All categories

3 years ago

25 - (-4) plzzzzz ijkfytwhdfvytwefvfevetvy

Mathematics

2 answers:

djyliett [7]3 years ago

3 0

Step-by-step explanation:

Given

25 - (- 4)

= 25 + 4

= 29

Hope it will help :)

SVETLANKA909090 [29]3 years ago

3 0

Answer:

the answer to this question is 29

You might be interested in

7(x+2)=2x-1 solve for the variable

Len [333]

Answer:

x = - 3

Step-by-step explanation:

Given

7(x + 2) = 2x - 1 ← distribute parenthesis on left side

7x + 14 = 2x - 1 ( subtract 2x from both sides )

5x + 14 = - 1 ( subtract 14 from both sides )

5x = - 15 ( divide both sides by 5 )

x = - 3

7 0

3 years ago

How to do this question

Dafna11 [192]

Answer:

$0$

Step-by-step explanation:

Given:

$f(x) = \begin{cases} -x -2 &, &x < 0 \\ x &, &0 \leqslant x < 3 \\ x^2 -1 &, &3 \leqslant x \end{cases}$

Solving for $f(0)$ :

$f(0) = x \text{ when } x = 0 \\ f(0) = 0$

6 0

2 years ago

Read 2 more answers

Name and describe all the angles in the drawing below

rjkz [21]

Acute is a angle of some sort I believe.

7 0

3 years ago

PLSSSSSS!!!! HELP ME I WILL MARK YOU!!!!!!! MATH!!!!!!!!!!!!!!

scoundrel [369]

Answer:

-24

Step-by-step explanation:

-1(3^2)+(-15)

-9+(-15)

-24

8 0

3 years ago

Read 2 more answers

Let X1, X2, ... , Xn be a random sample from N(μ, σ2), where the mean θ = μ is such that −[infinity] < θ < [infinity] and

Sliva [168]

Answer:

$l'(\theta) = \frac{1}{\sigma^2} \sum_{i=1}^n (X_i -\theta)$

And then the maximum occurs when $l'(\theta) = 0$ , and that is only satisfied if and only if:

$\hat \theta = \bar X$

Step-by-step explanation:

For this case we have a random sample $X_1 ,X_2,...,X_n$ where $X_i \sim N(\mu=\theta, \sigma)$ where $\sigma$ is fixed. And we want to show that the maximum likehood estimator for $\theta = \bar X$ .

The first step is obtain the probability distribution function for the random variable X. For this case each $X_i , i=1,...n$ have the following density function:

$f(x_i | \theta,\sigma^2) = \frac{1}{\sqrt{2\pi}\sigma} exp^{-\frac{(x-\theta)^2}{2\sigma^2}} , -\infty \leq x \leq \infty$

The likehood function is given by:

$L(\theta) = \prod_{i=1}^n f(x_i)$

Assuming independence between the random sample, and replacing the density function we have this:

$L(\theta) = (\frac{1}{\sqrt{2\pi \sigma^2}})^n exp (-\frac{1}{2\sigma^2} \sum_{i=1}^n (X_i-\theta)^2)$

Taking the natural log on btoh sides we got:

$l(\theta) = -\frac{n}{2} ln(\sqrt{2\pi\sigma^2}) - \frac{1}{2\sigma^2} \sum_{i=1}^n (X_i -\theta)^2$

Now if we take the derivate respect $\theta$ we will see this:

$l'(\theta) = \frac{1}{\sigma^2} \sum_{i=1}^n (X_i -\theta)$

And then the maximum occurs when $l'(\theta) = 0$ , and that is only satisfied if and only if:

$\hat \theta = \bar X$

6 0

3 years ago