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Arte-miy333 [17]
2 years ago
7

HELPPPPPP QUICKKKKKHELPPPPPP QUICKKKKKHELPPPPPP QUICKKKKK

Mathematics
2 answers:
Lilit [14]2 years ago
8 0

Answer:A

Step-by-step explanation: The first few square numbers are:

1

,

4

,

9

,

16

,

25

,

36

,

49

,

√

1

=

1

√

4

=

2

√

9

=

3

√

16

=

4

√

25

=

5

√

36

=

6

w

w

w

w

w

w

←

√

40

lies here

√

49

=

7

Therefore

√

40

is between

6

and

7

LenaWriter [7]2 years ago
8 0

Answer:

A

Step-by-step explanation:

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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.

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makvit [3.9K]
Root: (-4/3, 0)
vertical intercept: (0, -4)

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3 years ago
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Find the zeros of each function by using a graph and a table f(x)=x^2+3x-18.
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1) Finding the zeros of this function f(x) =x² +3x -18

f(x) = x²+3x-18 <em>Factoring this equation, and rewriting it</em>

<em />

<em>Which two numbers whose sum is equal to 3 and their product is equal to 18?</em>

<em>6 -3 = 3 and 6 *-3 = -18</em>

<em />

<em>So we can rewrite as (x +6) (x-3)</em>

<em> </em>

(x+6)(x-3)=0 <em>Applying the Zero product rule, to find the roots</em>

x+6=0,

x=-6

x-3=0,

x=3

S={3,-6}

2) Setting a table, plugging in the values of x into the factored form: (x-6)(x-3)

x | y |

1 | -14 (1 +6)(1-3) =-14

2 | -8 (2 +6)(2-3) =-8

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4 | 10

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3) Plotting the function:

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Find the mean, median, and mode of the data set. Round your answers to the nearest tenth if necessary.
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Answer:

Mean: 83.2

Median: 82.5

Mode: 72

Step-by-step explanation:

If you put them all in order, Its really straightforward from there.

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1. Point  is on the line . Find the measure of angle .
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Answer:

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Step-by-step explanation:

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