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Vesnalui [34]
3 years ago
15

Gun rights vs. gun control: In a December 2014 report, "For the first time in more than two decades of Pew Research Center surve

ys, there is more support for gun rights than gun control." According to a Pew Research survey, 52% of Americans say that protecting gun rights is more important than controlling gun ownership. Gun control advocates in an urban city believe that the percentage is lower among city residents and conduct a survey. They test the hypotheses H 0: p = 0.52 versus H a: p < 0.52. They calculate a P‐value of 0.078. Using a significance level of 0.05, which of the following is the best explanation for how to use the P‐value to reach a conclusion in this case? Group of answer choices Since the P‐value is greater than the significance level, we reject the null hypothesis. Since the P‐value is greater than the significance level, we fail to reject the null hypothesis. Since the P‐value is greater than the significance level, we accept the null hypothesis.
Mathematics
1 answer:
vagabundo [1.1K]3 years ago
3 0

Answer:

P-value is greater than alpha, the significance level thus we fail to reject the null hypothesis.                      

Step-by-step explanation:

We are given the following information in the question:

H_0 = 0.52\\H_a < 0.52

P-Value = 0.078

Alpha, ~\alpha = 0.05

We have to reach on a conclusion on the basis of p-value.

P-Value is the probability of finding an observation, basically, it could ne said the probability when the null hypothesis is true.

Alpha is the significance level.

So, if the p-value is greater than the chosen significance level, we fail to reject the null hypothesis and accept it and if the p-value is smaller than the significance level then we reject the null hypothesis and accept the alternate hypothesis.

Since P-Value > \alpha

P-value is greater than alpha, the significance level thus we fail to reject the null hypothesis.

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Help Please . its due in 5 hours and i cant find answers
dolphi86 [110]

Answer:

39 pi cubic inches

Step-by-step explanation:

The volume of a cone is (1/3)(area of the base)(height).

V=\frac{1}{3}Bh

Find the area of the circular base.  The circumference of a circle is C=2\pi r, so

2\pi r = 6\pi\\\\r=3

Now that you know the radius of the circular base, you can find its area:

A=\pi r^2 = \pi(3^2)=9\pi

The height of the cone is 13.  Time to use the volume formula.

V=\frac{1}{3}Bh\\\\V=\frac{1}{3}(9\pi)(13)\\\\V=39\pi

6 0
3 years ago
Solve 1.25x + 1.50y = 28.25.
fiasKO [112]

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4 0
3 years ago
On the blueprint for Joey's new house,his floor measures 9 square feet.The actual house will have an area of 1296 square feet.Wh
bonufazy [111]

Answer:

1:144 ft²

Step-by-step explanation:

9 : 1296

1 : x

9x = 1296

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Scale factor = 1:144

6 0
4 years ago
How many nanometers in 66cm
makvit [3.9K]
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4 0
3 years ago
Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered
inn [45]

Answer:

Step-by-step explanation:

2005 AMC 8 Problems/Problem 20

Problem

Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered clockwise, from 1 to 12. Both start on point 12. Alice moves clockwise and Bob, counterclockwise. In a turn of the game, Alice moves 5 points clockwise and Bob moves 9 points counterclockwise. The game ends when they stop on the same point. How many turns will this take?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 24$

Solution

Alice moves $5k$ steps and Bob moves $9k$ steps, where $k$ is the turn they are on. Alice and Bob coincide when the number of steps they move collectively, $14k$, is a multiple of $12$. Since this number must be a multiple of $12$, as stated in the previous sentence, $14$ has a factor $2$, $k$ must have a factor of $6$. The smallest number of turns that is a multiple of $6$ is $\boxed{\textbf{(A)}\ 6}$.

See Also

2005 AMC 8 (Problems • Answer Key • Resources)

Preceded by

Problem 19 Followed by

Problem 21

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

5 0
2 years ago
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