Answer: no because you get a decimal
Step-by-step explanation:
Answer:
a) 0.82
b) 0.18
Step-by-step explanation:
We are given that
P(F)=0.69
P(R)=0.42
P(F and R)=0.29.
a)
P(course has a final exam or a research paper)=P(F or R)=?
P(F or R)=P(F)+P(R)- P(F and R)
P(F or R)=0.69+0.42-0.29
P(F or R)=1.11-0.29
P(F or R)=0.82.
Thus, the the probability that a course has a final exam or a research paper is 0.82.
b)
P( NEITHER of two requirements)=P(F' and R')=?
According to De Morgan's law
P(A' and B')=[P(A or B)]'
P(A' and B')=1-P(A or B)
P(A' and B')=1-0.82
P(A' and B')=0.18
Thus, the probability that a course has NEITHER of these two requirements is 0.18.
Answer:
Kayla plays on average 0.2 more chess games than Becker
Step-by-step explanation:
The first thing we must do is calculate the mean of each one.
Becker:
(5 + 2 + 4 + 1 + 1 + 4 + 5 + 3 + 2 + 1) / 10 = 2.8
Kayla:
(2 + 3 + 1 + 1 + 4 + 1 + 5 + 3 + 5 + 5) / 10 = 3
if we subtract these averages, we have to:
3 - 2.8 = 0.2
which means that Kayla plays on average 0.2 more chess games than Becker. Since it is not even a complete game, the difference between the two is very small and almost irrelevant.
Answer:
the answer is b
Step-by-step explanation:
