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3 years ago

Two lines have the given equations. Use the LINEAR COMBINATION METHOD to solve.

Mathematics

1 answer:

Maurinko [17]3 years ago

8 0

Using linear combination method, the solution to given system of equations are (-7, -15)

<h3><u>Solution:</u></h3>

Linear combination is the process of adding two algebraic equations so that one of the variables is eliminated

Addition is used when the two equations have terms that are exact opposites, and subtraction is used when the two equations have terms that are the same.

<u><em>Given system of equations are:</em></u>

2x - y = 1 ---- eqn 1

3x - y = -6 ------ eqn 2

Subtract eqn 2 from eqn 1

2x - y = 1

3x - y = -6

(-) -------------

-x = 7

Substitute x = -7 in eqn 1

2(-7) - y = 1

-14 - y = 1

y = -14 - 1 = -15

Thus the solution to given system of equations are (-7, -15)

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20 points!! Type the correct answer in the box. Use numerals instead of words. What value of n makes the equation true? -1/5n+7=

vredina [299]

Hey there! :)

Answer:

$\huge\boxed{n = 25}$

-1/5n + 7 = 2

Start by subtracting 7 from both sides:

-1/5n + 7 - (7) = 2 - (7)

-1/5n = -5

Multiply both sides by the reciprocal of -1/5, or -5.

(-5) · (-1/5n) = (-5) · (-5)

n = 25

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3 years ago

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3 years ago

Read 2 more answers

Suppose that in a random selection of 100 colored candies, 21% of them are blue. The candy company claims that the percentage of

zhannawk [14.2K]

Answer:

The pvalue of the test is 0.0784 > 0.01, which means that at this significance level, we do not reject the null hypothesis that the percentage of blue candies is equal to 29%.

Step-by-step explanation:

The candy company claims that the percentage of blue candies is equal to 29%.

This means that the null hypothesis is:

$H_{0}: p = 0.29$

We want to test the hypothesis that this is true, so the alternate hypothesis is:

$H_{a}: p \neq 0.29$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.29 is tested at the null hypothesis:

This means that $\mu = 0.29, \sigma = \sqrt{0.29*0.71}$

Suppose that in a random selection of 100 colored candies, 21% of them are blue.

This means that $n = 100, X = 0.21$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.21 - 0.29}{\frac{\sqrt{0.29*0.71}}{\sqrt{100}}}$

$z = -1.76$

Pvalue of the test:

The pvalue of the test is the probability of the sample proportion differing at least 0.21 - 0.29 = 0.08 from the population proportion, which is 2 multiplied by the pvalue of Z = -1.76.

Looking at the z-table, z = -1.76 has a pvalue of 0.0392

2*0.0392 = 0.0784

The pvalue of the test is 0.0784 > 0.01, which means that at this significance level, we do not reject the null hypothesis that the percentage of blue candies is equal to 29%.

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Answer:

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

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