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2 years ago

I WILL GIVE BRAINLIEST

Mathematics

1 answer:

shusha [124]2 years ago

5 0

Answer:

x = -3

Step-by-step explanation:

$y\geq -\frac{2}{3} x-3\\
y\leq 3x+8$

$-\frac{2}{3} x-3=3x+8$

Add 2/3 x to both sides: $-3=\frac{11}{3}x+8$

Subtract 8 from both sides: $-11=\frac{11}{3}x$

Multiply both sides by 3: $-33=11x$

Divide both sides by 11: $x=-3$

Therefore, point of intersection between both lines is at point (-3, -1)

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A total of 8,644 people went to the football game. Of thos the visitors side. Of the people sitting on the visitor's side filled

lisov135 [29]

You divide 8644 by 8 which equals 1080.5, but since you can't have half of one person you approximate it to 1080 or 1081.

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3 years ago

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:

$\hat a = max(X_i)$

$E(a) - a= 0$ 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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3 years ago

Car rentals involve a $130 flat fee and an additional cost of $31.67 a day. What is the maximum number of days you can rent a ca

Elanso [62]

With a $500 budget you can rent a car for 11 days as $500-$130 is $370 which you then divide by $31.67 which gives you 11.682

3 0

3 years ago

HELP PLEASE Answer choices A). 12 B). 2 C). 6 D). 25/3

I am Lyosha [343]

That is 6 I believe

3 squared equals 9
9*2=18
18/3=6

4 0

2 years ago

Eduardo has a recipe that uses 2/3 cup of flour for each batch. If he makes 4 batches, how many cups of flour will he need? How

34kurt

Answer:

4 batches: $2\frac{2}{3}$ cups

7 batches: $4\frac{2}{3}$ cups

Step-by-step explanation:

If he uses 2/3 cup of flour for each batch, then that is 2/3 four times, or 2/3 · 4.

$\frac{2}{3}$ × $\frac{4}{1}$

= $\frac{8}{3}$ = $2\frac{2}{3}$

If he makes three more, than that will be 7 batches in total. We can multiply 2/3 by 3 to find out how much will be needed for those extra three, then add that product to $2\frac{2}{3}$ .

$\frac{2}{3}$ × $\frac{3}{1}$

= $\frac{6}{3}$ = 2

$2\frac{2}{3} +2$ = $4\frac{2}{3}$

Thus, $2\frac{2}{3}$ and $4\frac{2}{3}$ are the answers.

hope this helps!

8 0

1 year ago