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3 years ago

How to write 9,452,729 in expanded form?

Mathematics

2 answers:

murzikaleks [220]3 years ago

4 0

Answer:

$9,000,000+400,000+50,000+2,000+700+20+9$ .

Step-by-step explanation:

We are asked to write 9,452,729 in expanded form.

To write 9,452,729 in expanded form, we will write given number according to sum of place values as:

$9,452,729=9,000,000+400,000+50,000+2,000+700+20+9$

Therefore, our given number in expanded form would be $9,000,000+400,000+50,000+2,000+700+20+9$ .

hodyreva [135]3 years ago

3 0

Step-by-step explanation:

Expanded Notation Form:

9,000,000

+ 400,000

+ 50,000

+ 2,000

+ 700

+ 20

+ 9

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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:

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

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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:

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