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2 years ago

2x-y=52x-2y=4

3D%204" id="TexFormula1" title="2 \times - y = 5 \\ 2 \times - 2y = 4" alt="2 \times - y = 5 \\ 2 \times - 2y = 4" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

d1i1m1o1n [39]2 years ago

7 0

$\left\{\begin{array}{ccc}2x-y=5\\2x-2y=4&|\cdot(-1)\end{array}\right\\\underline{-\left\{\begin{array}{ccc}2x-y=5\\-2x+2y=-4\end{array}\right}\ \ \ \ |add\ both\ sides\\.\ \ \ \ \ \ \ \ y=1\\\\substitute\ the\ value\ of\ y\ to\ first\ equation\\\\2x-1=5\ \ \ |+1\\2x=6\ \ \ |:2\\x=3\\\\\left\{\begin{array}{ccc}x=3\\y=1\end{array}\right$

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3 years ago

Read 2 more answers

Data is collected to compare two different types of batteries. We measure the time to failure of 12 batteries of each type. Batt

-Dominant- [34]

Using the t-distribution, it is found that since the test statistic is greater than the critical value for the right-tailed test, it is found that there is enough evidence to conclude that Battery B outlasts Battery A by more than 2 hours.

At the null hypothesis, it is tested if it does not outlast by more than 2 hours, that is, the subtraction is not more than 2:

$H_0: \mu_B - mu_A \leq 2$

At the alternative hypothesis, it is tested if it outlasts by more than 2 hours, that is:

$H_1: \mu_B - \mu_A > 2$

The sample means are: $\mu_A = 8.65, \mu_B = 11.23$

The standard deviations for the samples are $s_A = s_B = 0.67$

Hence, the standard errors are:

$s_{Ea} = S_{Eb} = \frac{0.67}{\sqrt{12}} = 0.1934$

The distribution of the difference has mean and standard deviation given by:

$\overline{x} = \mu_B - \mu_A = 11.23 - 8.65 = 2.58$

$s = \sqrt{s_{Ea}^2 + s_{Eb}^2} = \sqrt{0.1934^2 + 0.1934^2} = 0.2735$

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{s}$

In which $\mu = 2$ is the value tested at the hypothesis.

Hence:

$t = \frac{\overline{x} - \mu}{s}$

$t = \frac{2.58 - 2}{0.2735}$

$t = 2.12$

The critical value for a right-tailed test, as we are testing if the subtraction is greater than a value, with a 0.05 significance level and 12 + 12 - 2 = 22 df is given by $t^{\ast} = 1.71$

Since the test statistic is greater than the critical value for the right-tailed test, it is found that there is enough evidence to conclude that Battery B outlasts Battery A by more than 2 hours.

A similar problem is given at brainly.com/question/13873630

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