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2 years ago

What is 25+30 pleas help me

Mathematics

1 answer:

olganol [36]2 years ago

3 0

Your answer would be 55

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For a polynomial p(x), the value of p(3) is −2. Which of the following must be true about p(x)?

mr Goodwill [35]

Answer:

Step-by-step explanation:

The factor theorem states if (x - h) is a factor of a polynomial p(x) then

p(h) = 0

However, p(h) ≠ 0 then the value obtained is the remainder.

Thus

p(3) = - 2

When p(x) is divided by (x - 3) then the remainder is - 2

8 0

2 years ago

0.2c for c=4<br> Please help me asap and provide explanation please I need help (its a quiz)

gladu [14]

Answer:

0.8

Step-by-step explanation:

Replace the c for 4, since you said that c=4

0.2(4) is what you'll get

Multiply 0.2•4 = 0.8

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2 years ago

What should you do to get both numeric terms on the right side of the equation?

Mariulka [41]

Answer:

you should take the numeric value _16 to right side.

6 0

2 years ago

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What is 56,392 rounded to the nearest ten thousand??

Dmitriy789 [7]

56392 rounded to the nearest ten thousand is 60,000

6 0

2 years ago

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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