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jeka57 [31]
3 years ago
5

How to factor 6n2 - 6n - 12

Mathematics
1 answer:
alexandr402 [8]3 years ago
7 0

We see that in these 3 terms, 6 is a common factor. So, let's factor out a 6:

6n^2-6n-12 = 6(n^2-n-2)

We can set this equal to 0 and factor by using the quadratic formula which is:

x = \frac{-b \pm \sqrt{b^2-4ac} }{2a}

So, let's do just that:

6(n^2-n-2) = 0

Note that the 6 goes away if you divide both sides by 6. In this case, a = 1, b = -1, and c = -2. Let's plug that into the quadratic equation:

\frac{1 \pm \sqrt{(-1)^2-4(1)(-2)} }{2(1)} = \frac{-1 \pm \sqrt{1-(-8)} }{2}

\frac{-1 \pm \sqrt{9} }{2} = {1, -2}

So, we can write this as:

6n^2 - 6n - 12 = 6(n-1)(n+2)

Notice that the 6 comes back because it was only temporarily mad. And that the roots have opposite signs in the parentheses because to find the roots, you need to set each parentheses equal to 0 and solve for n. With that in mind, your final answer is 6(n-1)(n+2). Hope I could help you!

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Answer:

(a) <em>H</em>₀: <em>μ ≥ 5.70 </em><em>vs. </em><em>Hₐ</em>:<em>μ < 5.70</em>

(b) The rejection region is (<em>t₀.₀₁,₉</em> <em>≤ -2.821</em>).

(c) The value of the test statistic is -4.33.

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Step-by-step explanation:

A hypothesis test should be conducted to determine that the if the true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa.

(a)

The hypothesis is:

<em>H</em>₀: The true mean breaking strength of the new bonding adhesive is not less than 5.70 Mpa, i.e. <em>μ ≥ 5.70</em><em>.</em>

<em>Hₐ</em>: The true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa, i.e. <em>μ < 5.70</em><em>.</em>

(b)

The alternate hypothesis indicates that the hypothesis test is left-tailed.

The rejection region for the left tailed test will be towards the lower tail of the t<em>-</em>distribution curve.

The significance level of the test is: <em>α</em> = 0.01.

The critical value is:

t_{\alpha ,(n-1)}=t_{0.01,(10-1)}=t_{0.01,9}

Use the <em>t-</em>table for the critical value.

t_{\alpha ,(n-1)}=t_{0.01,9}=-2.821

Since rejection region is in the lower tail the critical value will be negative.

Thus, the rejection region is (<em>t₀.₀₁,₉</em> <em>≤ -2.821</em>).

(c)

The test statistic value is:

t=\frac{\bar x-\mu}{s/\sqrt{n}}

Given:

\bar x=5.07\\s=0.46\\n=10\\\mu=5.70

Compute the value of the <em>t</em>-statistic as follows:

t=\frac{\bar x-\mu}{s/\sqrt{n}}=\frac{5.07-5.70}{0.46/\sqrt{10}} =-4.33

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This implies that the test statistic lies in the rejection region.

Hence the null hypothesis will be rejected at 1% significance level.

<u>Conclusion:</u>

As the null hypothesis is rejected it can be concluded that the true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa.

(e)

The conditions required for the <em>t-</em>test for single mean to be valid is:

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