The applicable identity is ...
... cos(a)cos(b) = (1/2)(cos(a+b) +cos(a-b))
Let a=36t, b=6t and substitute this into your expression. Then you have ...
... 20cos(36t)cos(6t) = 20·(1/2)(cos(36t+6t) + cos(36t-6t))
... = 10(cos(42t) + cos(30t))
... = 10cos(42t) +10cos(30t)
5/12 or approximately 0.417
I dont know sorry about that
Answer:
Step-by-step explanation:
Given that Z is a standard normal variate.
We are to calculate the probabilities as given
F(z) represents the cumulative probability i.e. P(Z<z)
a. P(z ≤ −1)
=F(-1)
= 0.158655
b. P(z > .95)
= 1-F(0.95)
= 0.1711
c. P(z ≥ −1.5)
= 1-F(-1.5)
= 0.9332
d. P(−.5 ≤ z ≤ 1.75)
=F(1.75)-F(-0.5)
= 0.6514
e. P(1 < z ≤ 3)
=F(3)-F(1)
=0.1573
-2 bc even tho there is a zero-2 will always be the lowest
