Answer:
Step-by-step explanation:
y = sin(t^2)
y' = 2tcos(t^2)
y'' = 2cos(t^2) - 4t^2sin(t^2)
so the equation become
2cos(t^2) - 4t^2sin(t^2) + p(t)(2tcos(t^2)) + q(t)sin(t^2) = 0
when t=0, above eqution is 2. That is, there does not exist the solution. so y can not be a solution on I containing t=0.
Answer:
The correct option is;
Low
Step-by-step explanation:
Given that the P-value of the linear correlation = 0.001, we have that the P-value is a demonstration that a linear correlation that has a value in the range of the given correlation is ,most arguably very low
From the z-table, a P-value of 0.001 corresponds to a z-value of -3.09, we have that in a normal distribution since 95% of the scores have a z-score of between -2 and 2, the z-score of -3.09 is very distant from the mean and having a low value, whereby the P-value shows that the likelihood of finding another linear correlation that is as far from the mean as the given correlation is very low.
Answer:
(a - b)(5x - 2y)
Step-by-step explanation:
5x(a - b) – 2y(a - b) =
The factors in parentheses (shown in bold) are a common factor.
Factor (a - b) out from both terms.
= (a - b)(5x - 2y)
11+i
that’s the answer hope it helps:)
(-2 + 0.8) ÷ (1²-1.3) =
(-1.2) ÷ (1-1.3)
(-1.2) ÷ (-0.3)
= 4
