Answer:
0.05
Step-by-step explanation:
checked via khan academy
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:
19.2
Step-by-step explanation:
9.6*2=19.2
Answer:
(3/4)a
Step-by-step explanation:
The angle at K is 120°, so the angle at L is its supplement: 60°. That makes triangle FKL an equilateral triangle with a base of FL = a. The vertex at K is centered over the base, so is a/2 from G.
The midsegement length is the average of GK and FL, so is ...
midsegment = (GK +FL)/2 = (a/2 +a)/2
midsegment = (3/4)a
Answer:
The answer in the attached figure
Step-by-step explanation:
Method 1
we have

Rewrite the equation as a compound system of equations
-----> equation A
-----> equation B
Using a graphing tool
The solution of the system of equations is the intersection point both graphs
see the attached figure
the solution is 
Method 2
we have

Group terms that contain the same variable and move the constant to the other side of equation

Combine like terms

Divide by
both sides
