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3 years ago

I need to know how to solve this using synthetic division

Mathematics

1 answer:

kati45 [8]3 years ago

8 0

Recall, first off, you sort the terms in descending order, in this case, they already are in descending order. Then you use the coefficients. first off you drop down the first value, and take it from there,

$\bf p(x)=\underline{3} x^3\underline{-4} x^2\underline{-5} x\underline{+1}\qquad a=2\\\\\\
\begin{array}{r|rrrrrrl}
2&&3&-4&-5&1\\
&&&6&4&-2\\
-&&--&--&--&--\\
&&3&2&-1&\boxed{-1}&\impliedby remainder
\end{array}
\\\\\\
3x^2+2x-1x+\boxed{\frac{-1}{3x^3-4x^2-5x+1}}
\\\\\\
3x^2+2x-x-\frac{1}{3x^3-4x^2-5x+1}$

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2 years ago

Find the exact value of cos(a+b) if cos a=-1/3 and cos b=-1/4 if the terminal side if a lies in quadrant 3 and the terminal side

maria [59]

Answer:

cos(a + b) = $\frac{1}{12}(1-2\sqrt{30})$

Step-by-step explanation:

cos(a + b) = cos(a).cos(b) - sin(a).sin(b) [Identity]

cos(a) = $-\frac{1}{3}$

cos(b) = $-\frac{1}{4}$

Since, terminal side of angle 'a' lies in quadrant 3, sine of angle 'a' will be negative.

sin(a) = $-\sqrt{1-(-\frac{1}{3})^2}$ [Since, sin(a) = $\sqrt{(1-\text{cos}^2a)}$ ]

= $-\sqrt{\frac{8}{9}}$

= $-\frac{2\sqrt{2}}{3}$

Similarly, terminal side of angle 'b' lies in quadrant 2, sine of angle 'b' will be negative.

sin(b) = $-\sqrt{1-(-\frac{1}{4})^2}$

= $-\sqrt{\frac{15}{16}}$

= $-\frac{\sqrt{15}}{4}$

By substituting these values in the identity,

cos(a + b) = $(-\frac{1}{3})(-\frac{1}{4})-(-\frac{2\sqrt{2}}{3})(-\frac{\sqrt{15}}{4})$

= $\frac{1}{12}-\frac{\sqrt{120}}{12}$

= $\frac{1}{12}(1-\sqrt{120})$

= $\frac{1}{12}(1-2\sqrt{30})$

Therefore, cos(a + b) = $\frac{1}{12}(1-2\sqrt{30})$

5 0

3 years ago

Two catalysts may be used in a batch chemical process. Twelve batches were prepared using catalyst 1, resulting in an average yi

aalyn [17]

Answer:

a) $t=\frac{(91-85)-(0)}{2.490\sqrt{\frac{1}{12}}+\frac{1}{15}}=6.222$

$df=12+15-2=25$

$p_v =P(t_{25}>6.222) =8.26x10^{-7}$

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

b) $(91-85) -2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =3.309$

$(91-85) +2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =8.691$

Step-by-step explanation:

Notation and hypothesis

When we have two independent samples from two normal distributions with equal variances we are assuming that

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}}+\frac{1}{n_2}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

This last one is an unbiased estimator of the common variance $\sigma^2$

Part a

The system of hypothesis on this case are:

Null hypothesis: $\mu_2 \leq \mu_1$

Alternative hypothesis: $\mu_2 > \mu_1$

Or equivalently:

Null hypothesis: $\mu_2 - \mu_1 \leq 0$

Alternative hypothesis: $\mu_2 -\mu_1 > 0$

Our notation on this case :

$n_1 =12$ represent the sample size for group 1

$n_2 =15$ represent the sample size for group 2

$\bar X_1 =85$ represent the sample mean for the group 1

$\bar X_2 =91$ represent the sample mean for the group 2

$s_1=3$ represent the sample standard deviation for group 1

$s_2=2$ represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

$\S^2_p =\frac{(12-1)(3)^2 +(15 -1)(2)^2}{12 +15 -2}=6.2$

And the deviation would be just the square root of the variance:

$S_p=2.490$

Calculate the statistic

And now we can calculate the statistic:

$t=\frac{(91-85)-(0)}{2.490\sqrt{\frac{1}{12}}+\frac{1}{15}}=6.222$

Now we can calculate the degrees of freedom given by:

$df=12+15-2=25$

Calculate the p value

And now we can calculate the p value using the altenative hypothesis:

$p_v =P(t_{25}>6.222) =8.26x10^{-7}$

Conclusion

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

Part b

For this case the confidence interval is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} S_p \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

For the 99% of confidence we have $\alpha=1-0.99 = 0.01$ and $\alpha/2 =0.005$ and the critical value with 25 degrees of freedom on the t distribution is $t_{\alpha/2}= 2.79$

And replacing we got:

$(91-85) -2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =3.309$

$(91-85) +2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =8.691$

7 0

3 years ago