What is the missing term in the factorization?

Mathematics

2 answers:

ludmilkaskok [199]3 years ago

8 0

3(2x+5)(2x-5), 5 is the missing term.

Verdich [7]3 years ago

8 0

ANSWER

The missing term is

$5$

EXPLANATION

The quadratic expression given to us is

$12 {x}^{2} - 75$

We factor 3 to obtain,

$12 {x}^{2} - 75 = 3(4 {x}^{2} - 25)$

We rewrite the expressions in the parenthesis to obtain,

$12 {x}^{2} - 75 = 3( ({2x})^{2} - {5}^{2} )$

Recall that,

${a}^{2} - {b}^{2} = (a + b)(a - b)$

We apply the difference of two squares formula for the expression in the parenthesis to obtain,

$12 {x}^{2} - 75 = 3(2x + 5)(2x - 5) )$

By comparing to,

$12 {x}^{2} - 75 = 3(2x + ?)(2x - 5) )$

The missing term is 5.

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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: