angle AOB = 132 and is also the sum of angles AOD and
DOB. Hence
angle AOD + angle DOB = 132° ---> 1
angle COD = 141 and is also the sum of angles COB and BOD. Hence
angle COB + angle DOB = 141° ---> 2
Now we add the left sides together and the right sides of equations 1 and 2
together to form a new equation.
angle AOD + angle DOB + angle COB + angle DOB = 132 + 141 ---> 3
We should also note that:
angle AOD + angle DOB + angle COB = 180°
Therefore substituting angle AOD + angle DOB + angle COB in equation 3 by 180
and solving for angle DOB:
180 + angle DOB = 132 + 141
angle DOB = 273 - 180 = 93°
Answer:
(SEE EXPLANATION FIRST)
Answer D is TRUE because on the graph if you were to keep going left even after the picture of the graph ends, the graph will still keep going up infinite
Step-by-step explanation:
If this is the graph you can read the answer
Answer:
the expected value of Xn , E(Xn) = 0 and the variance σ²(Xn) = n*(1-2n)
Step-by-step explanation:
If X1= number of tails when n fair coins are flipped , then X1 follows a binomial distribution with E(X1) = n*p , p=0,5 and the number of heads obtained is X2=n-X1
therefore
Xn =X1-X2 = X1- (n-X1) = 2X1-n
thus
E(Xn) =∑ (2*X1-n) p(X1) = 2*∑[X1 p(X1)] -n∑p(X1) = 2*E(X1)-n = 2*n*p--n= 2*n*1/2 -n = n-n =0
the variance will be
σ²(Xn) = ∑ [Xn - E(Xn)]² p(Xn) = ∑ [(2X1-n) - 0 ]² p(X1) = ∑ (4*X1²-4*X1*n+n²) p(X1) = = 4*∑ X1²p(X1) - 4n ∑X1 p(X1) - n²∑p(X1) = 2*E(X1²) -4n*E(X1)- n²
since
σ²(X1) = n*p*(1-p) = n*0,5*0,5=n/4
and
σ²(X1) = E(X1²) - [E(X1)]²
n/4 = E(X1²) - (n/2)²
E(X1²) = n(n+1)/4
therefore
σ²(Xn) = 4*E(X1²) -4n*E(X1)- n² = 4*n(n+1)/4 - 4*n*n/2 - n² = n(n+1) - 2n² - n²
= n - 2n² = n(1-2n)
σ²(Xn) = n(1-2n)
Answer:
83.79 kJ
Step-by-step explanation:
The potential energy is the product of weight and height, where weight is in newtons.
PE = Mgh = (190 kg)(9.8 m/s^2)(45 m) = 83,790 J