Answer:
Step-by-step explanation:
t1 = a1
t2 = a1 + d
t3 = a1 + 2d
The sum is t1 + t2 + t3 = a1 + a1 + d + a1 + 2d = 3a1 + 3d
Second question
t1 = a1
t2 = a1 + d
t3 = a1 + 2d
t4 = a1 + 3d
...
tn = a1 + (n - 1)*d
Sum = (a1 + a1 + (n- 1)*d) * n / 2
Sum = (2a1 + nd - d) * n / 2
Sum = (2a1*n + n^2*d - nd) / 2
<span>BECAUSE if it was small, white, and smooth it would be an aspirin, of course.
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1) 5:4 2) 4:5 3) 5:9 4) 2:3 5) 1:4 6)3:2 7) 5:13 8) 6:7 9) 7:5 10) 13:18 11) 18:5 12) 6:13. The others were weird, because it's been forever since I've had to do these.
The initial velocity is 19 ft/s.
This is in the form f(t) = -16t²+v₀t+h₀, where v₀ is the initial velocity. 19 is in the place of v₀, so it is the initial velocity.
The question is incomplete. The complete question is :
Let X be a random variable with probability mass function
P(X =1) =1/2, P(X=2)=1/3, P(X=5)=1/6
(a) Find a function g such that E[g(X)]=1/3 ln(2) + 1/6 ln(5). You answer should give at least the values g(k) for all possible values of k of X, but you can also specify g on a larger set if possible.
(b) Let t be some real number. Find a function g such that E[g(X)] =1/2 e^t + 2/3 e^(2t) + 5/6 e^(5t)
Solution :
Given :

a). We know :
![$E[g(x)] = \sum g(x)p(x)$](https://tex.z-dn.net/?f=%24E%5Bg%28x%29%5D%20%3D%20%5Csum%20g%28x%29p%28x%29%24)
So, 

Therefore comparing both the sides,


Also, 
b).
We known that ![$E[g(x)] = \sum g(x)p(x)$](https://tex.z-dn.net/?f=%24E%5Bg%28x%29%5D%20%3D%20%5Csum%20g%28x%29p%28x%29%24)
∴ 

Therefore on comparing, we get

∴ 