Ok, so:
For Part A, we have: P(Z|A)=P(Z and A)/P(A)
And if we replace, we got:
P(Z|A) = (0.15)/(0.25) and this is equal to 0.6.
For Part B, we have: P(A|Z)=P(Z and A)/P(Z)
P(A|Z) = (0.15)/(0.73) and this is equal to 0.205.
Answer:
43
Step-by-step explanation:
If you apply the or both
Only 1 of the students would need to know the "or both", therefore maximizing the remaining amount of students you can put in.
Gerald, let's call him, knows French AND German, so there's only one less student that knows french and german. Gerald is 1 student.
MAXIMUM:
There are now 14 monolinguistic French speakers and 16 monolinguistic German's, 30 students + Gerald=31.
Minimum:
As a bonus, the minimum is 15 students knowing french AND German and only 2 monolinguistic German speakers, so 17.
Answer:
The correct option is C). When it was purchased, the coin was worth $6
Step-by-step explanation:
Given function is f(t)=
Where t is number of years and f(t) is function showing the value of a rare coin.
A figure of f(t) shows that the graph has time t on the x-axis and f(t) on the y-axis.
Also y-intercept at (0,6)
hence, when time t was zero, the value of a rare coin is 6$
f(t)=
f(0)=
<em>f(0)=6</em>
Thus,
The correct option is C). When it was purchased, the coin was worth $6
