Answer:
a) ![P(X>2)= 1-P(X \leq 2) = 1-[P(X=0)+P(X=1)+P(X=2)]](https://tex.z-dn.net/?f=%20P%28X%3E2%29%3D%201-P%28X%20%5Cleq%202%29%20%3D%201-%5BP%28X%3D0%29%2BP%28X%3D1%29%2BP%28X%3D2%29%5D)
And we can find the individual probabilities like this:



And replacing we got:
![P(X>2)= 1-P(X \leq 2) = 1-[0.4493+0.3595+0.1438]=0.0474](https://tex.z-dn.net/?f=%20P%28X%3E2%29%3D%201-P%28X%20%5Cleq%202%29%20%3D%201-%5B0.4493%2B0.3595%2B0.1438%5D%3D0.0474%20)
b) 
Step-by-step explanation:
Let X the random variable that represent the number of hurricanes hitting the coast of Florida annualle. We know that
The probability mass function for the random variable is given by:
And f(x)=0 for other case.
For this distribution the expected value is the same parameter
Part a
For this case we want this probability: 
And for this case we can use the complement rule like this:
![P(X>2)= 1-P(X \leq 2) = 1-[P(X=0)+P(X=1)+P(X=2)]](https://tex.z-dn.net/?f=%20P%28X%3E2%29%3D%201-P%28X%20%5Cleq%202%29%20%3D%201-%5BP%28X%3D0%29%2BP%28X%3D1%29%2BP%28X%3D2%29%5D)
And we can find the individual probabilities like this:



And replacing we got:
![P(X>2)= 1-P(X \leq 2) = 1-[0.4493+0.3595+0.1438]=0.0474](https://tex.z-dn.net/?f=%20P%28X%3E2%29%3D%201-P%28X%20%5Cleq%202%29%20%3D%201-%5B0.4493%2B0.3595%2B0.1438%5D%3D0.0474%20)
Part b
Using the probability mass function we have:

So taking this as if there are 13 dozens of cookies. First we do 13 times 12 which would be 156 cookies in total. Then 156 cookies times 2.08 is 324.48
Answer:
Yes he has enough fence to go round the flower bed.
Step-by-step explanation:
1 yard = 3 feet
25 feet = 25/3 = 8.33 yards
Yes he has enough fence to go round the flower bed.
The answer that I got is C=-6
Answer:
160m/s
Step-by-step explanation:
The object can hit the ground when t = a; meaning that s(a) = s(t) = 0
So, 0 = -16a² + 400
16a² = 400
a² = 25
a = √25
a = 5 (positive 5 only because that's the only physical solution)
The instantaneous velocity is
v(a) = lim(t->a) [s(t) - s(a)]/[t-a)
Where s(t) = -16t² + 400
and s(a) = -16a² + 400
v(a) = Lim(t->a) [-16t² + 400 + 16a² - 400]/(t-a)
v(a) = Lim(t->a) (-16t² + 16a²)/(t-a)
v(a) = lim (t->a) -16(t² - a²)(t-a)
v(a) = -16lim t->a (t²-a²)(t-a)
v(a) = -16lim t->a (t-a)(t+a)/(t-a)
v(a) = -16lim t->a (t+a)
But a = t
So, we have
v(a) = -16lim t->a 2a
v(a) = -32lim t->a (a)
v(a) = -32 * 5
v(a) = -160
Velocity = 160m/s