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2 years ago

According to the Rational Root Theorem, which number is a potential root of f(x) = 9x8 + 9x6 – 12x + 7?

Mathematics

2 answers:

MrRa [10]2 years ago

5 0

7/3 is the answer to your question

telo118 [61]2 years ago

3 0

Answer:

All the potential root of f(x) are $\pm 1,\pm7, \pm \frac{1}{3},\pm \frac{7}{3}, \pm \frac{1}{9},\pm \frac{7}{9}$ .

Step-by-step explanation:

According to the rational root theorem, all the potential root of f(x) are defined as

$x=\pm\frac{p}{q}$

Where, p is factor of constant term and q is factor of leading coefficient.

The given function is

$f(x)=9x^8+9x^6-12x+7$

Here, constant term is 7 and leading coefficient is 9.

Factors of 7 are ±1, ±7 and the factors of 9 are ±1, ±3, ±9.

Using rational root theorem, all the potential root of f(x) are

$x=\pm 1,\pm7, \pm \frac{1}{3},\pm \frac{7}{3}, \pm \frac{1}{9},\pm \frac{7}{9}$

Therefore all the potential root of f(x) are $\pm 1,\pm7, \pm \frac{1}{3},\pm \frac{7}{3}, \pm \frac{1}{9},\pm \frac{7}{9}$ .

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If cosine theta almost-equals 0.3090, which of the following represents approximate values of sine theta and tangent theta, for

Zepler [3.9K]

Answer:

$\sin \theta = 0.9511$ and $\tan \theta = 3.078$

Step-by-step explanation:

Given that, $\cos \theta = 0.309$ .

Now, we have to calculate the values of $\sin \theta$ and $\tan \theta$ .

We know the identity that, $\sin^{2}\theta + \cos^{2}\theta = 1$

So, $\sin \theta = \sqrt{1 + \cos^{2}\theta} = \sqrt{1 - (0.309)^{2}} = 0.9511$ {Since $0 \leq \theta \leq 90^{\circ}$ }

Now, $\sec \theta = \frac{1}{\cos\theta} = \frac{1}{0.309} = 3.236$

Then, we know the identity, $\sec^{2}\theta - \tan^{2}\theta = 1$

So, $\tan \theta = \sqrt{\sec^{2}\theta - 1} = \sqrt{(3.236)^{2} - 1} = 3.078$ (Answer) {Since $0 \leq \theta \leq 90^{\circ}$ }

6 0

2 years ago

Read 2 more answers

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

8 0

2 years ago

A factory makes propeller drive shafts for ships. A quality assurance engineer at the factory needs to estimate the true mean le

borishaifa [10]

Answer:

$1000-1.96\frac{2}{\sqrt{4}}=998.04$

$1000+1.96\frac{2}{\sqrt{4}}=1001.96$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$