58,016 Square Kilometers</span>
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3 years ago
Read 2 more answers
Let $$X_1, X_2, ...X_n$$ be uniformly distributed on the interval 0 to a. Recall that the maximum likelihood estimator of a is $
Solnce55 [7]

Answer:

a) \hat a = max(X_i)  

For this case the value for \hat a is always smaller than the value of a, assuming X_i \sim Unif[0,a] So then for this case it cannot be unbiased because an unbiased estimator satisfy this property:

E(a) - a= 0 and that's not our case.

b) E(\hat a) - a= \frac{na}{n+1} - a = \frac{na -an -a}{n+1}= \frac{-a}{n+1}

Since is a negative value we can conclude that underestimate the real value a.

\lim_{ n \to\infty} -\frac{1}{n+1}= 0

c) P(Y \leq y) = P(max(X_i) \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n\leq y)

And assuming independence we have this:

P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) .... P(X_n \leq y) = [P(X_1 \leq y)]^n = (\frac{y}{a})^n

f_Y (Y) = n (\frac{y}{a})^{n-1} * \frac{1}{a}= \frac{n}{a^n} y^{n-1} , y \in [0,a]

e) On this case we see that the estimator \hat a_1 is better than \hat a_2 and the reason why is because:

V(\hat a_1) > V(\hat a_2)

\frac{a^2}{3n}> \frac{a^2}{n(n+2)}

n(n+2) = n^2 + 2n > n +2n = 3n and that's satisfied for n>1.

Step-by-step explanation:

Part a

For this case we are assuming X_1, X_2 , ..., X_n \sim U(0,a)

And we are are ssuming the following estimator:

\hat a = max(X_i)  

For this case the value for \hat a is always smaller than the value of a, assuming X_i \sim Unif[0,a] So then for this case it cannot be unbiased because an unbiased estimator satisfy this property:

E(a) - a= 0 and that's not our case.

Part b

For this case we assume that the estimator is given by:

E(\hat a) = \frac{na}{n+1}

And using the definition of bias we have this:

E(\hat a) - a= \frac{na}{n+1} - a = \frac{na -an -a}{n+1}= \frac{-a}{n+1}

Since is a negative value we can conclude that underestimate the real value a.

And when we take the limit when n tend to infinity we got that the bias tend to 0.

\lim_{ n \to\infty} -\frac{1}{n+1}= 0

Part c

For this case we the followng random variable Y = max (X_i) and we can find the cumulative distribution function like this:

P(Y \leq y) = P(max(X_i) \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n\leq y)

And assuming independence we have this:

P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) .... P(X_n \leq y) = [P(X_1 \leq y)]^n = (\frac{y}{a})^n

Since all the random variables have the same distribution.  

Now we can find the density function derivating the distribution function like this:

f_Y (Y) = n (\frac{y}{a})^{n-1} * \frac{1}{a}= \frac{n}{a^n} y^{n-1} , y \in [0,a]

Now we can find the expected value for the random variable Y and we got this:

E(Y) = \int_{0}^a \frac{n}{a^n} y^n dy = \frac{n}{a^n} \frac{a^{n+1}}{n+1}= \frac{an}{n+1}

And the bias is given by:

E(Y)-a=\frac{an}{n+1} -a=\frac{an-an-a}{n+1}= -\frac{a}{n+1}

And again since the bias is not 0 we have a biased estimator.

Part e

For this case we have two estimators with the following variances:

V(\hat a_1) = \frac{a^2}{3n}

V(\hat a_2) = \frac{a^2}{n(n+2)}

On this case we see that the estimator \hat a_1 is better than \hat a_2 and the reason why is because:

V(\hat a_1) > V(\hat a_2)

\frac{a^2}{3n}> \frac{a^2}{n(n+2)}

n(n+2) = n^2 + 2n > n +2n = 3n and that's satisfied for n>1.

8 0
3 years ago
Please help me: Ten athletes ran two races of the same length. The scatter plot shows their times. Identify each of these statem
Lemur [1.5K]

The scatter plot has been attached

Answer:

Options C, D & E are true

Step-by-step explanation:

Option A is wrong because from the scatter plot, only four athletes were faster in the second race than in the first one.

Option B is wrong because only 1 athlete had his second race time differing from the first race time by exactly 2 seconds.

Option C is true because exactly 9 of the times for the first race were at least 16 seconds

Option D is true because there are exactly 3 athletes who had the same time in both races

Option E is true because 8 of the times for the second race were less than 17 seconds

4 0
3 years ago
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