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jok3333 [9.3K]
3 years ago
8

As part of a​ sixth-grade class​ project, the teacher brings to class a large jar containing 200 gumballs of two different​ colo

rs: red and green. Andy is asked to draw a sample of his own choosing and estimate the number of red gumballs in the jar. Andy draws a sample of 40 ​gumballs, of which 12 are red and 28 are green. Use​ Andy's sample to estimate the number of red gumballs in the jar. There are approximately how many red gumballs in the jar?
Mathematics
1 answer:
34kurt3 years ago
3 0
About 100 I’m not really sure but I tried
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An umbrella you bought is shaped like a regular octagonal pyramid with a side length of four feet and a slant height of feet. Es
Angelina_Jolie [31]
The complete question in the attached figure

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[<span>amount of fabric]=8*[area of lateral triangle side]
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so
[amount of fabric]=8*[10]----> 80 ft²

the answer is
80 ft²

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2 years ago
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Step-by-step explanation:

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Suppose we want to choose 2 letters, without replacement, from the 4 letters A, B, C, and D.
Amiraneli [1.4K]

Answer:

Letters can be chosen in 12 different ways, if order matters, or 6 different ways, if order doesn't matter.

Step-by-step explanation:

Since we want to choose 2 letters, without replacement, from the 4 letters A, B, C, and D, to determine in how many ways can this be done, if the order of the choices matters, and in how many ways can this be done, if the order of the choices does not matter, the following calculations must be performed:

If order matters =

 

(4 x 3 x 2 x 1) / 2 = X

24/2 = X

12 = X

If the order doesn't matter =

12/2 = X

6 = X

Therefore, letters can be chosen in 12 different ways, if order matters, or 6 different ways, if order doesn't matter.

3 0
2 years ago
The following probability distributions of job satisfaction scores for a sample of information systems (IS) senior executives an
Mashcka [7]

Answer:

Step-by-step explanation:

To calculate ;

1) the expected value of the job satisfaction score for senior executives ;

expected value = Summation (Px)

= 1 x 0.05 + 2 x 0.09 + 3 x 0.03 + 4 x 0.42 + 5 x 0.41

= 4.05

2) the expected value of the job satisfaction score for middle managers;

= 1 x 0.04 + 2 x 0.10 + 3 x 0.12 + 4 x 0.46 + 5 x 0.28

= 3.84

c) the variance of job satisfaction scores for executives and middle managers (to 2 decimals).

Executives ; Variance = Summation(PX^2 - Summation(PX)^2

i) For Executive Managers = 1 x 0.05 + 2^2 x 0.09 + 3^2 x 0.03 + 4^2 x 0.42 + 5^2 x 0.41 - 4.05^2 = 1.246 = 1.25

ii) for middle managers ; 1 x 0.04 + 2^2 x 0.10 + 3^2 x 0.12 + 4^2 x 0.46 + 5^2 x 0.28 - 3.84^2 = 1.134 = 1.13

d) the standard deviation of job satisfaction scores for both probability distributions (to 2 decimals). Executives, Middle managers;

For Executives = square root [ 1 x 0.05 + 2^2 x 0.09 + 3^2 x 0.03 + 4^2 x 0.42 + 5^2 x 0.41 - 4.05^2] = 1.12

For Middle Managers ; Square root [1 x 0.04 + 2^2 x 0.10 + 3^2 x 0.12 + 4^2 x 0.46 + 5^2 x 0.28 - 3.84^2 ] = 1.06

e) from the values gotten for the variance of both executive and middle managers, the variance of the former is more than that of the latter as such higher satisfaction with the executive managers.

5 0
2 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
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