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kondaur [170]
3 years ago
13

52% of 1000 registered voters intend to vote for Steven Collins for mayor. What is the 95% confidence interval to describe the t

otal percentage of registered voters who intend to vote for Steven Collins?
A. (50.1%, 53.9%)

B. (48%, 56%)

C. (49.9%, 54.1%)

D. (48.9%, 55.1%)
Mathematics
1 answer:
Katen [24]3 years ago
3 0

Answer:

A. (49.9%, 54.1%)

B. (48%, 56%)

C. (50.1%, 53.9%)

D. (48.9%, 55.1%

Step-by-step explanation:

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Simplify the expression -1/2(-5/6 + 1/3)
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Answer: 1/4

Step-by-step explanation:

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Dafna1 [17]
Slope (y2-y1)/(x2-x1)
(0-4)/(-5-5) = -4/-10 = 2/5
Y = 2/5x + b
Plug in any point
4 = 2/5(5) + b
4 = 2 + b, b= 2
Final equation: y = 2/5x + 2
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Solve the system by substitution<br><br> -5x+3y=2<br> y=2x
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(x,y)=(0,0)

Step-by-step explanation:

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A forester studying the effects of fertilization on certain pine forests in the Southeast is interested in estimating the averag
ahrayia [7]

Answer:

P(-2 \leq \bar X -\mu \leq 2)

If we divide both sides by \frac{\sigma}{\sqrt{n}} we got:

P(\frac{-2}{\frac{4}{\sqrt{9}}}\leq Z \leq \frac{2}{\frac{4}{\sqrt{9}}})

And we can use the normal distribution table or excel to find the probabilites and we got:

P(-1.5 \leq Z \leq 1.5)= P(ZStep-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the area of a population, and for this case we know the distribution for X is given by:

X \sim N(M,4)  

Where \mu=M and \sigma=4

We select a a sample of n =4 and since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And we want to find this probability:

P(-2 \leq \bar X -\mu \leq 2)

If we divide both sides by \frac{\sigma}{\sqrt{n}} we got:

P(\frac{-2}{\frac{4}{\sqrt{9}}}\leq Z \leq \frac{2}{\frac{4}{\sqrt{9}}})

And we can use the normal distribution table or excel to find the probabilites and we got:

P(-1.5 \leq Z \leq 1.5)= P(Z

8 0
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