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aleksandrvk [35]
3 years ago
7

Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 2 times the number o

f marbles Mark has, how many does each boy have to sell if the total number of marbles is 73?
Mark has how many marbles to sell?
Don has how many marbles to sell?
Show your work and explain how you arrived at your answers.
Mathematics
1 answer:
Ksenya-84 [330]3 years ago
8 0
Let’s label Mark’s marbles as ‘x’ and don’s as ‘y’.
so Don has 1 more than 2 times ‘x’.
This is written as:
y=2x+1

73 is the sum of ‘x’ and ‘y’, so x+y=73.

To solve for x, substitute our equation for y:
x+(2x+1)=73
Simplify:
3x=72
x=24

Now that we have x, solve for y:
y+24=73
y=49

Mark has to sell 24 marbles,
Don has to sell 49 marbles

Hope this was helpful!
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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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