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3 years ago

What is the area of this figure?

Mathematics

2 answers:

guapka [62]3 years ago

6 0

The area is 81 sq cm

Fofino [41]3 years ago

4 0

The area will be 81 squared centimeters.

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I need help with these 2

VashaNatasha [74]

For the first one you need to do
x+2+x+2+2x+4+2x+4
=6x+12

When doing this kind of problem you should use FOIL
First, Outer, Inner, Last
(x+2)(2x+4)
First- x•2x= 2x^2
Outer- x•4= 4x
Inner- 2•2x= 4x
Last- 2•4= 8

Combined you have
=2x^2+8x+8

5 0

3 years ago

The Pacific halibut fishery has been modeled by the differential equation dy dt = ky 1 − y K where y(t) is the biomass (the tota

Reika [66]

Answer:

398.411

Step-by-step explanation:

Explanation has been given in the following attachments.

6 0

3 years ago

One number is 4 times as large another. The sum of their reciprocals is 45/4. Find the two numbers

Reptile [31]

Let,
the numbers be "x" and "y"
According to the question,
x = 4y ......................................................equation (1)

$\frac{1}{x} + \frac{1}{y}= \frac{45}{4}$ ...................................equation (2)

Taking equation (2)
$\frac{1}{x} + \frac{1}{y}= \frac{45}{4}$

Substituting the value of "x" from equation (1), we get,

$\frac{1}{(4y)} + \frac{1}{y}= \frac{45}{4}$

$\frac{1}{4y} + \frac{1}{y}* \frac{4}{4} = \frac{45}{4}$

$\frac{1}{4y} + \frac{4}{4y}= \frac{45}{4}$

$\frac{1+4}{4y} = \frac{45}{4}$

$\frac{5}{4y} = \frac{45}{4}$

Cross multiplying, we get,

$5*4 = 45*4y$

$20 = 180y$
<u />
$\frac{20}{180} = y$

$\frac{1}{9} = y$

$y = \frac{1}{9}$

Now,
Taking equation (1)
x = 4y
Substituting the value of "y", we get

$x = 4( \frac{1}{9} )$

$x= \frac{4}{9}$

So, the numbers are $\frac{1}{9}$ and $\frac{4}{9}$

7 0

3 years ago

Read 2 more answers

Please help select the equivalent representations of the interval written below

zhuklara [117]

Answer:

3rd option

Step-by-step explanation:

x < 7

means that x cannot equal 7 but all values less than 7

since not equal at either end of interval use parenthesis, not brackets , then

(- ∞ , 7 ) ← is the required interval

6 0

1 year 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