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ira [324]
2 years ago
10

The actual proportion of voters who plan to vote for a particular proposition in the next election is 70%, but LaShandra does no

t know this. LaShandra has been hired by the election campaign to conduct a sample survey to estimate the proportion of voters who will vote for the proposition. LaShandra decides to take a random sample of size 121 from all eligible voters. 31. Draw a picture of the sampling distribution of p-hat. Give a mathematical expression for the mean of your distribution as well as the number. Also label the x-axis on your picture. Upload a photo or scan of your picture, or else create an electronic file and upload the file.

Mathematics
1 answer:
Setler [38]2 years ago
4 0

Answer:

0.0417

Step-by-step explanation:

Given the following;

p = 0.7, n=121

The sampling distribution of sample proportion will be approximately normal with mean

\mu_{\hat{p}}=p=0.7

and standard deviation

\sigma_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.7\cdot 0.2}{121}}=0.0417

Check attachment for the curve diagram.

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Stephen spent $4 on milk, $6 on eggs, and $11 on cereal. He wrote the ratio 6/11 to describe some of his purchases. Explain why
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Attached the solution and work.

6 0
2 years ago
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Identify the variation as direct, inverse, joint or combined. A = r²
myrzilka [38]

Answer : Direct Variation

Question :

Identify the variation as direct, inverse, joint or combined. A = \pi r^2

We know direct variation is y = kx

Where k is the constant of proportionality

When value of x increases then y also increases. when x value decreases then y value decreases.

The value of y depends on value of x It means y is directly proportional to x.

Hence A = \pi r^2 is a direct variation.


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2 years ago
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PLEASE HELP. WILL GIVE BRAINLEST
swat32

Answer:

i would say A. The initial cost for renting a snowmobile is $75, with each hour of use costing an additional $25.

Step-by-step explanation:

5 0
2 years ago
What are the zeros of the function f(x) = x2 + 5x + 5 written in simplest radical form?
Pavel [41]

\boxed{x_{1}=\frac{-5 + \sqrt{5}}{2}} \\ \\ \\ \boxed{x_{2}=\frac{-5 - \sqrt{5}}{2}}

<h2>Explanation:</h2>

Using the quadratic formula:

x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ \\ Here: \\ \\ f(x) = x^2 + 5x + 5 \\ \\ \\ So: \\ \\ a=1 \\ \\ b=5 \\ \\ c=5 \\ \\ \\ x=\frac{-5 \pm \sqrt{5^2-4(1)(5)}}{2(1)} \\ \\ x=\frac{-5 \pm \sqrt{25-20}}{2} \\ \\ x=\frac{-5 \pm \sqrt{5}}{2} \\ \\ \\ Two \ solutions: \\ \\ \boxed{x_{1}=\frac{-5 + \sqrt{5}}{2}} \\ \\ \\ \boxed{x_{2}=\frac{-5 - \sqrt{5}}{2}}

<h2>Learn more:</h2>

Quadratic functions: brainly.com/question/12164750

#LearnWithBrainly

8 0
2 years ago
find the probability that a randomly selected automobile tire has a tread life between 42000 and 46000 miles
maria [59]
Given that in a national highway Traffic Safety Administration (NHTSA) report, data provided to the NHTSA by Goodyear stated that the mean tread life of a properly inflated automobile tires is 45,000 miles. Suppose that the current distribution of tread life of properly inflated automobile tires is normally distributed with mean of 45,000 miles and a standard deviation of 2360 miles.

Part A:

Find the probability that randomly selected automobile tire has a tread life between 42,000 and 46,000 miles.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is between two numbers, a and b is given by:
P(a \ \textless \  X \ \textless \  b) = P(X \ \textless \  b) - P(X \ \textless \  a) \\  \\ P\left(z\ \textless \  \frac{b-\mu}{\sigma} \right)-P\left(z\ \textless \  \frac{a-\mu}{\sigma} \right)
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles a standard deviation of 2360 miles.
The probability that randomly selected automobile tire has a tread life between 42,000 and 46,000 miles is given by:
P(42,000 \ \textless \ X \ \textless \ 46,000) = P(X \ \textless \ 46,000) - P(X \ \textless \ 42,000) \\ \\ P\left(z\ \textless \ \frac{46,000-45,000}{2,360} \right)-P\left(z\ \textless \ \frac{42,000-45,000}{2,360} \right) \\  \\ =P(0.4237)-P(-1.271)=0.66412-0.10183=\bold{0.5623}


b. Find the probability that randomly selected automobile tire has a tread life of more than 50,000 miles.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is greater than a numbers, a, is given by:
P(X \ \textgreater \  a) = 1-P(X \ \textless \ a)  \\  \\ =1-P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles a standard deviation of 2360 miles.
The probability that randomly selected automobile tire has a tread life of more than 50,000 miles is given by:
P(X \ \textgreater \  50,000) = 1 - P(X \ \textless \ 50,000) \\ \\ =1-P\left(z\ \textless \ \frac{50,000-45,000}{2,360} \right)=1-P(z\ \textless \ 2.1186) \\  \\ =1-0.98294=\bold{0.0171}


Part C:

Find the probability that randomly selected automobile tire has a tread life of less than 38,000 miles.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is less than a numbers, a, is given by:
P(X \ \textless \  a) =P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles a standard deviation of 2360 miles.
The probability that randomly selected automobile tire has a tread life of less than 38,000 miles is given by:
P(X \ \textless \  38,000) = P\left(z\ \textless \ \frac{38,000-45,000}{2,360} \right) \\  \\ =P(z\ \textless \ -2.966)=\bold{0.0015}


d. Suppose that 6% of all automobile tires with the longest tread life have tread life of at least x miles. Find the value of x.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is greater than a numbers, x, is given by:
P(X \ \textgreater \ x) = 1-P(X \ \textless \ a) \\ \\ =1-P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles and a standard deviation of 2360 miles and that the probability that all automobile tires with the longest tread life have tread life of at least x miles is 6%.

Thus:
P(X \ \textgreater \ x) =0.06 \\  \\ \Rightarrow1 - P(X \ \textless \ x)=0.06 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=1-0.06=0.94 \\  \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=P(z\ \textless \ 1.555) \\ \\ \Rightarrow \frac{x-45,000}{2,360}=1.555 \\  \\ \Rightarrow x-45,000=2,360(1.555)=3,669.8 \\  \\ \Rightarrow x=3,669.8+45,000=48,669.8
Therefore, the value of x is 48,669.8


e. Suppose that 2% of all automobile tires with the shortest tread life have tread life of at most x miles. Find the value of x.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is less than a numbers, x, is given by:
P(X \ \textless \ x) =P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles and a standard deviation of 2360 miles and that the probability that all automobile tires with the longest tread life have tread life of at most x miles is 2%.

Thus:
P(X \ \textless \ x)=0.02 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=1-0.02=0.98 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=P(z\ \textless \ 2.054) \\ \\ \Rightarrow \frac{x-45,000}{-2,360}=2.054 \\ \\ \Rightarrow x-45,000=-2,360(2.054)=-4,847.44 \\ \\ \Rightarrow x=-4,847.44+45,000=40,152.56
Therefore, the value of x is 40,152.56
4 0
2 years ago
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