Answer: $37.20
Step-by-step explanation:
Just add those 2 amounts together
Answer:
P(B|A)=0.25 , P(A|B) =0.5
Step-by-step explanation:
The question provides the following data:
P(A)= 0.8
P(B)= 0.4
P(A∩B) = 0.2
Since the question does not mention which of the conditional probabilities need to be found out, I will show the working to calculate both of them.
To calculate the probability that event B will occur given that A has already occurred (P(B|A) is read as the probability of event B given A) can be calculated as:
P(B|A) = P(A∩B)/P(A)
= (0.2) / (0.8)
P(B|A)=0.25
To calculate the probability that event A will occur given that B has already occurred (P(A|B) is read as the probability of event A given B) can be calculated as:
P(A|B) = P(A∩B)/P(B)
= (0.2)/(0.4)
P(A|B) =0.5
Answer:
FH ~ 10.02
Step-by-step explanation:
1. Approach
One should first find the circumference of the given circle. Then one should find how large the fraction of the circumference one is supposed to find is. Finally, one should multiply the fraction of the circumference one is supposed to find by the total circumference.
2. Circumference of the circle
The formula for circumference is;
π
Substitute in the given values;
It is given that the radius is, hence
2 (7) π
14π
3. Find the fraction of the circumference one is supposed to find
It is given that the angles over the measure of the total degrees of angles in a circle are equal to the arc surrounding the angles of the circumference. Essentially;

Substitute in the given information and solve;

arc = 
arc = 
arc ~ 10.02

is a right triangle with base length 1 and height 8, so the area of

is

.
The average value of

over

is given by the ratio

The denominator is just the area of

, which we already know. The average value is then simplified to

In the

-plane, we can describe the region

as all points

that lie between the lines

and

(the lines which coincide with the triangle's base and hypotenuse, respectively), taking

. So, the integral is given by, and evaluates to,



Answer:
-9.Try to caculate it, My friend