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Daniel [21]
4 years ago
7

A bag contains 9 red marbles and 18 blue marbles. If a representative sample contains 2 red ​marbles, then how many blue marbles

would you expect it to​ contain?
Mathematics
1 answer:
Ahat [919]4 years ago
4 0

Answer:

4

Step-by-step explanation:

9 x 2 Is 18 so you would multiply 2 x 2 as well

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The question is simply asking you to evaluate the function for an input value of -3. Meaning we plug in -3 for all cases of n. (-3)^3 + (-3). This gives us -27 -3 which is -30. So g(-3)= -30
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Answer this please tell me if I’m correct
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What is the value of x? Round to the nearest tenth.
kotykmax [81]

Answer:

27.5

Step-by-step explanation:

first you find what 30x30 is as A squared plus B squared is equal to C squared. your A is 12 and your C is 30.

so!

30x30 is 900

12x12 is 144

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5 0
3 years ago
Read 2 more answers
Which rectangular prism has the greatest volume?
Artyom0805 [142]

Answer:

you would get your answer by timesing all the numbers together, that it gives you.

Step-by-step explanation:

If my math is correct it would be the second one.


8 0
3 years ago
Read 2 more answers
Of 580580 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compa
Lubov Fominskaja [6]

Answer:

a) The 99% confidence interval would be given (0.204;0.296).

b) We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).  

c) No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

Step-by-step explanation:

Part a

Data given and notation  

n=580 represent the random sample taken    

X represent the seafood sold in the country that is mislabeled or misidentified by the people

\hat p=0.25 estimated proportion of seafood sold in the country that is mislabeled or misidentified by the people

\alpha=0.01 represent the significance level (no given, but is assumed)    

p= population proportion of seafood sold in the country that is mislabeled or misidentified by the people

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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 population proportion have the following distribution

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

0.25 - 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.204

0.25 + 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.296

And the 99% confidence interval would be given (0.204;0.296).

Part b

We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).  

Part c

A government spokesperson claimed that the sample size was too​ small, relative to the billions of pieces of seafood sold each​ year, to generalize. Is this criticism​ valid?

No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

4 0
4 years ago
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