Suppose the population proportion of American citizens who are in favor of gun control is .61. If a sampling distribution of size n= 50 was created from this population, what would be the mean of this sampling distribution?
Answers
Answer:
The Mean of the sampling distribution is μ = p = 0.61
Step-by-step explanation:
Given size of the sampling distribution (n) = 50
Suppose the population proportion of American citizens who are in favor of gun control is .61
That is p = 0.61
Sampling distribution of proportions:-
Let p be the probability of occurrence of an event (called its success) and q =1-p is the probability of non- occurrence (called its failure).Draw all possible samples of size n from an infinite population.
Compute the proportion P of success for each of these samples. Then
and variance sampling distributions are given by
and
variance
Standard deviation (S.D) =
Mean of the sampling distribution:-
The Mean of the sampling distribution is μ = p
Given data the proportion of American citizens who are in favor of gun control is 0.61
p = 0.61
The Mean of the sampling distribution is μ = p = 0.61
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C. Distance
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In this case, the number of rides in an amusement park would be discrete because you cannot have, for example, 5.6 rides.
The correct answer is A.
Answer:
C.banans in a bunch
Step-by-step explanation:
apexs
b. k = 1/3
c. k = 1/2
d. 2
Answers
The correct answer is:
C) k = 1/2
Explanation:
We will find the lengths of the horizontal segments of the pre-image and the image, and compare them to find the dilation factor.
For the pre-image, we have a horizontal segment from (2, 5) to (4, 5). This is a distance of 2.
The corresponding segment of the image goes from (2.5, 4) to (3.5, 4). This is a distance 1.
The segment of the image is 1/2 the size of the corresponding segment of the pre-image; this makes the dilation factor 1/2.
To verify, we have another horizontal segment of the pre-image that goes from (3, 2) to (6, 2). This is a distance of 3 units.
The corresponding horizontal segment of the image goes from (3, 2.5) to (4.5, 2.5). This is a distance of 1.5 units.
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Answers
Answer:
no
Step-by-step explanation:
its an obtuse triangle
Answers
Answer:
The length and width that maximize the area are:
W = 2*√8
L = 2*√8
Step-by-step explanation:
We want to find the largest area of a rectangle inscribed in a semicircle of radius 4.
Remember that the area of a rectangle of length L and width W, is:
A = L*W
You can see the image below to see how i will define the length and the width:
L = 2*x'
W = 2*y'
Where we have the relation:
4 = √(x'^2 + y'^2)
16 = x'^2 + y'^2
Now we can isolate one of the variables, for example, x'
16 - y'^2 = x^'2
√(16 - y'^2) = x'
Then we can write:
W = 2*y'
L = 2*√(16 - y'^2)
Then the area equation is:
A = 2*y'*2*√(16 - y'^2)
A = 4*y'*√(16 - y'^2)
If A > 1, like in our case, maximizing A is the same as maximizing A^2
Then if que square both sides:
A^2 = (4*y'*√(16 - y'^2))^2
= 16*(y'^2)*(16 - y'^2)
= 16*(y'^2)*16 - 16*y'^4
= 256*(y'^2) - 16*y'^4
Now we can define:
u = y'^2
then the equation that we want to maximize is:
f(u) = 256*u - 16*u^2
to find the maximum, we need to evaluate in the zero of the derivative:
f'(u) = 256 - 2*16*u = 0
u = -256/(-2*16) = 8
Then we have:
u = y'^2 = 8
solving for y'
y' = √8
And we know that:
x' = √(16 - y'^2) = √(16 - (√8)^2) = √8
And the dimensions was:
W = 2*y' = 2*√8
L = 2*y' = 2*√8
These are the dimensions that maximize the area.
Answers
Answer:
x = 12y
Step-by-step explanation: