Answer:
The coefficient matrix is singular, so there are no solutions or infinite solutions.
The two equations in the linear system are the same, so there are infinite solutions.
Yes, there are solutions
Step-by-step explanation:
Answer: A. 3 ways: k, DE, ED (both DE and ED have line markers over top)
To name a line, we just need two points on the line. We list them in any order because the line extends forever in both directions. Contrast this with a ray where order does matter. The little k is another way to name a line, potentially simplifying things.
Choice B is close, but it mentions ray DE instead of line DE. Choice C is missing line ED. Choice D is a similar story as choice B. These facts allow us to rule out B through D.
9r=3r+6
Subtract 3r from both sides
6r=6
Divide both sides by 6
r=1
Answer:
495 combinations of 4 students can be selected.
Step-by-step explanation:
The order of the students in the sample is not important. So we use the combinations formula to solve this question.
Combinations formula:
is the number of different combinations of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
How many combination of random samples of 4 students can be selected?
4 from a set of 12. So
![C_{n,x} = \frac{12!}{4!(8)!} = 495](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7B12%21%7D%7B4%21%288%29%21%7D%20%3D%20495)
495 combinations of 4 students can be selected.
Answer:
it would be 6
Step-by-step explanation: