Set Events:
T=tests positive~T=tests negativeP=subject is pregnant~P=subject is not pregnant
We are givenP(T n ~P)=0.02P(~T n P)=0.03P(P)=0.7
recall by definition of conditional probabilityP(A|B)=P(A n B)/P(B)
Need to find P(P|~T)
First step: make a contingency diagram of probabilities (intersection, n)
P ~P sum
T 0.67 0.02 0.69=P(T)
~T 0.03 0.28 0.31=P(~T)
sum 0.70 0.30 1.00
=P(P) =P(~P)
therefore
P(P|~T)=P(P n ~T)/P(~T)=0.03/0.31 [ both read off the contingency table ]
=0.0968
Answer:
is that even possible.
Step-by-step explanation:
Answer:
the coefficient are 9 and -8
the constant is 6........
For this case, we have to:
By definition, we know:
The domain of
is given by all real numbers.
Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root. Thus, it will always be defined.
So, we have:
with
:
is defined.
with
is also defined.
has a domain from 0 to ∞.
Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.
Thus, we observe that:
is not defined, the term inside the root is negative when
.
While
if it is defined for
.
Answer:

Option b
The answer is d which is 6/5