Find the mean and mode of the set of data. 8, 7, 4, 6, 6

Use matrix inversion to solve the collection of systems of linear equations. HINT [See Example 4.] (a) −x − 4y + 2z = −10 x + 2y

omeli [17]

Answer:

Step-by-step explanation:

given is a system of linear equations in 3 variables as

$-x - 4y + 2z = −10\\ x + 2y - z = 11 \\x + y - z = 14$

This can be represented in matrix form as

AX=B Or

$\left[\begin{array}{ccc}-1&-4&2\\1&2&-1\\1&1&-1\end{array}\right] *\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}-10\\11\\14\end{array}\right]$

So solution set

X would be $A^{-1} B$

|A|=-1(-1)+4(0)+2(-1)=--1

Cofactors of A are

-1 0 -1

-2 -1 -3

0 1 2

So inverse of A is

1 2 0

0 1 -1

1 3 -2

Solution set would be

x=12

y=-3

z=-5

8 0

3 years ago

Colin has a pad with x pieces of paper on it. For his first class, he wrote on 5 fewer than half of the pieces of paper in the p

PIT_PIT [208]

Answer:

Colin has 8 sheets left for his third class.

Step-by-step explanation:

Given that:

Total Number of pieces of papers = $x$

Number of pieces of papers used for 1st class = 5 fewer than half of the pieces in the pad

Writing the equation:

$\text{Number of pieces of papers used for 1st class =} \dfrac{x}{2} -5 ...... (1)$

Also, Given that number of pieces of papers used for the 2nd class are 2 more than that of papers used in the 1st class.

$\text{Number of pieces of papers used for 2nd class =} \dfrac{x}{2} -5+2 = \dfrac{x}2 -3 ...... (2)$

Now, number of pieces of papers left for the third class = Total number of pieces of papers in the pad - Number of pieces of papers used in the first class - Number of pieces of papers used in the first class

$\text{number of pieces of papers left for the third class = }x-(\dfrac{x}{2}-5)-(\dfrac{x}{2}-3)\\\Rightarrow x-\dfrac{x}2-\dfrac{x}2+5+3\\\Rightarrow x-x+5+3\\\Rightarrow 8$