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finlep [7]
4 years ago
15

there are 103 students eating lunch in the cafeteria. each table seats 4 students. all the tables are full except for 1 table. h

ow many students are sitting at the table that is not full?
Mathematics
2 answers:
amm18124 years ago
8 0
You're looking for the reminder, or "left overs", of a division problem. 103/4=25 r3 or 25 full tables and people left over, so your answer is 3 students
natta225 [31]4 years ago
6 0
4*25= 100. There are 26 tables. At each of the 25 tables 4 people are sitting. Which means 100 people are sitting in complete tables of 4 people. That means that at the 26th table 3 people are sitting. 103-100= 3. Hope that helps. :) 
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Answer:

73 units

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Which is greater 0.1 or 3.7
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Answer:

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Suppose we wish to demonstrate that there is a difference between the proportions of wives and husbands who do laundry at home.
maxonik [38]

Answer:

Step-by-step explanation:

From this study:

The null hypothesis:

H_o : p_1 =p_2

The altenative is:

H_a : p_1 \ne p_2

This test is a two-tailed test.

However; we are told that the wives have 44 success out of 66, then the number of failures will be 22.

Then;

\hat p_1 = \dfrac{44}{66}

\hat p_1 = 0.6667

Similarly, the husbands have 18 success out of 46, then the number of failures will be 28

Then:

\hat p_2 = \dfrac{18}{46}

\hat p_2 = 0.3913

The pooled proportion p = \dfrac{18+44}{66+46}

p = \dfrac{62}{112}

p = 0.55357

The estimated standard error S.E is:

= \sqrt{\dfrac{ \bar p(1- \bar p)}{n_1} +\dfrac{ \bar p(1-\bar p)}{n_2}}

= \sqrt{ 0.55357(1-0.55357) \Big( \dfrac{1}{66} + \dfrac{1}{46} \Big)}

=\sqrt{ 0.55357(0.44643) \Big(0.01515 + 0.021739 \Big)}

=\sqrt{ 0.00911638798}

= 0.0955

The Z test statitics can now be computed as:

Z = \dfrac{ \hat p_1 - \hat p_2}{\sqrt{\dfrac{ \bar p(1- \bar p)}{n_1} +\dfrac{ \bar p(1-\bar p)}{n_2}}}

Z = \dfrac{0.6667 -0.3913}{0.0955}

Z = 2.88

Th p -value from the test statistics is:

p-value = 2P(Z > 2.88)

p- value = 2 P (1 - Z < 2.88)

p-value = 2 ( 1 - 0.998)

p-value = 2 ( 0.002)

p -alue = 0.004

Decision Rule:

Thus, at 0.01 significance level, we reject the null hypothesis because, p-value is less than that (i.e. significance level)

Conclusion:

We conclude that there is a significant  difference between the proportions.

5 0
3 years ago
What is 117.456 rounded to the nearest tenth
mylen [45]
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4 years ago
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~~~20 POINTS~~~<br> Please answer this question!
dusya [7]
So,

We have 10 mayflies.

Their mean lifespan is 4 hours.  That means that if you add the lifespans of all 10 mayflies and then divide the result by 10, you will get 4.

They have a MAD of 2 hours.  MAD stands for Mean Absolute Deviation (From the Mean, so it's really MADFM :P).  Once you get the mean, which we know is 4, you take the absolute value of the difference between the mean and each mayfly's lifespan, add those differences up, and divide the result by the number of mayflies (10).  Kind of complicated.

So, we can start with the first and second criteria, which is that there are 10 mayflies and their mean is 4 hours.  Therefore, all of their lifespans could be 4 hours.

4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 <span>+ 4 = 40
40/10 = 4
4 is the mean.

Of course, it could also be:
3 + 3</span> + 3 + 3 + 3 + 5 + 5 + 5 + 5<span> + 5 = 40
40/10 = 4
4 is still the mean.

Now that we know the solution set (infinite) for the first and second criteria, we are ready to factor in the third criterion, which is that the MAD is 2 hours.

Once again, to find the MAD, </span><span>you take the absolute value of the difference between the mean and each mayfly's lifespan, add those differences up, and divide the result by the number of mayflies (10).

The MAD of the first set of solutions I gave is 0, because all of the numbers were exactly on the mean.

Let's find the MAD of the second set of solutions I gave.

</span>3 + 3 + 3 + 3 + 3 + 5 + 5 + 5 + 5<span> + 5
</span>
All ten numbers are exactly 1 hour away from the mean.  Therefore, we will add up those differences and divide those differences by the number of mayflies.

1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1<span> + 1 = 10
10/10 = 1
The MAD for this set is 1.  However, we need the MAD to be 2 hours.  To do this, we just need to make all the differences 1 more.

2 + 2</span> + 2 + 2 + 2 + 6 + 6 + 6 + 6<span> + 6

This time, each number is 2 hours away from the mean.

2 + 2</span> + 2 + 2 + 2 + 2 + 2 + 2 + 2<span> + 2 = 20
20/10 = 2

This is a valid solution set that meets all three criteria.

The second question is actually just asking us to add another criterion: make one of the mayflies live for 1 day, or 24 hours.  So we just need to change the last one to 24, right?  Wrong.  You see, we will need to adjust the other numbers so that the mean and the MAD stay the same.  Since all of the numbers added to 40 in our solutions, we will just change the other numbers so that the mean will stay the same.

40 - 24 = 16

That means that the other 9 mayflies will live a total of 16 hours.  Dividing 16 by 9 tells us how long each mayfly has to live.

So our new set is:
</span>\frac{16}{9}+\frac{16}{9}+\frac{16}{9}+\frac{16}{9}+\frac{16}{9}+\frac{16}{9}+\frac{16}{9}+\frac{16}{9}+\frac{16}{9}+24=40
<span>
40/10 = 4

Amazing!  Our mean is still 4.

Now, what is the MAD?

MAD = </span>2\ or\  \frac{18}{9}

Let's see if it still meets this criterion.

\frac{2}{9}+\frac{2}{9}+\frac{2}{9}+\frac{2}{9}+\frac{2}{9}+\frac{2}{9}+\frac{2}{9}+\frac{2}{9}+\frac{2}{9}+ \frac{162}{9} = 20

There IS a problem.  Can we increase some of the numbers so that they are closer to the mean (decreasing the MAD) while decreasing some of the other numbers (to keep the mean constant)?  No.  That will change the mean.  So the answer to that question is no.

NO!!!!!!
6 0
3 years ago
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