A "solution" would be a set of three numbers ... for Q, a, and c ... that
would make the equation a true statement.
If you only have one equation, then there are an infinite number of triplets
that could do it. For example, with the single equation in this question,
(Q, a, c) could be (13, 1, 2) and they could also be (16, 2, 1).
There are infinite possibilities with one equation.
In order to have a unique solution ... three definite numbers for Q, a, and c ...
you would need three equations.
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.
The new price would be $19.99
29.99 x .6666 = $19.99
(Old price multiplied by discount percentage = new price)
There are a few definitions depending on your context ( which is not that great btw) but some possible definitions are - Congruent, Equivalent, Proportionate