Complete question :
A random sample of n = 83 measurements is drawn from a binomial population with probability of success 0.4 . Complete parts a through d below. a. Give the mean and standard deviation of the sampling distribution of the sample proportion, . The mean of the sampling distribution of is nothing. The standard deviation of the sampling distribution of is nothing.
Answer:
Mean = 33.2000
Standard deviation = 4.4632
Step-by-step explanation:
Given that :
Sample size (n) = 83
Probability of success (p) = 0.4
q = p' = (1 - p) = 1 - 0.4 = 0.6
The mean of the sampling distribution :
Sample size * probability of success
n * p = 83 * 0.4 = 33.2000
The standard deviation of the sampling distribution :
σ=√(sample size * probability of success * (1 - p))
σ = √n * p * (1 - p)
σ = √(83 * 0.4 * 0.6)
σ = √19.92
σ = 4.46318
σ = 4.4632
Answer:
56
Step-by-step explanation:
You subtract how long the shadow is from how long it is from the tree. This is true because the shadow is shorter than the tree because of the sun's ability to create it
By simplifying the number given which is 8 2/3, we get an improper fraction of 26/3. Also, when we try to divide 26 by 3, we get an answer of 8.6667, with 6 as a repeating number after the decimal place. Therefore, this number is an example of real and rational number. Thus the answer is the second choice.
Explanation:
The N stands for Pi so when you type in the answers to Khan Academy use the Pi symbol when theres an N! These are the common questions and common radius Khan Academy gives when they ask to find the radius of the circle.
Answers:
For 4 - 267.95
For 1/4 - 1/48n
For 1 - 4/3n
For 9 - 972n
For 7 - 1372/3n
For 6 - 288n
For 3 - 36n
For 10 - 4000/3n
For 8 - 2048/3n
For 1/2 - 1/6n
For 5 - 500/3n
For 2 - 33.49
Step-by-step explanation:
<u><em>I did the go math so all of these rights and the majority of the answers are the questions Khan Academy will ask you to find the radius of the circle.</em></u>
The answer is c ≥ 1
Subtract 10 on both sides.
11-10= 1
and that’s how u get ur answer, hope this helped :)
