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lora16 [44]
3 years ago
13

4. The figure below consists of a square and 2 semicircles, with dimensions as shown. What is the outside perimeter, in centimet

ers, Show all work or explain how you found your answer. of the figure?

Mathematics
1 answer:
andriy [413]3 years ago
6 0

Answer:

41.13 centimeters (rounded to 2 decimal places)

Step-by-step explanation:

The perimeter is the sum of all sides of the figure.

This figure consists of two semicircular lines (right and left) and two lines (top and bottom). We add up all of them and find the perimeter of the figure.

<u>Left:</u>

Semicircle perimeter = half of circle perimeter (aka circumference)

Circumference = 2πr, where r is the radius (half of diameter)

The diameter of the circle is 8, so radius is 8/2 = 4

So the circumference is:

C = 2πr = 2π(4) = 8π

Semicircle perimeter is half of that, so, 8π/2 = 4π

<u>Right:</u>

This is exactly same semicircle as the left one, so its perimeter is also 4π

<u>Top:</u>

The top is a side with length 8

<u>Bottom:</u>

Same as top, so the length is 8

Total perimeter = 4π + 4π + 8 + 8 = 16 + 8π = 41.13 centimeters

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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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If f(x)=5x^3 and g(x)=x+1,find(f*g)(x)
natta225 [31]

Answer:

f * g)(x)=(5x^3)*(x+1)=5x^4 +5x^3

Step-by-step explanation:

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3 years ago
¿Cuántas cifras tiene el número quinientos cincuenta y cinco mil? Respuesta: ____________ CM DM UM C D U NÚM. DE CIFRAS
Burka [1]

Answer:

6 cifras.

Step-by-step explanation:

Siempre que tengamos un numero que termina en "mil", entonces la primer parte la escribimos y luego ira una coma:

por ejemplo, en "ciento once mil trecientos catorce"

lo separamos en:

"ciento once mil" y "trecientos catorce"

y lo escribimos como:

"ciento once" , "trecientos catorce"

111,314

Ahora que definimos esto, vamos a nuestro problema.

Tenemos el numero "quinientos cincuenta y cinco mil"

El cual podemos escribir como:

555,000

Ahora simplemente podemos contar las cifras, tenemos 6.

las cuales son:

Cientos de miles = CM = 5

Decenas de miles = DM = 5

Un miles = UM = 5

Cientos = C = 0

Decenas = D = 0

Unidades = U = 0

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3 years ago
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lana [24]

Hey there!

8(3x - 6) = 6(4x + 8)

8(3x) + 8(-6) = 6(4x) + 6(8)

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SUBTRACT 24x to BOTH SIDES

24x - 48 - 24x = 24x + 48 - 24

SIMPLIFY IT!

NEW EQUATION: -48 = 48

ADD 48 to BOTH SIDES

-48 + 48 = 48 + 48

NEW EQUATION: 0 = 96

Thus this makes your equation has

NO SOLUTION.

Therefore, your answer is:

Option A. Because this is a false

statement, the equation has no solution.

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

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3 years ago
How do i do theese problems
lyudmila [28]
These should be correct
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