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Murljashka [212]

2 years ago

13

What is the basis of statistical inference

1 answer:

Sphinxa [80]2 years ago

3 0

Answer:

Statistical inference is the process of using data analysis to deduce properties of an underlying probability distribution. Inferential statistical analysis infers properties of a population, for example by testing hypotheses and deriving estimates.

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Write each missing number 7+2=2+

Firdavs [7]

7+2=2+6 because 7+2=8 and 2+6 is = to 8 as well

5 0

2 years ago

Read 2 more answers

serious [3.7K]

Answer:

10.6 oz

Step-by-step explanation:

X × 0.25 = 0.265

x= 2.65/0.25= 10.6 oz

8 0

2 years ago

Help I will be marking brainliest!!!

expeople1 [14]

Answer:

A. 51°

Step-by-step explanation:

By the theorem of intersecting secants.

$m\angle B=\frac{1}{2} (m\widehat{LW} - m\widehat{MU})$

$44.5\degree =\frac{1}{2} (140\degree - m\widehat{MU})$

$2\times 44.5\degree = 140\degree - m\widehat{MU}$

$89\degree = 140\degree - m\widehat{MU}$

$m\widehat{MU} = 140\degree -89\degree$

$m\widehat{MU} = 51\degree$

8 0

2 years ago

Translate the following words to algebra symbols, using "n" as the variable. The product of six and a number is 24.

sveta [45]

So n will be the unknown, and the expression is:
6n = 24
from where we can find the number n:
n = 24/6
n = 4

7 0

2 years ago

Use the power series for 1 1−x to find a power series representation of f(x) = ln(1−x). What is the radius of convergence? (Note

Viktor [21]

a. Recall that

$\displaystyle\int\frac{\mathrm dx}{1-x}=-\ln|1-x|+C$

For $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n$

By integrating both sides, we get

$\displaystyle-\ln(1-x)=C+\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}$

If $x=0$ , then

$\displaystyle-\ln1=C+\sum_{n=0}^\infty\frac{0^{n+1}}{n+1}\implies 0=C+0\implies C=0$

so that

$\displaystyle\ln(1-x)=-\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}$

We can shift the index to simplify the sum slightly.

$\displaystyle\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}n$

b. The power series for $x\ln(1-x)$ can be obtained simply by multiplying both sides of the series above by $x$ .

$\displaystyle x\ln(1-x)=-\sum_{n=1}^\infty\frac{x^{n+1}}n$

c. We have

$\ln2=-\dfrac\ln12=-\ln\left(1-\dfrac12\right)$

$\displaystyle\implies\ln2=\sum_{n=1}^\infty\frac1{n2^n}$

4 0

2 years ago