Answer:
XY 
Represent a distance √
Represent a segment √
Represent a numerical value √
Represent a geometric figure √
Can be found by applying the ruler postulate √
Can be be duplicated with a compass and a straight edge √
Step-by-step explanation:
The difference between XY and are that the points XY represents the distance between two points having a numerical value whose value can be found by using a ruler to find difference in the numbers on the ruler that coincides with the points, while
Answer:
(a) PC(C)= 
(b) E[C] = 24 cents
Step-by-step explanation:
Given:
Cost to receive a photo = 20 cents
Cost to send a photo = 30 cents
Probability of receiving a photo = 0.6
Probability of sending a photo = 0.4
We need to find
(a) PC(c)
(b) E[C]
Solution:
(a)
PC(C)= 
(b)
Expected value can be calculated by multiplying probability with cost.
E[C] = Probability × cost
E[C] = 
A is the correct letter name for D1 i believe
Answer:
9 teams
Step-by-step explanation:
If the total games played was 36 and no team played each other twice, we need to ensure there isn't any double counting.
36 = (n-1) + (n-2) + (n-3) ... + (n-(n-1))
using this knowledge, we can then count up:
1+2+3+4+5+6+7+8 = 36
If our highest number is 8, then we know there must be 9 teams, because no team can play themselves.
Answer:
=(k−1)*P(X>k−1) or (k−1)365k(365k−1)(k−1)!
Step-by-step explanation:
First of all, we need to find PMF
Let X = k represent the case in which there is no birthday match within (k-1) people
However, there is a birthday match when kth person arrives
Hence, there is 365^k possibilities in birthday arrangements
Supposing (k-1) dates are placed on specific days in a year
Pick one of k-1 of them & make it the date of the kth person that arrives, then:
The CDF is P(X≤k)=(1−(365k)k)/!365k, so the can obtain the PMF by
P(X=k) =P (X≤k) − P(X≤k−1)=(1−(365k)k!/365^k)−(1−(365k−1)(k−1)!/365^(k−1))=
(k−1)/365^k * (365k−1) * (k−1)!
=(k−1)*(1−P(X≤k−1))
=(k−1)*P(X>k−1)
