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3 years ago

What is the value of n in the equation –1/2(2n + 4) + 6 = –9 + 4(2n + 1)?

Mathematics

1 answer:

ioda3 years ago

7 0

Answer:

Step-by-step explanation:

$-\dfrac{1}{2}(2n-4)+6=-9+4(2n+1)\qquad\text{use the distributive property}\\\\\left(-\dfrac{1}{2}\right)(2n)+\left(-\dfrac{1}{2}\right)(-4)+6=-9+(4)(2n)+(4)(1)\\\\-n-2+6=-9+8n+4\qquad\text{combine like terms}\\\\-n+(-2+6)=8n+(-9+4)\\\\-n+4=8n-5\qquad\text{subtract 4 from both sides}\\\\-n=8n-9\qquad\text{subtract}\ 8n\ \text{from both sides}\\\\-9n=-9\qquad\text{divide both sides by (-9)}\\\\n=1$

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Vladimir [108]

Answer:

The value of t test statistics is 5.9028.

We conclude that the true mean is greater than 10 at the .01 level of significance.

Step-by-step explanation:

We are given that a firm issued a guideline that transmissions of 10 pages or more should be sent by 2-day mail instead.

The firm examined 35 randomly chosen fax transmissions during the next year, yielding a sample mean of 14.44 with a standard deviation of 4.45 pages.

<em />

<em>Let </em> $\mu$ <em> = true mean transmission of pages.</em>

So, Null Hypothesis, $H_0$ : $\mu$ $\leq$ 10 pages {means that the true mean is smaller than or equal to 10}

Alternate Hypothesis, $H_A$ : $\mu$ > 10 pages {means that the true mean is greater than 10}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = 14.44 pages

s = sample standard deviation = 4.45 pages

n = sample of fax transmissions = 35

So, <u><em>test statistics</em></u> = $\frac{14.44-10}{\frac{4.45}{\sqrt{35} } }$ ~ $t_3_4$

= 5.9028

(a) The value of t test statistics is 5.9028.

Now, at 0.01 significance level the z table gives critical values of 2.441 at 34 degree of freedom for right-tailed test.

<em>Since our test statistics is more than the critical values of z as 5.9028 > 2.441, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis.</u></em>

Therefore, we conclude that the true mean is greater than 10.

(b) Now, P-value of the test statistics is given by the following formula;

P-value = P( $t_3_4$ > 5.9028) = Less than 0.05% {using t table)

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