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Mariulka [41]
3 years ago
13

You weigh six packages and find the weights to be 20, 12, 52, 16, 48, and 44 ounces. If you include a package that weighs 60 oun

ces, which will increase more, the median or the mean?
Mathematics
1 answer:
kvv77 [185]3 years ago
3 0

Answer:

median

Step-by-step explanation:

The first step is to sort the numbers from largest to smallest (you could go the other way as well).

60 52 48 44 20 16 12

The median is the middle number 44 in this case. The median has 3 numbers on its left and 3 on its right (in this case). 44 has 60 52 48 or its left and 20 16 and 12 on its right.

The mean is the average. We'll call it the new average

The mean is 60 + 52 + 48 + 44 + 20 + 16 + 12 = 252

The new mean is 252 / 7 = 36

=====================

The old median is the average between the middle 2 numbers (before you added 60 you had 6 numbers)

52 + 48 + 44 + 20 + 16 + 12 = the average of 44 + 20 which 64/2 = 32

The old mean was (52 + 48 + 44 + 20 + 16 + 12)/6 = 32.

=======================

The median increased from 44 to 32 = 12

The mean went from 32 to 36 which is 4

Answer: the median increased more  

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Answer:

A) Verified

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Step-by-step explanation:

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A^{-1}=\frac{1}{a.d-b.c} \left[\begin{array}{ccc}d&-b\\-c&a\end{array}\right]

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B^{-1}A^{-1}=\frac{1}{(a.e+b.g)(c.f+d.h)-(a.f+b.h)(c.e+d.g)}\left[\begin{array}{ccc}c.f+d.h&-(a.f+b.h)\\-(c.e+d.g)&a.e+b.g\end{array}\right]

So it is proved that the results are same.

b) Now, let's prove it for any n x n matrix.

(AB)(AB)^{-1}=I\\\\A^{-1}(AB)(AB)^{-1}=A^{-1}I\\\\IB(AB)^{=1}=A^{-1}I\\\\B(AB)^{=1}=A^{-1}\\\\B^{-1}B(AB)^{=1}=B^{-1}A^{-1}\\\\I(AB)^{=1}=B^{-1}A^{-1}\\\\(AB)^{=1}=B^{-1}A^{-1}

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</span>
The correct result would be -18 * i.
4 0
3 years ago
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