Answer:
Answer: Yes, There is a linear correlation between the weights of the bears and their chest sizes because the absolute value of the test statistics 0.961 exceeds the critical value
Explanation:
Claim: There is a linear correlation between the weights of the bears and their chest sizes
Null hypothesis, H₀ : p=0 (there is no significant correlation)
Alternative hypothesis, H₁ : p ≠0 (there is no significant correlation)
Level of significance, α = 0.05
Decision rule: Reject H₀ if robserved ≥ rcritical
Sample correlation coefficient r = robserved = 0.961
Yes, There is a linear correlation between the weights of the bears and their chest sizes because the absolute value of the test statistics 0.961 exceeds the critical value
Answer:
v(t) = (2t + 1)i + 3t²j + 4t³k
r(t) = (t² + t)i + (t³ + 7)j + (t⁴ - 4)k
Explanation:
a(t) = 2i + 6tj + 12t²k
v(t) = ∫a(t)dt
= ∫(2i + 6tj + 12t²k)dt
= 2ti + (6t²/2)j + (12t³/3)k + c
= 2ti + 3t²j + 4t³k + c
v(0) = i
i = 0i + 0j + 0k + c
c = i
∴ v(t) = 2ti + 3t²j + 4t³k + i
v(t) = (2t + 1)i + 3t²j + 4t³k
r(t) = ∫ v(t)dt
= i ∫ (2t + 1)dt + 3j ∫ t²dt + 4k ∫ t³dt
= i (2t²/2 + t) + 3j(t³/3) + 4k(t⁴/4) + d
= i (t² + t) + jt³ + t⁴k + d
r(0) = 7j - 4k
0i + 0j + 0k + d = 7j - 4k
d = 7j - 4k
∴ r(t) = (t² + t)i + t³j + t⁴k + 7j - 4k
r(t) = (t² + t)i + (t³ + 7)j + (t⁴ - 4)k
Answer:
In this growing economy and competitive world, it is important for any business to maintain a good customer relationship by providing value to the customers. However, focusing only on profit maximization will not benefit the business in the long run.
Focusing on profit maximizing the profit will benefit the organisation in the short term and the company will only think about business interest keeping the costumers and society interest aside.
Explanation:
The likely result of having to have a discarded computer that was fully depreciated and the residual value is discarded as it is not present, the transaction will likely cause a loss equal in regards to the residual value that may be recognized.
