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3 years ago

4x + 2y = 8 16x - y = 14

1 answer:

7 0

Equation 1) 4x + 2y = 8
Equation 2) 16x - y = 14

Multiply all of equation 2 by 2.

2) 2(16x - y = 14)

2) 32x - 2y = 28
1) 4x + 2y = 8

Add equations together.

36x = 36

Divide both sides by 36.

x = 1

Plug in 1 for x in the first equation.

4x + 2y = 8

4(1) + 2y = 8

4 + 2y = 8

Subtract 4 from both sides.

2y = 8 - 4

2y = 4

Divide both sides by 2.

y = 4/2

y = 2

So, x = 1, y = 2

~Hope I helped!~

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3 years ago

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3 years ago

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Answer:

SAS postulate

Step-by-step explanation:

AD (common)

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3 years ago

Suppose that a recent article stated that the mean time spent in jail by a first-time convicted burglar is 2.5 years. A study wa

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Using the z-distribution, it is found that since the test statistic is greater than the critical value, it can be concluded that the mean length of jail time has increased.

At the null hypothesis, it is tested if the mean length of jail time is still of 2.5 years, that is:

$H_0: \mu = 2.5$

At the alternative hypothesis, it is tested if it has increased, that is:

$H_1: \mu > 2.5$

We have the standard deviation for the population, thus, the z-distribution is used. The test statistic is given by:

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

The parameters are:

$\overline{x}$ is the sample mean.
$\mu$ is the value tested at the null hypothesis.
$\sigma$ is the standard deviation of the sample.
n is the sample size.

For this problem, the values of the parameters are: $\overline{x} = 3, \mu = 2.5, \sigma = 1.5, n = 26$

Hence, the value of the test statistic is:

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

$z = \frac{3 - 2.5}{\frac{1.5}{\sqrt{26}}}$

$z = 1.7$

The critical value for a right-tailed test, as we are testing if the mean is greater than a value, with a significance level of 0.05, is of $z^{\ast} = 1.645$

Since the test statistic is greater than the critical value, it can be concluded that the mean length of jail time has increased.

A similar problem is given at brainly.com/question/24166849

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2 years ago

A 32 foot tall tree casts a shadow that is 48 feet long. How far away from the tree is a

sveticcg [70]

Answer:

39 feet

Step-by-step explanation:

take the tree as AB , the distance between the man and the tree as BC, the distance between the man and the tip of the shadow CD and the point intersecting the hypotenuse CE

since CE parallel to AB we can BPT

CE/AB = CD/BD

6/32= CD/48

CD = 9 feet

since CD is feet

BC is 48-9 = 39 feet

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2 years ago