Answer: 1/5, 1/2, 0.
Step-by-step explanation:
given data:
no of cameras = 6
no of cameras defective = 3
no of cameras selected = 2
Let p(t):=P(X=t)
p(2)=m/n,
m=binomial(3,2)=3!/2!= 3
n=binomial(6,2)=6!/2!/4! = 15
p(3)= 3/15
= 1/5.
p(1)=m/n,
m=binomial(6,1)*binomial(2,2)=6!/1!/4!*2!/2!/0!= 7.5
n=binomial(6,2)= 15
p(2)= 7.5/15
= 1/2
p(0)=m/n,
m=0
p(0)=0
Answer:
it depends
Step-by-step explanation:
If 17 is the hypotenuse
if you are looking for the hypotenuse
Step-by-step explanation:
Left hand side:
4 [sin⁶ θ + cos⁶ θ]
Rearrange:
4 [(sin² θ)³ + (cos² θ)³]
Factor the sum of cubes:
4 [(sin² θ + cos² θ) (sin⁴ θ − sin² θ cos² θ + cos⁴ θ)]
Pythagorean identity:
4 [sin⁴ θ − sin² θ cos² θ + cos⁴ θ]
Complete the square:
4 [sin⁴ θ + 2 sin² θ cos² θ + cos⁴ θ − 3 sin² θ cos² θ]
4 [(sin² θ + cos² θ)² − 3 sin² θ cos² θ]
Pythagorean identity:
4 [1 − 3 sin² θ cos² θ]
Rearrange:
4 − 12 sin² θ cos² θ
4 − 3 (2 sin θ cos θ)²
Double angle formula:
4 − 3 (sin (2θ))²
4 − 3 sin² (2θ)
Finally, apply Pythagorean identity and simplify:
4 − 3 (1 − cos² (2θ))
4 − 3 + 3 cos² (2θ)
1 + 3 cos² (2θ)
Answer:
0.1994 is the required probability.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 166 pounds
Standard Deviation, σ = 5.3 pounds
Sample size, n = 20
We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.
Formula:
Standard error due to sampling =
P(sample of 20 boxers is more than 167 pounds)
Calculation the value from standard normal z table, we have,
0.1994 is the probability that the mean weight of a random sample of 20 boxers is more than 167 pounds
