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4 years ago

How many 5/6 are in 1 ?

Mathematics

1 answer:

AlexFokin [52]4 years ago

6 0

Answer:

no there are not any of them

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Let $\mu$ and $\sigma^2$ denote the mean and variance of the random variable x. determine $e[(x-\mu)/\sigma]$ and $e{[((x-\mu)/\

Nezavi [6.7K]

$\mathbb E\left(\dfrac{X-\mu}{\sigma}\right)=\dfrac1\sigma\mathbb E(X)-\dfrac\mu\sigma=\dfrac{\mu-\mu}\sigma=0$

$\mathbb E\bigg(\left(\dfrac{X-\mu}\sigma\right)^2\bigg)=\dfrac1{\sigma^2}\mathbb E\left(X^2-2\mu X+\mu^2\right)=\dfrac{\mathbb E(X^2)-2\mu\mathbb E(X)+\mu^2}{\sigma^2}$
$=\dfrac{\mathbb E(X^2)-2\mu^2+\mu^2}{\sigma^2}=\dfrac{\mathbb E(X^2)-\mu^2}{\sigma^2}=\dfrac{\mathbb E(X^2)-\mathbb E(X)^2}{\sigma^2}$
$=\dfrac{\mathbb V(X)}{\sigma^2}=\dfrac{\sigma^2}{\sigma^2}=1$

3 0

4 years ago

Please answer correctly !!!!!!!!!! Will make brainliest !!!!!!!!!!!!!

frosja888 [35]

Answer:

Step-by-step explanation:

Let's call their ages m and b.

We have:

m = b + 12

m - 17 = 4(b - 17) → m = 4b - 51

b + 12 = 4b - 51

3b = 63

b = 21

m = 21 + 12 = 33

6 0

3 years ago

Read 2 more answers

In , there were immigrants admitted to a country. In 1960, the number was . a. Assuming that the change in immigration is lin

Arisa [49]

Answer:

The answer is below

Step-by-step explanation:

The complete question is:

In 1960, there were 237,794 immigrants admitted to a country. In 2001, the number was 1,150,729.

a. Assuming that the change in immigration is linear, write an equation expressing the number of immigrants, y, in terms of t, the number of years after 1900.

b. Use your result in part a to predict the number of immigrants admitted to the country in 2013.

c. Considering the value of the y-intercept in your answer to part a, discuss the validity of using this equation to model the number of immigrants throughout the entire 20th century.

Answer:

a) From the question, we can get two ordered pairs which are $(t_1,y_1)=(60,237794)\ and\ (t_2,y_2)=(101,1150729)$

Using the equation of a line given two points:

$y-y_1=\frac{y_2-y_1}{t_2-t_1}(t-t_1)\\ \\y-237794=\frac{1150729-237794}{101-60}( t-60)\\\\y-237797=22266.71(t-60)\\\\y=22266.71t-1098205.44$

b) In 2013, t = 2013 - 1900 = 113.

Hence:

$y=22266.71t-1098205.44\\\\y=22266.71(113)-1098205.44\\\\y=1417932.488\\\\y=1417933$

$y=22266.71t-1098205.44\\\\the\ y\ intercept\ is\ at\ t=0,hence:\\\\y=22266.71(0)-1098205.44\\\\y=-1098205.44$

Since the y intercept is negative, that is in 1900 the number of immigrants was -1098206 which can not be possible. Hence this equation is not valid and the growth may or may not be linear.

7 0

4 years ago

To the nearest tenth -4 -2 1,3 find PQ

Triss [41]

Not really understanding what you are asking

4 0

3 years ago

Simplify the given expression below:

Svetllana [295]

The answer would be C. 12+8i/13

7 0

3 years ago