Which inequalities are true? Check all that apply.

~~~20 POINTS~~~ 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 + 4 = 40
40/10 = 4
4 is the mean.

Of course, it could also be:
3 + 3 + 3 + 3 + 3 + 5 + 5 + 5 + 5 + 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, 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.

3 + 3 + 3 + 3 + 3 + 5 + 5 + 5 + 5 + 5

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 + 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 + 2 + 2 + 2 + 6 + 6 + 6 + 6 + 6

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

2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 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:
 $\frac{16}{9}+\frac{16}{9}+\frac{16}{9}+\frac{16}{9}+\frac{16}{9}+\frac{16}{9}+\frac{16}{9}+\frac{16}{9}+\frac{16}{9}+24=40$

40/10 = 4

Amazing! Our mean is still 4.

Now, what is the MAD?

MAD = $2\ or\ \frac{18}{9}$

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

A physicist examines 10 water samples for iron concentration. The mean iron concentration for the sample data is 0.711 cc/cubic

Ratling [72]

Answer:

Critical value: T = 1.8331

The 90% confidence interval for the population mean iron concentration is between 0.561 cc/m³ and 0.861 cc/m³

Step-by-step explanation:

We have the standard deviation for the sample. So we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 10 - 1 = 9

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.95$ . So we have T = 1.8331, which is the critical value.