use m a t h w a y
Step-by-step explanation:
In order to do mean you have to add all of the numbers in the list together. then divide them by the total amount of numbers (ex. 7,9,3,2,5. you would add all of these up 7+9+3+2+5=26 then you would divide that by 5 because there are five numbers in the list.
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Answer: Choice C) </h3><h3>
The system can only be independent and consistent</h3>
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Explanation:
Let's go through the answer choices
- A) This isn't possible. Either a system is consistent or inconsistent. It cannot be both at the same time. The term "inconsistent" literally means "not consistent". It's like saying a cup is empty and full at the same time. We can rule out choice A.
- B) This is similar to choice A and we cannot have a system be both independent and dependent. Either a system is independent or dependent, but not both. Independence means that the two equations are not tied together, while dependent equations are some multiple of each other. We can rule out choice B.
- C) We'll get back to this later
- D) The independence/dependence status is unknown without the actual equations present. However, we know 100% that this system is not inconsistent. This is because the system has at least one solution. Inconsistent systems do not have any solutions at all (eg: parallel lines that never cross). We can rule out choice D because of this.
Going back to choice C, again we don't have enough info to determine if the system is independent or dependent, but we at least know it's consistent. Consistent systems have one or more solutions. So part of choice C can be confirmed. It being the only thing left means that it has to be the final answer.
If it were me as the teacher, I'd cross out the "independent" part of choice C and simply say the system is consistent.
Answer:
See proof below
Step-by-step explanation:
An equivalence relation R satisfies
- Reflexivity: for all x on the underlying set in which R is defined, (x,x)∈R, or xRx.
- Symmetry: For all x,y, if xRy then yRx.
- Transitivity: For all x,y,z, If xRy and yRz then xRz.
Let's check these properties: Let x,y,z be bit strings of length three or more
The first 3 bits of x are, of course, the same 3 bits of x, hence xRx.
If xRy, then then the 1st, 2nd and 3rd bits of x are the 1st, 2nd and 3rd bits of y respectively. Then y agrees with x on its first third bits (by symmetry of equality), hence yRx.
If xRy and yRz, x agrees with y on its first 3 bits and y agrees with z in its first 3 bits. Therefore x agrees with z in its first 3 bits (by transitivity of equality), hence xRz.
Answer:
What is the Question so I can answer?
Step-by-step explanation:
