Answer:
You know that you can look it up then you can see the answer right away
Step-by-step explanation:
:)
Answer:
I think elimination with subtraction would be the best answer choice, since all the numbers are positive.
Answer:
m=1/3
equation y=1/3x+10
Step-by-step explanation:
Find the rise and run.
1. -6--9=run=3
2. 8-7=rise=1
Slope intercept form of a line is y=mx+b where m= slope and b=y-intercept.
Now compare this form with the tiven equation y=4x-9 to get the value of m.
So, m= 4.
Slope of all parallel lines are always equal. Which means the line which is parallel to this has also the same slope: m=4.
Now we need to find the value of b to get the equation. That parallel line passes through (-5,3).
The point intercept form of a line is,
(y-y1)=m(x-x1)
So, we can plug in the m=4, x1=-5 and y1=3 in the above equation. Hence,
y-3=4(x-(-5))
y-3=4(x+5) Since -(-5)=+5
y-3=4x+20 By distributing property.
y=4x+20+3 Adding 3 to each sides.
y=4x+23
So, the slope-intercept form of that parallel line is y=4x+23.
Answer:
a

b

Step-by-step explanation:
From the question we are told that
The number of students in the class is N = 20 (This is the population )
The number of student that will cheat is k = 3
The number of students that he is focused on is n = 4
Generally the probability distribution that defines this question is the Hyper geometrically distributed because four students are focused on without replacing them in the class (i.e in the generally population) and population contains exactly three student that will cheat.
Generally probability mass function is mathematically represented as

Here C stands for combination , hence we will be making use of the combination functionality in our calculators
Generally the that he finds at least one of the students cheating when he focus his attention on four randomly chosen students during the exam is mathematically represented as

Here




Hence


Generally the that he finds at least one of the students cheating when he focus his attention on six randomly chosen students during the exam is mathematically represented as

![P(X \ge 1) =1- [ \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}]](https://tex.z-dn.net/?f=P%28X%20%20%5Cge%201%29%20%3D1-%20%5B%20%20%5Cfrac%7B%5E%7Bk%7DC_x%20%2A%20%5E%7BN-k%7DC_%7Bn-x%7D%7D%7B%5E%7BN%7DC_n%7D%5D%20)
Here n = 6
So
![P(X \ge 1) =1- [ \frac{^{3}C_0 * ^{20 -3}C_{6-0}}{^{20}C_6}]](https://tex.z-dn.net/?f=P%28X%20%20%5Cge%201%29%20%3D1-%20%5B%20%20%5Cfrac%7B%5E%7B3%7DC_0%20%2A%20%5E%7B20%20-3%7DC_%7B6-0%7D%7D%7B%5E%7B20%7DC_6%7D%5D%20)
![P(X \ge 1) =1- [ \frac{^{3}C_0 * ^{17}C_{6}}{^{20}C_6}]](https://tex.z-dn.net/?f=P%28X%20%20%5Cge%201%29%20%3D1-%20%5B%20%20%5Cfrac%7B%5E%7B3%7DC_0%20%2A%20%5E%7B17%7DC_%7B6%7D%7D%7B%5E%7B20%7DC_6%7D%5D%20)
![P(X \ge 1) =1- [ \frac{1 * 12376}{38760}]](https://tex.z-dn.net/?f=P%28X%20%20%5Cge%201%29%20%3D1-%20%5B%20%20%5Cfrac%7B1%20%20%2A%20%2012376%7D%7B38760%7D%5D%20)

