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lana66690 [7]
3 years ago
5

PLSSSSSSSSHELPMEPLSSSSSS

Mathematics
2 answers:
andreev551 [17]3 years ago
8 0
23 miles



15-2.35 = 12.65
12.65/ .55= 23
Fittoniya [83]3 years ago
4 0
It has to be 23 miles fam
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Sam was given a bag of 100 red and blue marbles. He believes
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(D) There are probably more blue marbles than red marbles in the bag.

Step-by-step explanation:

There are a total of 100 marbles in the bag.

In the experiment of 50 trials, Sam had the following outcome:

Blue=35

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Relative Frequency of Blue Marbles =35/50=0.7

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Solve the system of equations: <br> X+y+z=5<br> -3x-4y+4z=15<br> 2x-y-4z=-8
atroni [7]

Answer:

  (x, y, z) = (3, -2, 4)

Step-by-step explanation:

The attachment shows a solution using a graphing calculator where the first equation is solved for z, and that expression is substituted into each of the other two equations. The solution to that system is (x, y) = (3, -2). Using these values in the expression for z, we find ...

  z = 5 -(3 +(-2)) = 4

The solution is (x, y, z) = (3, -2, 4).

__

Your graphing calculator and/or online tools can give you the reduced row echelon form of the augmented matrix representing the system:

  \left[\begin{array}{ccc|c}1&1&1&5\\-3&-4&4&15\\2&-1&-4&-8\end{array}\right]

__

If you're solving by hand, you can do the substitution described above to eliminate z from one equation. You can eliminate z from another equation by adding the last two:

  -3x -4y +4(5 -x -y) = 15   ⇒   -7x -8y = -5

  (-3x -4y +4z) +(2x -y -4z) = (15) +(-8)   ⇒   -x -5y = 7

This pair of equations can be solved for x and y in any of the usual ways. Using the second equation to substitute for x, for example, gives you ...

  -7(-5y -7) -8y = -5   ⇒   27y = -54   ⇒   y = -2

  x = -5(-2) -7 = 3

  z = 5 -(3) -(-2) = 4

_____

<em>Additional comment</em>

If all that is needed is a solution, we prefer a graphical or matrix approach. These take about the same amount of time. Both require a little additional effort to determine exact values when the solutions are not integers.

6 0
2 years ago
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