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jenyasd209 [6]
3 years ago
9

Write the system of equations represented by the following matrices and matrix equation

Mathematics
1 answer:
MaRussiya [10]3 years ago
8 0
We have:

A=\left[\begin{array}{cc}3&-5\\4&1\end{array}\right]\qquad X=\left[\begin{array}{c}a\\b\end{array}\right]\qquad C=\left[\begin{array}{c}2\\10\end{array}\right]

so:

A\cdot X=C\\\\\\
\left[\begin{array}{cc}3&-5\\4&1\end{array}\right]\cdot\left[\begin{array}{c}a\\b\end{array}\right]=\left[\begin{array}{c}2\\10\end{array}\right]\\\\\\
\left[\begin{array}{c}3a-5b\\4a+1b\end{array}\right]=\left[\begin{array}{c}2\\10\end{array}\right]\\\\\\
\boxed{\begin{cases}3a-5b=2\\4a+b=10\end{cases}}

And the solution of the system of equations:

\begin{cases}3a-5b=2\\4a+b=10\quad|\cdot5\end{cases}\\\\\\
\begin{cases}3a-5b=2\\20a+5b=50\end{cases}\\---------(+)\\\\3a-5b+20a+5b=2+50\\\\3a+20a=52\\\\23a=52\quad|:23\\\\\boxed{a=\dfrac{52}{23}=2\dfrac{6}{23}}\\\\\\
4a+b=10\\\\b=10-4a\\\\b=10-4\cdot\dfrac{52}{23}\\\\\\b=\dfrac{230}{23}-\dfrac{208}{23}\\\\\\\boxed{b=\dfrac{22}{23}}\\\\\\
\boxed{\begin{cases}a=2\dfrac{6}{23}\\\\b=\dfrac{22}{23}\end{cases}}
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The personnel files of all eight employees at the Pawnee location of Acme Carpet Cleaners Inc. revealed that during the last six
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Answer:

a) For Pawnee:

Range = 11, Mean = 3.13, Mean deviation = 2.44

For Chickpee:

Range =  7, Mean = 4.25, Mean deviation = 1.59

b) Pawnee Location has fewer lost days.

c) variation of the Chickpee is less than Pawnee location

Step-by-step explanation:

Data given:

For Pawnee:

3 1 0 0 2 11 5 3

Max = 11

Min = 0

Range = Max - Min

Range = 11-0 = 11

Range = 11

For Chickpee:

3 6 8 4 4 5 3 1

Max = 8

Min = 1

Range = Max - Min

Range = 8 - 1

Range = 7

For mean of Pawnee:

Mean = (3 + 1 + 0 + 0 + 2 + 11 + 5 + 3)/8

Mean = 25/8

Mean = 3.125

For mean deviation, we find the difference from the mean of each point.

Note: all the negative values will be taken as 0.

Differences from the mean of each point =

(3-3.13)=0.13, (1-3.31) = 2.13 ,

(0-3.13)=0, (0-3.13)=3.13,

(2-3.13)=1.13, (11-3.13)=7.87,

(5-3.13)=1.87, (3-3.13)=0.13

So, our differences are:

(0.13, 2.13, 0, 3.13, 1.13, 7.87, 1.87, 0.13 )

Mean deviation =  Sum of all differences/8

Mean deviation = (0.13+2.13+3.13+3.13+ 1.13+ 7.87+ 1.87+ 0.13)/8

Mean deviation = 2.44

For mean of Chickpee:

Mean = (3+6+8+4+4+5+3+1)/8

Mean = 34/8

Mean = 4.25

Similarly, we need to find mean deviation for Chickpee, for that we need to find the differences first as done above for Pawnee.

Note: all the negative values will be taken as 0

Differences from the mean:

(3-4.25) = 1.25, (6-4.25) =1.75,

(8-4.25) = 3.75, (4-4.25)=0.25,

(4-4.25)=0.25, (5-4.25)= 0.75,

(3-4.25)=1.25, (1-4.25) =3.25

So, the differences are:

Differences = (1.25,1.75,3.75,0.25,0.25,0.75,1.25,3.25)

Mean deviation = (1.25+ 1.75+ 3.75+ 0.25+ 0.25+ 0.75+ 1.25+ 3.25)/8

Mean deviation = 1.59

b) which location has fewer lost days:

This can be found out by the total number of days of Pawnee and chickpee.

Pawnee = Sum of values = (3 + 1 + 0 + 0 + 2 + 11 + 5 + 3) = 25 days

Chickpee = Sum of values = (3+6+8+4+4+5+3+1) = 34 days

Hence, Pawnee Location has fewer lost days.

C) which location has less variation?

Formula for variation = ∑(\frac{x^{2} }{8}) - (mean^{2})

where x = values of the set.

For Pawnee:

mean^{2} = 3.13^{2} = 9.8

Sum of square of all the data points of Pawnee = 169

Variation = 169/8 - 98 = 11.325

Similarly,

For Chickpee:

mean^{2}  = 4.25^{2} = 18.06

Sum of square of all the data points of Chickpee = 176

Variation = 176/8 - 18.06

Variation = 22-18.06

Variation = 4

Hence, variation of the Chickpee is less than Pawnee location

8 0
3 years ago
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