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2 years ago

5(10-6) 6(4+10) Help me, I find the bracket confusing

Mathematics

2 answers:

ira [324]2 years ago

4 0

Answer:

Step-by-step explanation:

Do 5 times ten.

5 times -6

Gives you 20

Then do 6 times 4, 6 times 10, to get the final answer of 30.

pentagon [3]2 years ago

3 0

A. 5X4= 20
B. 6X14=84

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Need help with Algebra 2! Thank you! :)

Olin [163]

Answer:

Consider the parent logarithm function f(x) = log(x)

Now,

Let us make transformations in the function f(x) to get the function g(x)

•On streching the graph of f(x) = log(x) , vertically by a factor of 3, the graph of y = 3log(x) is obtained.

•Now, shrinking the graph of y = 3log(x) horizontally by a fctor of 2 to get the grpah of y = 3log(x/2) i.e the graph of g(x)

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g(x) = 3 log(x/2) (Option-B).

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2 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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2 years ago

Write the values that make the denominators zero. then solve the equation?

Olin [163]

Answer:

denominators are zero for x=1

there is no solution

Step-by-step explanation:

We suspect your equation is supposed to be ...

10/(5x -5) +1/5 = 2/(x -1) . . . . . parentheses are required on denominators

The denominators will be zero for x=1.

This version of the equation has no solution. It simplifies to ...

2/(x-1) + 1/5 = 2/(x -1)

1/5 = 0 . . . . . subtract 2/(x-1) from both sides

There is no value of x that will make this equation true.

_____

<em>Alternative interpretation</em>

The way your equation is written, it must be interpreted according to the order of operations to be ...

(10/5)x -5 +1/5 = (2/x) -1 . . . . x=0 makes the denominator zero

2x -3.8 = 2/x

2x^2 -3.8x = 2 . . . . multiply by x

2(x^2 -1.9x +.95^2) = 2 +2(0.95^2)

(x -0.95)^2 = 1.9025

x = 0.95 ± √1.9025 ≈ {-0.4293, 2.3293}

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2 years ago

How do you put the equation 2x+y=2 into slope intercept form

Dahasolnce [82]

Answer:

Step-by-step explanation:

you put the equation in slope and intercept form by making y the subject of the formula

i.e

y = -2x + 2

-2 is the slope and 2 is the intercept

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2 years ago

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