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2 years ago

Mse of gamma distribution scale parameter

1 answer:

3 0

MSE is an abbreviation for mean-squared-error and it typically encompasses both variance and bias of a statistical population.

<h3>What is a gamma distribution?</h3>

A gamma distribution can be defined as a two-parameter family of continuous probability distributions which can be described or modeled with a finite number of parameters (α and β).

In probability theory and statistics, MSE is an abbreviation for mean-squared-error and it typically encompasses both variance and bias of a statistical population.

Mathematically, the mean-squared-error (MSE) of a given parameter ˆθ is given by Eθ[(ˆθ - θ)²].

Read more on gamma distribution here: brainly.com/question/14281632

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Mum am i adopted what does the mom say<br> A-yeah<br> B-no<br> C-screw off<br> D-wii sports

qwelly [4]

It depends if you are or not so either b or a

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2 years ago

Select the correct answer. it was not the first time for me to experience the river's charm. this time i saw the tupelo, cypress

Mademuasel [1]

Answer:

This time I saw the tupelo, cypress, and yaupon trees through the eyes of a man who has lived his entire life along the river.

Explanation:

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1 year ago

Which of the following is a true statement about diffusion.

Murljashka [212]

Answer:

D) Molecules move from areas of high concentration to areas of low concentration.

Explanation:

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2 years ago

If sin(1/x^2+1) is an antiderivative for f(x) then.

STatiana [176]

The limit from 1 to 2 of the given antiderivative is; -0.19865

<h3>What is the Limit of the Integral?</h3>

We are given the antiderivative of f(x) as sin(1/(x² + 1)). Thus, to find the limit from 1 to 2, we will solve as;

$\int\limits^2_1 {(sin\frac{1}{x^{2} + 1} )} \, dx = {(sin\frac{1}{2^{2} + 1} )} - {(sin\frac{1}{1^{2} + 1} )}$

⇒ (sin ¹/₅) - (sin ¹/₂)

⇒ 0.19866 - 0.47942

⇒ -0.19865

Complete Question is;

If sin(1/(x² + 1)) is an anti derivative for f(x), then what is the limit of f(x)dx from 1 to 2?

Read more about integral limits at; brainly.com/question/10268976

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2 years ago

Newton’s law of cooling states that for a cooling substance with initial temperature t0, the temperature t(t) after t minutes ca

antoniya [11.8K]

The cooling rate of the substance is approximately 0.0732.

According to the statement, the Newton's law of cooling is defined by the following formula:

$T(t) = T_{s} + (T_{o}-T_{s})\cdot e^{-k\cdot t}$ (1)

Where:

$T_{s}$ - Final temperature, in degrees Celsius.
$T_{o}$ - Initial temperature, in degrees Celsius.
$t$ - Time, in minutes.
$k$ - Cooling rate, in $\frac{1}{min}$ .
$T(t)$ - Current temperature, in degrees Celsius.

Please notice that substance reaches thermal equilibrium when $T(t) = T_{s}$ , that is when temperature of the substance is equal to the temperature of surrounding air.

If we know that $T_{o} = 80\,^{\circ}C$ , $t = 15\,min$ , $T_{s} = 50\,^{\circ}C$ and $T(15) = 60\,^{\circ}C$ , then the cooling rate of the substance is: