Response:
Thorough analysis:
I believe that option 4 is the right choice
Answer:
0.40
Step-by-step explanation:
The percentage of members who engage only in long-distance running is 8%
Therefore, the probability that a member focuses solely on long-distance running is P(A) = 0.08
The percentage of members who participate exclusively in field events is 32%
Thus, the probability of a member competing only in field events is P(B) = 0.32
The percentage of members acting as sprinters is 12%
So, the probability that a member is a sprinter is P(C) = 0.12
We need to determine the probability that a team member is either an exclusive long-distance runner or an only field event competitor, which equates to finding P(A or B). Since these two events cannot occur simultaneously, we can express this as:
P(A or B) = P(A) + P(B)
Substituting the known values results in:
P(A or B) = 0.08 + 0.32 = 0.40
Thus, the likelihood that a randomly selected team member runs exclusively long-distance or participates solely in field events stands at 0.40
The equation of the perpendicular line can be identified by determining its slope and applying the given point within the standard formula.
Standard equation: y-y1 = m(x-x1)
m*m'=-1
where m' indicates the slope of the perpendicular line
m denotes the slope of the original line
m = -coefficient of x/coefficient of y = -4/-3 = 4/3
m' = -3/4
Substituting the point (3, -2):
y+2 = -3/4*(x-3)
4y+8 = -3x+9
Thus, the equation of the perpendicular line is: 3x+4y-1=0
c) Step-by-step breakdown: The collision rate is 1.2 incidents per 4 months, which can be expressed as 0.3 incidents monthly. Therefore, the Poisson distribution for the variable X representing monthly collisions is defined as P(X = x) =... for x ∈ N ∪ {0} = 0 otherwise. (1) Where X = 0 denotes no collisions during a 4-month timeframe, substituting gives P(X = 0) =... (2). For a 4-month period, P(No collision in 4 month period) =... (3). Two collisions in a 2-month span translate to 1 per month, thus P(X =1) =... (4). Over 2 months, P(2 collisions in a 2 month period) =... (5). One collision over a 6-month period equates to P(1 collision in 6 months period) =... (6). Consequently, P(1 collision in 6 month period) results in... (7). For no collisions in a 6-month period, P(No collision in 6 months period) =... (8). Finally, the probability of 1 or fewer collisions over six months is P(1 or fewer collision in 6 months period) = (8) + (7) = 0.0785 + 0.1653.
