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3 years ago

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What is the absolute davatation round to the nearest tenth for 6.3, 6.4, 6.5, 6.6, 6.8. 6.8, 7.5?

2 answers:

WITCHER [35]3 years ago

8 0

Let's find the mean of all the values first.
6.3+6.4+6.5+6.6+6.8+6.8+7.5= 46.9
Let's divide by the number of values.
46.9÷7=
6.7
Now let's find the distance that each number is away from 6.7 and find the mean of those numbers.
0.4+0.3+0.2+0.1+0.1+0.1+0.8=
.2857
≈ 0.3
So, the absolute deviation is 0.3.

bija089 [108]3 years ago

6 0

Answer:

6.2

Step-by-step explanation:

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andreyandreev [35.5K]

Answer:

The given system of equations has solutions below:

1) The solution is (2,3)

2) The solution is ( $\frac{-8}{7}, \frac{5}{7}$ )

3) The solution is infinitely many solutions

4) No solution

Step-by-step explanation:

Given system of equation are

$-x+2y=4\hfill (1)$

$-4x+y=-5\hfill (2)$

To solve equation by using elimination method

Multiply eqn (2) into 2

$-8x+2y=-10\hfill (3)$

Now subtracting (1) and (3)

$-x+2y=4$

$-8x+2y=-10$

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7x=14

x= $\frac{14}{7}$

$x=2$

Substitute x=2 in equation (1)

-2+2y=4

2y=4+2

$y=\frac{6}{2}$

y=3

Therefore the solution is (2,3)

2) Given equation is

$-2x+y=3\hfill (1)$

4y-4=x

Rewritting as below

$x-4y=-4\hfill (2)$

To solve equation by using elimination method

multiply (2) into 2

$2x-8y=-8\hfill (3)$

Adding (1) and (3)

-2x+y=3

2x-8y=-8

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-7y=-5

$y=\frac{5}{7}$

substitute $y=\frac{5}{7}$ in (1)

$-2x+\frac{5}{7}=3$

$-2x=3-\frac{5}{7}$

$-2x=\frac{21-5}{7}$

$x=-\frac{8}{7}$

Therefore the solution is ( $\frac{-8}{7},\frac{5}{7}$ )

3) Given equation is $6x+2y=10\hfill (1)$

$3x+y=5\hfill (2)$

equation (1) can be written as

2(3x+y)=10

$3x+y=\frac{10}{2}$

3x+y=5

Therefore equations (1) and (2) are same therefore it has infinitely many solutions

4) Given equation is $-x-2y=14\hfill (1)$

$-2x-4y=12\hfill (2)$

multiply equation (1) into 2

$-2x-4y=28\hfill (3)$

To solve equation by using elimination method

subtracting (2) and (3)

-2x-4y=28

-2x-4y=12

_______

$28\neq -12$

therefore it has no solution

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3 years ago

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Tanzania [10]

Answer:

I am not sure, ig it's D...

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2 years ago

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First of all, we can observe that

$x^2-6x+9 = (x-3)^2$

So the expression becomes

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This means that the expression is defined for every $x\neq 3$

Now, since the denominator is always positive (when it exists), the fraction can only be positive if the denominator is also positive: we must ask

$x^2-4 \geq 0 \iff x\leq -2 \lor x\geq 2$

Since we can't accept 3 as an answer, the actual solution set is

$(-\infty,-2] \cup [2,3) \cup (3,\infty)$

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3 years ago

Help, how would I get my answer for this question?

Maslowich

Answer:

$(-4,32)$

Step-by-step explanation:

Given points are $(-1,-8)$

And given transformation is $D_4$ $r_{x-axis}$

We will start from left to right.

First transformation is reflection about x-axis.

When we reflect about x-axis $(x,y)\ became\ (x,-y)$

So, $(-1,-8)=[-1,-(-8)]=(-1,8)$

Now next transformation is dilation with a factor 4.

If we do dilation with a factor $'k'$ to the point $(x,y)$

New co-ordinates after dilation became $(kx,ky)$

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Answer:

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Step-by-step explanation:

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