Answer:
B. One of the populations is normally distributed.
Step-by-step explanation:
To test a claim about two population standard deviation or variance, it is imperative that the data meets certain requirements which include :
Randomness : Data must not be biased as such it must be drawn as a random sample from a larger group.
The data must be independent. That is not related to one another, the outcome of one should not rely on the outcome or value of another.
Both groups must be drawn From a population which is normally distributed.
One group being normally distributed by stribuyed while the other isn't a requirement for hypothesis testing in this scenario.
Interesting factoid: Only perfect squares have an odd number of factors. So we know to try 4, 9, and 16, and ignore the rest. It's 16, whose factors are 1, 2, 4, 8, and. 16 .
Answer:
Sₙ = √5/2 n (n + 1)
Step-by-step explanation:
<u>Given AP:</u>
<u>Rewriting as:</u>
- √5,√4*5,√9*5,√16*5,...
- √5, 2√5, 3√5, 4√5,..., n√5
<u>Common difference:</u>
<u>Sum of n terms:</u>
- Sₙ = 1/2n (a₁ + aₙ) =
- 1/2n (√5 + √5 + (n-1)√5) =
- 1/2n (n + 1)√5=
- √5/2 n (n + 1)
Sₙ = √5/2 n (n + 1)
NO
16 ounce = 1 pound
22 ounce = 1.375 pound ≠ 6 pound
