Answer:
21u-16v+18w
Step-by-step explanation:
(7u-4v+4w) 14u-12v+14w
21u-16v+18w
Not an expertise on infinite sums but the most straightforward explanation is that infinity isn't a number.
Let's see if there are anything we missed:
∞
Σ 2^n=1+2+4+8+16+...
n=0
We multiply (2-1) on both sides:
∞
(2-1) Σ 2^n=(2-1)1+2+4+8+16+...
n=0
And we expand;
∞
Σ 2^n=(2+4+8+16+32+...)-(1+2+4+8+16+...)
n=0
But now, imagine that the expression 1+2+4+8+16+... have the last term of 2^n, where n is infinity, then the expression of 2+4+8+16+32+... must have the last term of 2(2^n), then if we cancel out the term, we are still missing one more term to write:
∞
Σ 2^n=-1+2(2^n)
n=0
If n is infinity, then 2^n must also be infinity. So technically, this goes back to infinity.
Although we set a finite term for both expressions, the further we list the terms, they will sooner or later approach infinity.
Yep, this shows how weird the infinity sign is.
The answer will be (-2,-7)
Answer:
Use the normal distribution if the population standard deviation is known.
Use the student's t distribution when the population standard deviation is unknown.
Explanation:
A mound-shaped distribution refers to the normal distribution.
A good sample size for testing against the normal distribution should be
n >= 30.
The condition for the sample size is satisfied.
However, we are not given the population standard deviation, therefore it is assumed to be unknown.
Therefore the student's t distribution should be used.
