Commutative property of multiplication
Lets see the answer to this question the fraction of the entire load of dirt in each mound is b.1/2 this is how i got my answer you multiply 3*1/6=1/2 so the fraction of the entire load of dirt that was in each mound is 1/2
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:
60 feets
Step-by-step explanation:
Given that :
Using Pythagoras rule :
Hypotenus = sqrt(Adjacent^2 + opposite^2)
From the attached picture :
x² = 87^2 - 63^2
x^2 = 7569 - 3969
X^2 = 3600
Take the square root of both sides :
X = sqrt(3600)
X = 60 feets
Hence, Katy and Tony are 60 feets apart
Answer:
The population of Hawks and other species would likely start to edge off.
Step-by-step explanation:
Commonly, a Food Chain is dependent on all the parts of it sataying at a perfect point. If the food that the Snakes eat starts to run out, more of them will die off. With less Snakes, the Hawks will then have less food, and that continues throughout the chain. The same thing would happen if there was too much food, except instead of the populations of species dying off, the Snake population would spike.