Answer:
1/6
Step-by-step explanation:
3 percent i think just double check
Hello !
cos (a+b) = cos a cos b - sin a sin b
sin (a+b) = sin a cos b + sin b cos a
cos (a+b+c) = cos (a+(b+c))
cos (a+b+c) = cos a cos (b+c) - sin a sin (b+c)
cos (a+b+c) = cos a (cos b cos c - sin b sin c) - sin a (sin b cos c + sin c cos b)
cos (a+b+c)=cos a cos b cos c - cos a sin b sin c - sin a sin b cos c - sin a cos b sin c
I don’t really know exactly to what degree you need to put in your solution. But I rounded to the nearest tenth degree.
A = 112 degrees
B= 28 degrees (180-112-40 bc all sides of a triangle must equal 180)
C= 40
a= 27.6 (this is the side opposite of angle A)
b= 14 (side opposite b)
c= 19.2 (side opposite c)
HOW TO SOLVE:
c= law of sines, so c/sin(40) = 14/sin(28), multiply both sides by sin(40) so c can be isolated and solved for. c = 14sin(40)/sin(28). Plug into calculator then get answer. c is approximately 19.2.
a = law of sines again, so a/sin(112) = 14/sin(28). Multiply both sides again by sin(112) then solve. a = 14sin(112)/sin(28). Calculator again. a is around 27.6
Answer:
a. P(x = 0 | λ = 1.2) = 0.301
b. P(x ≥ 8 | λ = 1.2) = 0.000
c. P(x > 5 | λ = 1.2) = 0.002
Step-by-step explanation:
If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:
a. What is the probability of selecting a carton and finding no defective pens?
This happens for k=0, so the probability is:
b. What is the probability of finding eight or more defective pens in a carton?
This can be calculated as one minus the probablity of having 7 or less defective pens.
c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?
We can calculate this as we did the previous question, but for k=5.
