The best way to solve this is by giving a visual idea of the glass ornaments. so We will have letter G represent Green, O for Orange, B for Brown, P for Purple.
Sample 1:G,O,B,P
Sample 2:G,P,O,B
Sample 3:P,G,O,B
Sample 4:O,G,P,B
Sample 5:G,O,P,B
This is what you can do this should be the right amount of ways to arrange them.
You would multiply the power by the powers of both exponents
2 × 3 = 6 so a^6
3 × 4 = 12 so b^12
Answer is
a^6b^12
<u>Answer-</u>
<em>A. strong negative correlation.</em>
<u>Solution-</u>
<u>Direction of a relationship</u>
- Positive- If one variable increases, the other tends to also increase. If one decreases, the other tends to also. It is represented by positive numbers(i.e 0 to 1).
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Negative- If one variable increases, the other tends to decrease, and vice-versa. It is represented by negative numbers(i.e 0 to -1)
<u>Strength of a relationship</u>
- Perfect Relationship- When two variables are linearly related, the correlation coefficient is either 1 or -1. They are said to be perfectly linearly related, either positively or negatively.
- No relationship- When two variables have no relationship at all, their correlation is 0.
As in this case, correlation coefficient was found to be -0.91, which is negative and close to -1, so it is a strong negative correlation.
Answer:
See answer below
Step-by-step explanation:
The statement ‘x is an element of Y \X’ means, by definition of set difference, that "x is and element of Y and x is not an element of X", WIth the propositions given, we can rewrite this as "p∧¬q". Let us prove the identities given using the definitions of intersection, union, difference and complement. We will prove them by showing that the sets in both sides of the equation have the same elements.
i) x∈AnB if and only (if and only if means that both implications hold) x∈A and x∈B if and only if x∈A and x∉B^c (because B^c is the set of all elements that do not belong to X) if and only if x∈A\B^c. Then, if x∈AnB then x∈A\B^c, and if x∈A\B^c then x∈AnB. Thus both sets are equal.
ii) (I will abbreviate "if and only if" as "iff")
x∈A∪(B\A) iff x∈A or x∈B\A iff x∈A or x∈B and x∉A iff x∈A or x∈B (this is because if x∈B and x∈A then x∈A, so no elements are lost when we forget about the condition x∉A) iff x∈A∪B.
iii) x∈A\(B U C) iff x∈A and x∉B∪C iff x∈A and x∉B and x∉C (if x∈B or x∈C then x∈B∪C thus we cannot have any of those two options). iff x∈A and x∉B and x∈A and x∉C iff x∈(A\B) and x∈(A\B) iff x∈ (A\B) n (A\C).
iv) x∈A\(B ∩ C) iff x∈A and x∉B∩C iff x∈A and x∉B or x∉C (if x∈B and x∈C then x∈B∩C thus one of these two must be false) iff x∈A and x∉B or x∈A and x∉C iff x∈(A\B) or x∈(A\B) iff x∈ (A\B) ∪ (A\C).
Answer:
$0.75
Step-by-step explanation:
if Felix wants 1/4 kilogram of marmalade and it cost $3 per kilogram, you divide 3/4= 0.75
