∑x = 1 + 2 + 3 + 4 + 5 + 6 = 21
∑y = 8 + 3 + 0 + 1 + 2 + 1 = 15
∑x^2 = 1 + 4 + 9 + 16 + 25 + 36 = 91
∑y^2 = 64 + 9 + 0 + 1 + 4 + 1 = 79
∑xy = 8 + 6 + 0 + 4 + 10 + 6 = 34
r
= (n∑xy - ∑x∑y)/(sqrt(n∑x^2 - (∑x)^2)*sqrt(n∑y^2 - (∑y)^2)) = (6(34) -
21(15))/(sqrt(6(91) - (21)^2)*sqrt(6(79) - (15)^2)) = (204 -
315)/(sqrt(546 - 441)*sqrt(474 - 225)) = -111/(sqrt(105)*sqrt(249)) =
-111/(10.25*15.78) = -111/161.7 = -0.68
For the answer to the question above, we will use this formula to solve this problem
<span>d = kv^2 </span>
<span>plugging in, </span>
<span>4.2 = k*10^2 </span>
<span>k = 0.042 </span>
<span>d = 0.042v^2 </span>
<span>the revoised equation will be </span>
<span>d = 0.9*0.042v^2 , i.e. </span>
<span>d = 0.0378v^2
I hope my answer helped you</span>
Yes. As long as it is linear, it will be continuous no matter the numbers. Now, I have no idea what a "real" number is, but I hope this helped.
<span>Toni can carry five books in her backpack. You must start by subtracting the weight of the other required items. Toni's lunch which is one pound and gym clothes which are two pounds. This can be calculated by 18-1-2=15. Toni has 15 pounds left to carry books. She should divide this by the weight of each book, which is three pounds. 15/3=5. Therefore, Toni can carry five books.</span>
Answer:
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