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3 years ago

What is the area of a regual octagon with sides five times as long?

Mathematics

1 answer:

Ugo [173]3 years ago

7 0

Answer:

A=2(1+2)a exponet 2

a Side

Step-by-step explanation:

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What is the Mean Average deviation from the Mean?<br> Numbers: 6, 2, 3, 3, 2, 2

Irina-Kira [14]

Answer:

2.42

Step-by-step explanation:

standard deviation= sqrt(variance)

Variance=E[X2] - E[X]2

E[X] = sum(X)/n=35/9, so E[X]2 = 1225/81

E[X2] =sum(X2)/n = 189/9 = 170/81

Variance= 476/81

Standard deviation = sqrt(476/81) ~2.42

6 0

3 years ago

Which of the following predictions is most likely true

Sunny_sXe [5.5K]

Your answer is D because 10*2 is 20 and 6*2 is 12 so 12 out of the next 20 customers will purchase a sandwich.

5 0

3 years ago

Read 2 more answers

You have 5 Red Marbles, 4 Green Marbles, and 3 Blue Marbles in a Bowl. You randomly pick 4 marbles from the bowl (without replac

makvit [3.9K]

Answer:

Step-by-step explanation:

In this case each one is an independent event, therefore, the multiplication of each one would be the final probability

We have 5 Red Marbles, 4 Green Marbles, and 3 Blue Marbles, that is, there are 5 + 4 + 3 12 Marbles in total.

Now if I draw a red one, the probability would be: 5/12

When drawing another red, the probability would be: 4/11

When taking the green: 4/10

When removing the blue: 3/9

Finally, the final probability is:

P = (5/12) * (4/11) * (4/10) * (3/9)

P = 0.020

In other words, the probability of this happening is 2%

7 0

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.

8 0

4 years ago

Determine the domain for the graphed function:<br> Please help meeee

Rom4ik [11]

Answer:

Step-by-step explanation:

First, domain refers to the x axis. So you would find the lowest number on the X axis, -2, and then you look at the kind of dot on that number. It is an open dot, which means that it is all the numbers up to -2, but does not include -2. Then you find the highest number, in this case 2. Looking at the dot that is marking it, it is a closed dot, meaning it includes the number 2. So the domain would be numbers between -2 and 2, but does not include -2. all numbers greater than -2, x, all numbers less than and equal to 2.

7 0

3 years ago