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3 years ago

Simplify each exponential expression using the properties of exponents and match it to the correct answer

Mathematics

2 answers:

jekas [21]3 years ago

7 0

3^-3*2^-3*6^3 = 1

last one=2

2nd one =1/2

mr_godi [17]3 years ago

6 0

3^-3 * 2^-3 *6^3 / (4^0)^2

= 6^-3 * 6^3 / (1)^2

= 1 / 1

= 1

---------------------------------------

2^4 * 3^5 / (2*3)^5

= 2^4 * 3^5 / 2^5 *3^5

= 1/ 2^1

= 1/2

---------------------------

(3 * 2)^4 * 3^-3 / 2^3 * 3

=3^4 * 2^4 * 3^-3 / 2^3 * 3

= 3^1 * 2^4 / 2^3 * 3

= 2^1

= 2

-----------------------

3^2 * 4^3 * 2^-1 / (3 * 4)^2

= 3^2 * 4^3 / 3^2 * 4^2 * 2^1

= 4^1 / 2^1

= 4/2

= 2

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Let X1, . . . ,Xn ∈ R be independent random variables with a common CDF F0. Let Fn be their ECDF and let F be any CDF. If F = Fn

Georgia [21]

Answer:

See the proof below.

Step-by-step explanation:

For this case we need to proof that: Let $X_1, X_2, ...X_n \in R$ be independent random variables with a common CDF $F_0$ . Let $F_n$ be their ECDF and let F any CDF. If $F \neq F_n$ then $L(F)$

Proof

Let $z_a$ different values in the set { $X_1,X_2,...,X_n}$ } and we can assume that $n_j \geq 1$ represent the number of $X_i$ that are equal to $z_j$ .

We can define $p_j = F(z_j) +F(z_j-)$ and assuming the probability $\hat p_j = \frac{n_j}{n}$ .

For the case when $p_j =0$ for any $j=1,....,m$ then we have that the $L(F) =0< L(F_n)$

And for the case when all $p_j >0$ and for at least one $p_j \neq \hat p_j$ we know that $log(x) \leq x-1$ for all the possible values $x>0$ . So then we can define the following ratio like this: