Is there anything to look at?
We can proceed in solving the problem since all information are given such as 2*4*5costheta.
we have a=4, b=5, and C=theta
let us solve for "c" using Pythagorean
c²=a²+b²
c²=4²+5²
c=6.4
Solving for theta or C
c²=a²+b²-2abcosC
6.4²=4²+5²-2*4*5*cosC
C=90
I think AVM= 35 degrees
Hope this helps
Answer:
0.0903
Step-by-step explanation:
Given that :
The mean = 1450
The standard deviation = 220
sample mean = 1560



P(X> 1560) = P(Z > 0.5)
P(X> 1560) = 1 - P(Z < 0.5)
From the z tables;
P(X> 1560) = 1 - 0.6915
P(X> 1560) = 0.3085
Let consider the given number of weeks = 52
Mean
= np = 52 × 0.3085 = 16.042
The standard deviation =
The standard deviation = 
The standard deviation = 3.3306
Let Y be a random variable that proceeds in a binomial distribution, which denotes the number of weeks in a year that exceeds $1560.
Then;
Pr ( Y > 20) = P( z > 20)


From z tables
P(Y > 20)
0.0903