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ladessa [460]
2 years ago
10

Miss Riley's class has 24 students. She has the students in groups. Each group can have no more than 5 students. if every studen

t has to be in groups how many does she have left?
Mathematics
2 answers:
netineya [11]2 years ago
8 0

Answer:

if there can be no more than 5 students, then anything less than 5 students will suffice. However we don’t want any left over ones cause we don’t want that one kid sitting alone during lunch so we can simply put <u><em>4 into each group.</em></u>

Step-by-step explanation:

Hope this helped :DDD

77julia77 [94]2 years ago
3 0

Answer:

6 groups, 4 students in each one

Step-by-step explanation:

The factors of 24 include 6 and 4

That means that you can break 24<em> into</em> those numbers

You are able to put the students up into groups of 6, with 4 students in each

However, you cannot place the students into groups of 4 (with 6 students in each) because no more than 5 students are allowed in each group

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Answer:

(21.9-19.8) -1.966\sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 1.422

(21.9-19.8) -1.966 \sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 2.778

And we are 95% confident that the true difference means are between 1.422 \leq \mu_1 -\mu_2 \leq 2.778

Step-by-step explanation:

We know the following info:

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\bar X_2 = 19.8 sample mean for group 2

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We want to find a confidence interval for the difference of means and the correct formula to do this is:

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Now we just need to find the critical value. The confidence level is 0.95 then the significance is 1-0.95 =0.05 and \alpha/2 =0.025. The degrees of freedom are given by:

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The critical value for this case would be :t_{\alpha/2}=1.966  

And replacing into the confidence interval formula we got:

(21.9-19.8) -1.966\sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 1.422

(21.9-19.8) -1.966 \sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 2.778

And we are 95% confident that the true difference means are between 1.422 \leq \mu_1 -\mu_2 \leq 2.778

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