Answer:
-24 - 57i
Step-by-step explanation:
i'm assuming that this is a question about imaginary numbers and that i²= -1
we can expand the binomial by using the FOIL method (see attached)
(6 − 7i)(3 − 6i)
= (6)(3) + (6)(-6i) + (-7i)(3) + (-7i)(-6i)
=18 - 36i - 21i + 42i² (recall that i²= -1)
= 18 - 57i + 42(-1)
= 18 - 57i - 42
= -24 - 57i
Answer:
(a) P (Y = 3) = 0.0844, P (Y ≤ 3) = 0.8780
(b) The probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.
Step-by-step explanation:
The random variable <em>Y</em> is defined as the number of consecutive time intervals in which the water supply remains below a critical value <em>y₀</em>.
The random variable <em>Y</em> follows a Geometric distribution with parameter <em>p</em> = 0.409<em>.</em>
The probability mass function of a Geometric distribution is:
(a)
Compute the probability that a drought lasts exactly 3 intervals as follows:
Thus, the probability that a drought lasts exactly 3 intervals is 0.0844.
Compute the probability that a drought lasts at most 3 intervals as follows:
P (Y ≤ 3) = P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3)
Thus, the probability that a drought lasts at most 3 intervals is 0.8780.
(b)
Compute the mean of the random variable <em>Y</em> as follows:
Compute the standard deviation of the random variable <em>Y</em> as follows:
The probability that the length of a drought exceeds its mean value by at least one standard deviation is:
P (Y ≥ μ + σ) = P (Y ≥ 1.445 + 1.88)
= P (Y ≥ 3.325)
= P (Y ≥ 3)
= 1 - P (Y < 3)
= 1 - P (X = 0) - P (X = 1) - P (X = 2)
Thus, the probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.