Answer:
The slope of the line of best fit is 
⇒ 2nd option
Step-by-step explanation:
The formula of the slope of a line is 
<em>To find the slope of the best line fit choose two points their positions make the number of the points over the line equal to the number of the points below the line</em>
From the attached graph points (1 , 9) and (8 , 3) are the best choice
∵ The line passes through points (1 , 9) and (8 , 3)
∴ 
= 1 and 
= 8
∴ 
= 9 and 
= 3
- Substitute them in the formula of the slope
∴ 
∴ The slope of the line of best fit is
Answer:
<em>(9,-5)</em>
Step-by-step explanation:
When you go down, the number will decrease. The vertical position is the Y axis. -3-2=5.
When you go right, the number will increase. The horizontal position is the X axis. 5+4=9
<em>(9,-5)</em>
<u>Hope this helps :-)</u>
Answer:
The answer is
<h2>( 3 , - 1)</h2>
Step-by-step explanation:
The midpoint M of two endpoints of a line segment can be found by using the formula
where
(x1 , y1) and (x2 , y2) are the points
From the question the points are
(2, -3) and (4, 1)
The midpoint is
We have the final answer as
<h3>( 3 , - 1)</h3>
Hope this helps you
Answer:
The students should request an examination with 5 examiners.
Step-by-step explanation:
Let <em>X</em> denote the event that the student has an “on” day, and let <em>Y</em> denote the
denote the event that he passes the examination. Then,
The events (
) follows a Binomial distribution with probability of success 0.80 and the events (
) follows a Binomial distribution with probability of success 0.40.
It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,
Then,
⇒
Then,
Compute the probability that the students passes if request an examination with 3 examiners as follows:
![=[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}](https://tex.z-dn.net/?f=%3D%5B%5Csum%5Climits%5E%7B3%7D_%7Bx%3D2%7D%7B%7B3%5Cchoose%20x%7D%280.80%29%5E%7Bx%7D%281-0.80%29%5E%7B3-x%7D%7D%5D%5Ctimes%5Cfrac%7B2%7D%7B3%7D%2B%5B%5Csum%5Climits%5E%7B3%7D_%7Bx%3D2%7D%7B%7B3%5Cchoose%20x%7D%280.40%29%5E%7B3%7D%281-0.40%29%5E%7B3-x%7D%7D%5D%5Ctimes%5Cfrac%7B1%7D%7B3%7D)
The probability that the students passes if request an examination with 3 examiners is 0.715.
Compute the probability that the students passes if request an examination with 5 examiners as follows:
![=[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}](https://tex.z-dn.net/?f=%3D%5B%5Csum%5Climits%5E%7B5%7D_%7Bx%3D3%7D%7B%7B5%5Cchoose%20x%7D%280.80%29%5E%7Bx%7D%281-0.80%29%5E%7B5-x%7D%7D%5D%5Ctimes%5Cfrac%7B2%7D%7B3%7D%2B%5B%5Csum%5Climits%5E%7B5%7D_%7Bx%3D3%7D%7B%7B5%5Cchoose%20x%7D%280.40%29%5E%7Bx%7D%281-0.40%29%5E%7B5-x%7D%7D%5D%5Ctimes%5Cfrac%7B1%7D%7B3%7D)
The probability that the students passes if request an examination with 5 examiners is 0.734.
As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.
The answer is 4 units. The distance between the two points of B and C is equals 6-4=2, because the y is the same. According to the question, the BC is equals 0.5*B'C'. So the length of B'C' is 2/0.5=4 units.
