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Anastasy [175]
3 years ago
13

During the cold winter months, a sheet of ice covers a lake near the Arctic Circle. At the beginning of spring, the ice starts t

o melt.
The variable sss models the ice sheet's thickness (in meters) t weeks after the beginning of spring.
s=-0.25t+4
By how much does the ice sheet's thickness decrease every 6 weeks?

____ meters
Mathematics
1 answer:
kobusy [5.1K]3 years ago
8 0

Answer:

2.5 meters each week.

Step-by-step explanation:

Plug in 6 weeks for <em>t </em>and your equation would be s = -2.5(6) + 4. After solving you would get 2.5 meters.

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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
A landscaper has enough cement to make a patio with an area of 150 Sq ft. The homeowner wants the length of the patio to be 6 ft
Natasha_Volkova [10]
A=lw=150\\l=w+6\\(w+6)w=150\\w^2+6w=150\\w^2+6w-150=0\\w=\frac{-b+-\sqrt{b^2-4ac}}{2a}\\w=\frac{-(6)+-\sqrt{(6)^2-4(1)(-150)}}{2(1)}\\w=\frac{-6+-\sqrt{36+600}}{2}\\w=\frac{-6+-\sqrt{636}}{2}\\w=\frac{-6+-2\sqrt{159}}{2}\\w=-3+-\sqrt{159}>0\\w=-3+\sqrt{159}\\\\l=(-3+\sqrt{159})+6\\l=3+\sqrt{159}

Width = -3 + sqrt(159)
Length = 3 + sqrt(159)
4 0
3 years ago
Solve this inequality: j/4 – 8 &lt; 4.
VLD [36.1K]
J/4 - 8 < 4
×4 ×4
j - 8 < 16
+8 +8
j < 24
8 0
3 years ago
Sam bought shoes for $36.94 and x pairs of socks for $1.92 each. Which expression shows the total money spent? 36.94x + 1.92 36.
NISA [10]
It is 36.94 + 1.92x because you already know how much money he spent on shoes but not for socks
5 0
4 years ago
Read 2 more answers
The greatest common factor of 54x and 96xy is
Fittoniya [83]
54 = 2*3*3*3 96 = 2*2*2*2*2*3 common factors = 2*3 = 6 common factor = 6x
3 0
4 years ago
Read 2 more answers
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