I'm willing to help you with both parts
Answer:
y=-1/2x-3
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:
5y+6x-40=0
Step-by-step explanation:
y-2=-6/5(x-5)
5y-10=-6(x-5)
5y-10=-6x+30
5y+6x-10-30=0
5y+6x-40=0
