Part (a)
P(A) = 0.5
P(B) = 0.4
P(B/A) = 0.6
P(A and B) = P(A)*P(B/A)
P(A and B) = 0.5*0.6
P(A and B) = 0.3
<h3>Answer: 0.3</h3>
==========================================
Part (b)
We'll use the result from part (a)
P(A or B) = P(A) + P(B) - P(A and B)
P(A or B) = 0.5 + 0.4 - 0.3
P(A or B) = 0.6
<h3>Answer: 0.6</h3>
===========================================
Part (c)
A and B are not independent since P(B) does not equal P(B/A). The fact that event A happens changes the probability P(B). Recall that P(B/A) means "probability P(B) based on event A already happened". A and B are independent if P(B) = P(B/A).
Events A and B are not mutually exclusive since P(A or B) is not zero.
<h3>Answer: Neither</h3>
Answer:
5(x+2)
Step-by-step explanation:
If tan0=-3/4 and 0 is in quadrant IV, cos20= (33/25, -17/25, 32/25, 7/25, 24/25?) and tan20= (24/7, -24/7, 7/25, -7/25, 13/7, -1
Oksana_A [137]
Answer:
- cos(2θ) = 7/25
- tan(2θ) = -24/7
Step-by-step explanation:
Sometimes, it is easiest to let a calculator do the work. (See below)
__
The magnitude of the tangent is less than 1, so the reference angle will be less than 45°. Then double the angle will be less than 90°, so will remain in the 4th quadrant, where the cosine is positive and the tangent is negative.
You can also use the identities ...
cos(2θ) = (1 -tan(θ)²)/(1 +tan(θ)²)
cos(2θ) = (1 -(-3/4)²)/(1 +(-3/4)²) = ((16-9)/16)/((16+9)/16)
cos(2θ) = 7/25
__
tan(2θ) = 2tan(θ)/(1 -tan(θ)²) = 2(-3/4)/((16-9/16) = (-6/4)(16/7)
tan(2θ) = -24/7
2km=2000m
25m=25m
2500cm=25m
3000mm=3m
Hence the longest is 2km=2000m
You can get a vertical asymptote at x=1 using y = 1/(x-1)
You can generate a hole at x=3 by multiplying by (x - 3/(x - 3) which is undefined at x=3 but otherwise equals 1
You can move the horizontal asymptote up to y=2 by adding 2
y = (x - 3)/((x - 1)(x - 3)) + 2
