Answer:
The mean of the sampling distribution of the sample proportions is 0.82 and the standard deviation is 0.0256.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean 
and standard deviation 
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation 
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For proportions, the mean is 
and the standard deviation is 
In this problem, we have that:
.
So
The mean of the sampling distribution of the sample proportions is 0.82 and the standard deviation is 0.0256.
Answer:
21
Step-by-step explanation:
plug in 4 where you have the j
6(4)-3=
24-3=
21
The answer is 32.5%
and if you need it in decimal form, it would be,
Answer:
Step-by-step explanation:
Given the equation
Sin(5x) = ½
5x = arcSin(½)
5x = 30°
Then,
The general formula for sin is
5θ = n180 + (-1)ⁿθ
Divide through by 5
θ = n•36 + (-1)ⁿ30/5
θ = 36n + (-1)ⁿ6
The range of the solution is
0<θ<2π I.e 0<θ<360
First solution
When n = 0
θ = 36n + (-1)ⁿθ
θ = 0×36 + (-1)^0×6
θ = 6°
When n = 1
θ = 36n + (-1)ⁿ6
θ = 36-6
θ = 30°
When n = 2
θ = 36n + (-1)ⁿ6
θ = 36×2 + 6
θ = 78°
When n =3
θ = 36n + (-1)ⁿ6
θ = 36×3 - 6
θ = 102°
When n=4
θ = 36n + (-1)ⁿ6
θ = 36×4 + 6
θ = 150
When n =5
θ = 36n + (-1)ⁿ6
θ = 36×5 - 6
θ = 174°
When n = 6
θ = 36n+ (-1)ⁿ6
θ = 36×6 + 6
θ = 222°
When n = 7
θ = 36n + (-1)ⁿ6
θ = 36×7 - 6
θ = 246°
When n =8
θ = 36n + (-1)ⁿ6
θ = 36×8 + 6
θ = 294°
When n =9
θ = 36n + (-1)ⁿ6
θ = 36×9 - 6
θ = 318°
When n =10
θ = 36n + (-1)ⁿ6
θ = 36×10 + 6
θ = 366°
When n = 10 is out of range of θ
Then, the solution is from n =0 to n=9
So the equation have 10 solutions in the range 0<θ<2π
