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2 years ago

Show how to solve (3X+2)-(2X-1)

Mathematics

1 answer:

lukranit [14]2 years ago

3 0

(3x+2)-(2x-1)
3x+2-2x-1
x+3
remove parentheses then collect the like terms to calculate the sum

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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:

$log (\frac{L(F)}{L(F_n)}) = \sum_{j=1}^m n_j log (\frac{p_j}{\hat p_j})$

$log (\frac{L(F)}{L(F_n)}) = n \sum_{j=1}^m \hat p_j log(\frac{p_j}{\hat p_j})$

$log (\frac{L(F)}{L(F_n)}) < n\sum_{j=1}^m \hat p_j (\frac{p_j}{\hat p_j} -1)$

So then we have that:

$log (\frac{L(F)}{L(F_n)}) \leq 0$

And the log for a number is 0 or negative when the number is between 0 and 1, so then on this case we can ensure that $L(F) \leq L(F_n)$

And with that we complete the proof.

8 0

3 years ago

PLEASE HELP I NEED HELP!!!

Setler79 [48]

Answer:

$2=1

Step-by-step explanation:

6 =$3

6 0

2 years ago

Help 6.5.AP-23 There were 675 people in line for a rock concert. Only 486 people got into the concert. Select the percent of peo

mina [271]

Answer:

D. 28%

Step-by-step explanation:

675 people - 486 people that got in = 189 people left out

189 people out of 675 people = 189 / 675 = 0.28 or 28%

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3 years ago

Tell someone a story about "1 more and then 1 more

gayaneshka [121]

Once upon a time there was a guy named Jeff. he went to a new school where he moved. so he was all by himself with no friends at all.. so then he met this guy named Zac so it was 1+1=2. and they were bff so they lived happily ever after together being bffs

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3 years ago

Write an exponential function in the form y=ab^xy=ab x that goes through points (0, 20)(0,20) and (6, 1280)(6,1280).

Phoenix [80]

Answer:

$y=20(2)^x$