Answer:
In the long run, ou expect to lose $4 per game
Step-by-step explanation:
Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n^2 if heads comes up first on the nth toss.
Assuming X be the toss on which the first head appears.
then the geometric distribution of X is:
X
geom(p = 1/2)
the probability function P can be computed as:

where
n = 1,2,3 ...
If I agree to pay you $n^2 if heads comes up first on the nth toss.
this implies that , you need to be paid 

![\sum \limits ^{n}_{i=1} n^2 P(X=n) =Var (X) + [E(X)]^2](https://tex.z-dn.net/?f=%5Csum%20%5Climits%20%5E%7Bn%7D_%7Bi%3D1%7D%20n%5E2%20P%28X%3Dn%29%20%3DVar%20%28X%29%20%2B%20%5BE%28X%29%5D%5E2)
∵ X
geom(p = 1/2)








Given that during the game play, You pay me $10 , the calculated expected loss = $10 - $6
= $4
∴
In the long run, you expect to lose $4 per game
Answer:
y = -3x + 3
Step-by-step explanation:
Perpendicular slope: -3
y -(-3)= -3(x-2)
y+3= -3x + 6
y = -3x + 3
Answer:
Option C: 47 tickets
Step-by-step explanation:
Given that:
Tickets given to employees = 1 part
Tickets given to listeners = 2 parts
This means 141 tickets are totally divided into 3 parts so:
1 part = 141/3
1 part = 47
So now,
Tickets given to employees = 1 part = 47 tickets
Tickets given to listeners = 2 parts = 47 *2 tickets = 94 tickets
I hope it will help you!
Answer:
1st option
Step-by-step explanation:
10 + 12 ← factor out 2 ( the GCF of 10 and 12 ) from each term
= 2(5 + 6)
Answer:
0.9%
Step-by-step explanation:
Given that:
A nest of ant is represented by the exponential growth function :
P(w)=24,000(1.063)^w
Where w = number of weeks
Recall :
Growth function is generally represented as :
P(t) = A(1 + r)^t
Where ; A is the initial population figure; r = growth rate and t = time
Hence, the growth rate per week of the function given is :
(1 + r) = 1.063
r = 1.063 - 1
r = 0.063
Number of days in a week = 7
Hence, growth rate per day = r /7
0.063 / 7
= 0.009
(0.009 * 100%) = 0.9%