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3 years ago

Which expression is equivalent to (x superscript four-thirds baseline x superscript two-thirds Baseline ) superscript one-third

Mathematics

2 answers:

Andrei [34K]3 years ago

5 0

<h3>The equivalent expression is:</h3>

$(x^{\frac{4}{3}}x^{\frac{2}{3}})^{\frac{1}{3}} = x^{\frac{2}{3}$

Solution:

Given expression is:

$\displaystyle (x^{\frac{4}{3}}x^{\frac{2}{3}})^{\frac{1}{3}}$

We have to find the equivalent expression

We can simplify the above expression using law of exponents

Use the following law of exponents:

$a^m \times a^n = a^{m+n}$

Therefore,

$\displaystyle (x^{\frac{4}{3}}x^{\frac{2}{3}})^{\frac{1}{3}} = (x^{\frac{4}{3}+\frac{2}{3}})^{\frac{1}{3}}\\\\Simplify\\\\\displaystyle (x^{\frac{4}{3}}x^{\frac{2}{3}})^{\frac{1}{3}} = (x^2)^\frac{1}{3}$

Use another law of exponent

$(a^m)^n = a^{mn}$

Therefore,

$(x^{\frac{4}{3}}x^{\frac{2}{3}})^{\frac{1}{3}} = x^{\frac{2}{3}$

Thus the equivalent expression is found

Roman55 [17]3 years ago

5 0

Answer:

B on edg

Step-by-step explanation:

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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:

$\hat a = max(X_i)$

$E(a) - a= 0$ 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

What number is equivalent to the fraction 8/9?

Marizza181 [45]

Answer:

0.88888889, 88.888889%,

Step-by-step explanation:

Convert to decimal, and convert decimal to percentage.

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3 years ago

What method can you use to find the product of 70×55

IceJOKER [234]

You can use the regrouping

6 0

3 years ago

Write an equation of a line that is perpendicular to the given line and passes through the given point.

weeeeeb [17]

Answer:

• y = -6x-5

Step-by-step explanation:

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2 years ago

X-5=5x-13 i need the steps

posledela

Answer:

x=2

Step-by-step explanation:

x-5=5x-13

5x-x=-5+13

4x=8

x=8/4

x=2

4 0

3 years ago

Read 2 more answers