Answer:
(-2.4, 37.014)
Step-by-step explanation:
We are not told how to approach this problem.
One way would be to graph f(x) = x^5 − 10x^3 + 9x on [-3,3] and then to estimate the max and min of this function on this interval visually. A good graph done on a graphing calculator would be sufficient info for this estimation. My graph, on my TI83 calculator, shows that the relative minimum value of f(x) on this interval is between x=2 and x=3 and is approx. -37; the relative maximum value is between x= -3 and x = -2 and is approx. +37.
Thus, we choose Answer A as closest approx. values of the min and max points on [-3,3]. In Answer A, the max is at (-2.4, 37.014) and the min at (2.4, -37.014.
Optional: Another approach would be to use calculus: we'd differentiate f(x) = x^5 − 10x^3 + 9x, set the resulting derivative = to 0 and solve the resulting equation for x. There would be four x-values, which we'd call "critical values."
Answer:
480.18
Step-by-step explanation:
453+6%
Youre welcome
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.