Answer:
a)0.024
b)0.148
Explanation:
Let 's represent the set of deer ticks Carrying Lyme disease with L and the set of deer ticks carrying Human Granulocytic Ehrlichiosis with H
Given:
P(L) = 0.16
P(H) = 0.10
P(L n H) = 0.1 ·P( L u H )
Hence, P( L u H) = 10 ·P( L nH)
(a)
Hence. using the equation. P(L U H) = P(L) + P(H) - P(L n H)
Hence, 10 · P(L n H ) = 0.16 + 0.1 - P(L n H )
Hence, 11 · P(L n H) = 0.16 + 0.1 = 0.26
Hence, P(L n H) =
0.26/11=0.024
(b)
We know that condition probability P(H ║ L) = p(L n H)/P(L)
hence, P(H ║ L) =(0.26/11)/0.16 =0.148
Answer:
The answer to your question is : vf = 15.18 m/s
Explanation:
Data
vo = 24 m/s
d = 120 m
vf = ? when d = 60.0 m
Formula
vf² = vo² + 2ad
For d =100m
a = (vf² - vo²) / 2d
a = (0 -24²) / 2(100)
a = -576/200
a = 2.88 m/s²
Now, when d = 60
vf² = (24)² - 2(2.88)(60)
vf² = 576 - 345.6
vf² = 230.4
vf = 15.18 m/s
