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3 years ago

Write a matrix to represent the system of equations.

Mathematics

2 answers:

ioda3 years ago

7 0

Answer:

7 -1 6 | -8

7 -5 7 | 1

0 1 -4 | -5

Step-by-step explanation:

Note that there are four columns in this system, representing the variables x, y and z and constants. So the given matrix pattern is appropriate: It has three columns for the coefficients of x, y and z and one column for the constants.

We need only identify the coefficients and transfer them into the appropriate boxes of the matrix pattern.

We get:

7 -1 6 -8

7 -5 7 1

0 1 -4 -5

ozzi3 years ago

5 0

7 - 1 + 6 | 8

7 - 5 + 7 | 1

3 - 4 | -5

mark my answer as brainlist please

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-3(x + 5) = 6 - 3(2x + 4) solve for x and show work

salantis [7]

Your answer is x = 3.

-3(x + 5) = 6 - 3(2x + 4)
-3x - 15 = 6 -6x - 12
-3x - 15 = -6 - 6x
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3x = -6 + 15
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3 years ago

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KengaRu [80]

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3 years ago

Simplify |–9.23 – 1.79| – |–7.34 – 2.57|.

vampirchik [111]

Answer:

11.02-9.91= 1.11

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3 years ago

Ninety-one percent of products come off the line within product specifications. Your quality control department selects 15 produ

Allisa [31]

Answer:

Probability of stopping the machine when $X < 9$ is 0.0002

Probability of stopping the machine when $X < 10$ is 0.0013

Probability of stopping the machine when $X < 11$ is 0.0082

Probability of stopping the machine when $X < 12$ is 0.0399

Step-by-step explanation:

There is a random binomial variable $X$ that represents the number of units come off the line within product specifications in a review of $n$ Bernoulli-type trials with probability of success $0.91$ . Therefore, the model is ${15 \choose x} (0.91) ^ {x} (0.09) ^ {(15-x)}$ . So:

$P (X < 9) = 1 - P (X \geq 9) = 1 - [{15 \choose 9} (0.91)^{9}(0.09)^{6}+...+{ 15 \choose 15}(0.91)^{15}(0.09)^{0}] = 0.0002$

$P (X < 10) = 1 - P (X \geq 10) = 1 - [{15 \choose 10}(0.91)^{10}(0.09)^{5}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0013$

$P (X < 11) = 1 - P (X \geq 11) = 1 - [{15 \choose 11}(0.91)^{11}(0.09)^{4}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0082$

$P (X < 12) = 1- P (X \geq 12) = 1 - [{15 \choose 12}(0.91)^{12}(0.09)^{3}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0399$

Probability of stopping the machine when $X < 9$ is 0.0002

Probability of stopping the machine when $X < 10$ is 0.0013

Probability of stopping the machine when $X < 11$ is 0.0082

Probability of stopping the machine when $X < 12$ is 0.0399

8 0

3 years ago