Answer:
SU=5 the first option.
Step-by-step explanation:
I did the assignment.
Answer:
center = 98.6, variability = 0.08
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean
and standard deviation
, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
and standard deviation ![s = \frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
The center is the mean.
So ![\mu = 98.6](https://tex.z-dn.net/?f=%5Cmu%20%3D%2098.6)
The standard deviation of the sample of 50 adults is the variability, so
![s = \frac{0.6}{\sqrt{50}} = 0.08](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B0.6%7D%7B%5Csqrt%7B50%7D%7D%20%3D%200.08)
So the correct answer is:
center = 98.6, variability = 0.08
Take some you time to focus on yourself, do calming activities wether it’s a walk, puzzle, or anything along those lines
The ratio of the distance between the foci and the length of the <em>major</em> axis is called eccentricity.
<h3>
Definitions of dimensions in ellipses</h3>
Dimensionally speaking, an ellipse is characterized by three variables:
- Length of the <em>major</em> semiaxis (
). - Length of the <em>minor</em> semiaxis (
). - Distance between the foci and the center of the ellipse (
).
And there is the following relationship:
(1)
Another variable that measure how "similar" is an ellipse to a circle is the eccentricity (
), which is defined by the following formula:
,
(2)
The greater the eccentricity, the more similar the ellipse to a circle.
Therefore, the ratio of the distance between the foci and the length of the <em>major</em> axis is called eccentricity. ![\blacksquare](https://tex.z-dn.net/?f=%5Cblacksquare)
To learn more on ellipses, we kindly invite to check this verified question: brainly.com/question/19507943
Consider such events:
A - slip with number 3 is chosen;
B - the sum of numbers is 4.
You have to count ![Pr(A|B).](https://tex.z-dn.net/?f=Pr%28A%7CB%29.)
Use formula for conditional probability:
![Pr(A|B)=\dfrac{Pr(A\cap B)}{Pr(B)}.](https://tex.z-dn.net/?f=Pr%28A%7CB%29%3D%5Cdfrac%7BPr%28A%5Ccap%20B%29%7D%7BPr%28B%29%7D.)
1. The event
consists in selecting two slips, first is 3 and second should be 1, because the sum is 4. The number of favorable outcomes is exactly 1 and the number of all possible outcomes is 5·4=20 (you have 5 ways to select 1st slip and 4 ways to select 2nd slip). Then the probability of event
is
![Pr(A\cap B)=\dfrac{1}{20}.](https://tex.z-dn.net/?f=Pr%28A%5Ccap%20B%29%3D%5Cdfrac%7B1%7D%7B20%7D.)
2. The event
consists in selecting two slips with the sum 4. The number of favorable outcomes is exactly 2 (1st slip 3 and 2nd slip 1 or 1st slip 1 and 2nd slip 3) and the number of all possible outcomes is 5·4=20 (you have 5 ways to select 1st slip and 4 ways to select 2nd slip). Then the probability of event
is
![Pr(B)=\dfrac{2}{20}=\dfrac{1}{10}.](https://tex.z-dn.net/?f=Pr%28B%29%3D%5Cdfrac%7B2%7D%7B20%7D%3D%5Cdfrac%7B1%7D%7B10%7D.)
3. Then
![Pr(A|B)=\dfrac{\frac{1}{20} }{\frac{1}{10} }=\dfrac{1}{2}.](https://tex.z-dn.net/?f=Pr%28A%7CB%29%3D%5Cdfrac%7B%5Cfrac%7B1%7D%7B20%7D%20%7D%7B%5Cfrac%7B1%7D%7B10%7D%20%7D%3D%5Cdfrac%7B1%7D%7B2%7D.)
Answer: ![\dfrac{1}{2}.](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D.)