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irina [24]
3 years ago
13

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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What is −3 1/3⋅(−2 7/10) in simplest form
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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
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