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DiKsa [7]
3 years ago
14

A band selects 404040 ticket-holders at random from each concert to win t-shirts. The band knows that 55\%55%55, percent of the

ticket-holders buy their tickets at least a month before the concert. Suppose the band calculated the proportion \hat p
p
^
​ of winners from each concert who bought their tickets at least a month before the concert.
Which of the following distributions is the best approximation of the sampling distribution of \hat p
p
^
​ ?
Each distribution uses the same scale.
Choose 1 answer:
Choose 1 answer:

(Choice A, Checked)
A


(Choice B)
B


(Choice C)
C


(Choice D)
D
Mathematics
1 answer:
drek231 [11]3 years ago
3 0

Answer:

They would be normally distributed.

Step-by-step explanation:

Because there are more than 30 samples, the distribution would closely resemble the normal distribution.

40 ticketholders are winning the contest. 40 ≥ 30. This is enough to make the distribution approximately normal.

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What is 15 girls to 6 boys as a fraction in simplest form
Rasek [7]
2 1/2 this is the answer
4 0
3 years ago
The delivery times for all food orders at a fast-food restaurant during the lunch hour are normally distributed with a mean of m
UkoKoshka [18]

Answer:

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

X \sim N(14.7,3.7)  

Where \mu=14.7 and \sigma=3.7

Since the distribution of 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 have;

\mu_{\bar X}= 14.70

\sigma_{\bar X} =\frac{3.7}{\sqrt{40}}= 0.59

Step-by-step explanation:

Assuming this question: The delivery times for all food orders at a fast-food restaurant during the lunch hour are normally distributed with a mean of 14.7 minutes and a standard deviation of 3.7 minutes. Let R be the mean delivery time for a random sample of 40 orders at this restaurant. Calculate the mean and standard deviation of \bar X Round your answers to two decimal places.

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 delivery times of a population, and for this case we know the distribution for X is given by:

X \sim N(14.7,3.7)  

Where \mu=14.7 and \sigma=3.7

Since the distribution of 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 have;

\mu_{\bar X}= 14.70

\sigma_{\bar X} =\frac{3.7}{\sqrt{40}}= 0.59

4 0
3 years ago
Solve the oblique triangle where side a has length 10 cm, side c has length 12 cm, and angle beta has measure thirty degrees. Ro
Ugo [173]

Answer:

Side\ B = 6.0

\alpha = 56.3

\theta = 93.7

Step-by-step explanation:

Given

Let the three sides be represented with A, B, C

Let the angles be represented with \alpha, \beta, \theta

[See Attachment for Triangle]

A = 10cm

C = 12cm

\beta = 30

What the question is to calculate the third length (Side B) and the other 2 angles (\alpha\ and\ \theta)

Solving for Side B;

When two angles of a triangle are known, the third side is calculated as thus;

B^2 = A^2 + C^2 - 2ABCos\beta

Substitute: A = 10,  C =12; \beta = 30

B^2 = 10^2 + 12^2 - 2 * 10 * 12 *Cos30

B^2 = 100 + 144 - 240*0.86602540378

B^2 = 100 + 144 - 207.846096907

B^2 = 36.153903093

Take Square root of both sides

\sqrt{B^2} = \sqrt{36.153903093}

B = \sqrt{36.153903093}

B = 6.0128115797

B = 6.0 <em>(Approximated)</em>

Calculating Angle \alpha

A^2 = B^2 + C^2 - 2BCCos\alpha

Substitute: A = 10,  C =12; B = 6

10^2 = 6^2 + 12^2 - 2 * 6 * 12 *Cos\alpha

100 = 36 + 144 - 144 *Cos\alpha

100 = 36 + 144 - 144 *Cos\alpha

100 = 180 - 144 *Cos\alpha

Subtract 180 from both sides

100 - 180 = 180 - 180 - 144 *Cos\alpha

-80 = - 144 *Cos\alpha

Divide both sides by -144

\frac{-80}{-144} = \frac{- 144 *Cos\alpha}{-144}

\frac{-80}{-144} = Cos\alpha

0.5555556 = Cos\alpha

Take arccos of both sides

Cos^{-1}(0.5555556) = Cos^{-1}(Cos\alpha)

Cos^{-1}(0.5555556) = \alpha

56.25098078 = \alpha

\alpha = 56.3 <em>(Approximated)</em>

Calculating \theta

Sum of angles in a triangle = 180

Hence;

\alpha + \beta + \theta = 180

30 + 56.3 + \theta = 180

86.3 + \theta = 180

Make \theta the subject of formula

\theta = 180 - 86.3

\theta = 93.7

5 0
3 years ago
Solve 5x ≥ 25. Graph the solution.
Zina [86]

Answer:

Step-by-step explanation:

Alright so we are going to solve

5x ≥ 25  Divide both sides by 5

x≥5

4 0
3 years ago
Upvote an will mark brainlest
Irina18 [472]

Answer:heyyy there...the answer is b that is -12x^3+12x^2-20

Step-by-step explanation:

<h3>f(x)= -12x^3+19x^2-5</h3><h3>g(x)= 7x^2+15</h3><h3>f(x)-g(x)=(-12x^3+19x^2-5)-( 7x^2+15)...{while opening the bracket the sign of the second polynomial changes accordingly}</h3><h3>it becomes -12x^3+19x^2-5-7x^2-15</h3><h3>=-12x^3+12x^2-20</h3><h3>HOPE IT HELPED UUUU</h3>
7 0
3 years ago
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