<span>The length of an arc, L = theta/360 * 2 * pie * r
Where theta = 81 and r = 10ft = 3.048m
So length = 81/360 * 2 * pie * 3.048
L = * 0.225 * 2 * pie * 3.048
L = 1.37 * pie
L = pie. Option A</span>
D because a rational number can be converted into a fraction and the square root of 5 cannot be covert into a fraction because it equals 2.2
The mean of the given sample data is 210, and the standard deviation is 7.937.
Given size 'n' = 300
The population proportion 'p' = 0.7
Let 'x' be the random variable of the binomial distribution
a) mean of the binomial distribution = n p = 300 × 0.7
μ = 210
b) variance of the binomial distribution
⇒ n p q
⇒ 300 × 0.7 ×0.3
⇒ σ² = 63
The standard deviation of the binomial distribution:
⇒ √n p q = √63 = 7.937
Thus, the mean of the given sample data is 210, and the standard deviation is 7.937.
Learn more about the standard deviation here:
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The question seems to be incomplete the correct question would be:
describe the sampling of p hat. Assume that the size of the population is 25000 n= 300 p=0.7 a) Determine the mean of the sampling distributionb) Dtermine the standard deviation of the sampling distribution
Answer:
a. 7/22
b. 28/33
c. 7/22
d. dependent events
Step-by-step explanation:
The total number of cards is given as 7 + 5 = 12.
The probability if a card being green on the first pick, P(G) = 7/12
The probability of a card being yellow on the first pick, P(Y) = 5/12
Because there is no replacement, the card are going to be short by one card. Therefore, the second pick will be like this:
P(GG) = ( 7/12×6/11)
P(GY) = ( 7/12 × 5/11)
P(YG) = ( 5/12×7/11)
P(RR) = ( 5/12 ×4/11)
a. P (G1 and G2) = ( 7/12×6/11)
= 7/22
b. P( At least one is green) = ( 7/12×6/11) + ( 7/12 × 5/11) + ( 5/12×7/11)
= 28/33
c. P(G2G1) = 7/22
d. the events are dependent. They do not give the same result.
