Find p(-2), for p(t) = 2 + t + 2t3 – t2
1 answer:
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Answer:
Part A
The rotation of the point (x, y) 180° about the origin gives the point (-x, -y)
Therefore, we have the points, ΔLMN, L(1, 1), and M(2, 2), and N(3, 3)
The coordinates of the vertices of the image, ΔL'M'N' are L'(-1, -1), and M'(-2, -2), and N'(-3, -3)
Please find attached the graph of triangle ΔLMN and ΔL'M'N'
Part B
The lines drawn through L and L' and through M and M' are colinear
Part C
A line is defined by two points, such as <em>L </em>and <em>M.</em> Rotation of a line through 180° about the origin will give an image location on the same path as the extension of the original line.
The third point, <em>N</em>, however, defines a plane, and the rotation of a plane by 180° will give an image which is turned upside down with regards to the preimage
Therefore, a trough in the preimage becomes a peak in the image and the lines drawn through N and N' crosses the colinear lines drawn throgh M and M' and L and L'
Step-by-step explanation:
Answer:
a) 0.96
b) 0.016
c) 0.018
d) 0.982
e) x = 2
Step-by-step explanation:
We are given with the Probability density function f(x)= 2/x^3 where x > 1.
<em>Firstly we will calculate the general probability that of P(a < X < b) </em>
P(a < X < b) = =
= { Because
}
= =
= =
a) Now P(X < 5) = P(1 < X < 5) {because x > 1 }
Comparing with general probability we get,
P(1 < X < 5) = =
= 0.96 .
b) P(X > 8) = P(8 < X < ∞) = 1/ - 1/∞ = 1/64 - 0 = 0.016
c) P(6 < X < 10) = =
= 0.018 .
d) P(x < 6 or X > 10) = P(1 < X < 6) + P(10 < X < ∞)
= + (1/
- 1/∞) = 1 - 1/36 + 1/100 + 0 = 0.982
e) We have to find x such that P(X < x) = 0.75 ;
⇒ P(1 < X < x) = 0.75
⇒ = 0.75
⇒ = 1 - 0.75 = 0.25
⇒ =
⇒
= 4 ⇒ x =
Therefore, value of x such that P(X < x) = 0.75 is 2.