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3 years ago

Calculate 6.75 tax on 23.05

Mathematics

1 answer:

Ksenya-84 [330]3 years ago

3 0

The answer is 12dollas

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A home has a triangular backyard. The second angle of the triangle is 6º more than the first angle. The third angle is 14º more

Leto [7]

Answer:

The three angles of the triangle are 40, 45, and 95. Done.

4 0

3 years ago

Write an expression for twelve divided by a number

Rom4ik [11]

12 ÷ 3= 4. Hope this helps. ;)

8 0

3 years ago

The quadrilateral shown is rotated 90° clockwise about the origin. In which quadrant is the image of the quadrilateral located?

Viktor [21]

Answer:

Option (2). 1

Step-by-step explanation:

Coordinates of point A, B, C and D are,

A(-4, 4), B(-2, 4), C(-2, 1) and D(-4, 3).

Quadrilateral ABCD when rotated 90° clockwise about the origin,

Rule for the rotation of the vertices,

(x, y) → (y, -x)

Following the rule of rotation coordinates of the image points,

A(-4, 4) → A'(4, 4)

B(-2, 4) → B'(4, 2)

C(-2, 1) → C'(1, 2)

D(-4, 3) → D'(3, 4)

Since all image points have the positive coordinates (x and y coordinates), image quadrilateral A'B'C'D' will be located in 1st quadrant.

Option (2) is the correct option.

4 0

3 years ago

Calculate the Laplace transforms of the following from the definition. 1. y = t^2. y = t^3

Likurg_2 [28]

Answer:

1) $L(y)=\frac{2}{s^{3}}$

2) $L(y)=\frac{6}{s^{4}}$

Step-by-step explanation:

To find : Calculate the Laplace transforms of the following from the definition ?

Solution :

We know that,

Laplace transforms of $t^n$ is given by,

$L(t^n)=\frac{n!}{s^{n+1}}$

1) $y=t^2$

Laplace of y,

$L(y)=L(t^2)$ here n=2

$L(y)=\frac{2!}{s^{2+1}}$

$L(y)=\frac{2}{s^{3}}$

2) $y=t^3$

Laplace of y,

$L(y)=L(t^3)$ here n=3

$L(y)=\frac{3!}{s^{3+1}}$

$L(y)=\frac{3\times 2}{s^{4}}$

$L(y)=\frac{6}{s^{4}}$

5 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

3 years ago