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STALIN [3.7K]
3 years ago
7

Dr. Stein bought 30 notebooks, 60 pencils and 300 erasers to make identical packages with some notebooks, some pencils and some

erasers for his students. He used everything he bought, and every student got a package. What is the largest number of students Dr. Stein can have in his class?
Mathematics
2 answers:
SashulF [63]3 years ago
7 0
M.m.c (30, 60, 300)
30, 60, 300|3
10, 20, 100|2
 5, 10,   50 |5
 1,  2,    10 | ===> 3*2*5 = 30 
30 stundents
Murljashka [212]3 years ago
5 0
The key word is "identical." If all students received identical packages, then Dr. Stein could have handed out 30 packages. He has a maximum of 30 notebooks to hand out, so if each package has one notebook, he can hand out 30 packages. Each of those 30 packages will also have 2 pencils and 10 erasers.


ANSWER: There are 30 packages maximum so he can have a maximum of 30 students.

Hope this helps! :)
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Express the confidence interval 0.555 less than 0.777 in the form Modifying above p with caret plus or minus Upper E.
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Complete Question

Express the confidence interval 0.555 less than p less than  0.777 in the form Modifying above p with caret plus or minus Upper E.

Answer:

The modified representation is \r p \pm  E  = 0.666 \pm  0.111

Step-by-step explanation:

From the question we are told that

    The  confidence interval interval is  0.555 <  p <  0.777

Now looking at the values that make up  the up confidence interval we see that this is a symmetric  confidence interval(This because the interval covers 95%  of the area under the normal curve which mean that the probability of a value falling outside the interval is 0.05 which is divided into two , the first half on the left -tail and the second half on the right tail  as shown on the figure in the first uploaded image(reference - Yale University ) ) which means

     Now  since the confidence interval is  symmetric , we can obtain the sample proportion as follows

             \r p  =  \frac{0.555 + 0.777}{2}

            \r p  =0.666

Generally the margin of error is mathematically represented as

            E  =  \frac{1}{2}  *  K

Where  K is the length of the confidence interval which iis mathematically represented as

           K  =  0.777 -0.555

         K  = 0.222

Hence  

          ME  =  \frac{1}{2}  *  0.222

           ME  =  0.111

So the confidence interval can now be represented as

        \r p \pm  E  = 0.666 \pm  0.111

     

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