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10 months ago

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Rewrite the expression (19x3 – 5x2 + 6x – 2)/(x + 1) in the form q(x) + r(x)/b(x) where q(x) = quotient, r(x) = remainder, and b

(x) = divisor, using the long division method.

1 answer:

wolverine [178]10 months ago

7 0

We are asked to perform the long division on the following expression.

$\frac{(19x^3-5x^2+6x-2)}{(x+1)}$

Let us perform the long division.

So, we have

Quotient: q(x) = 19x^2 - 24x + 30

Remainder: r(x) = -32

Divisor: b(x) = x + 1

Therefore, the equivalent expression is

$19x^2-24x+30-\frac{32}{(x+1)}$

The 3rd option is the correct answer.

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Cos ( α ) = √ 6/ 6 and sin ( β ) = √ 2/4 . Find tan ( α − β )

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Answer:

$\purple{ \bold{ \tan( \alpha - \beta ) = 1.00701798}}$

Step-by-step explanation:

$\cos( \alpha ) = \frac{ \sqrt{6} }{6} = \frac{1}{ \sqrt{6} } \\ \\ \therefore \: \sin( \alpha ) = \sqrt{1 - { \cos}^{2} ( \alpha ) } \\ \\ = \sqrt{1 - \bigg( {\frac{1}{ \sqrt{6} } \bigg )}^{2} } \\ \\ = \sqrt{1 - {\frac{1}{ {6} }}} \\ \\ = \sqrt{ {\frac{6 - 1}{ {6} }}} \\ \\ \red{\sin( \alpha ) = \sqrt{ { \frac{5}{ {6} }}} } \\ \\ \tan( \alpha ) = \frac{\sin( \alpha ) }{\cos( \alpha ) } = \sqrt{5} \\ \\ \sin( \beta ) = \frac{ \sqrt{2} }{4} \\ \\ \implies \: \cos( \beta ) = \sqrt{ \frac{7}{8} } \\ \\ \tan( \beta ) = \frac{\sin( \beta ) }{\cos( \beta ) } = \frac{1}{ \sqrt{7} } \\ \\ \tan( \alpha - \beta ) = \frac{ \tan \alpha - \tan \beta }{1 + \tan \alpha . \tan \beta} \\ \\ = \frac{ \sqrt{5} - \frac{1}{ \sqrt{7} } }{1 + \sqrt{5} . \frac{1}{ \sqrt{7} } } \\ \\ = \frac{ \sqrt{35} - 1 }{ \sqrt{7} + \sqrt{5} } \\ \\ \purple{ \bold{ \tan( \alpha - \beta ) = 1.00701798}}$

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2 years ago

Complete the sentence.<br> 9 is 18% of

vfiekz [6]

Answer: 50%

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In

lord [1]

Answer:

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

P.Q. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of males having blood disorder= $\frac{250}{1000}$ = 0.25

$\hat p_2$ = sample proportion of females having blood disorder = $\frac{275}{1000}$ = 0.275

$n_1$ = sample of males = 1000

$n_2$ = sample of females = 1000

$p_1$ = population proportion of males having blood disorder

$p_2$ = population proportion of females having blood disorder

<em>Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.</em>

<u>So, 95% confidence interval for the difference between the population proportions, </u><u>(</u> $p_1-p_2$ <u>)</u><u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ${(\hat p_1-\hat p_2)-(p_1-p_2)}$ < $1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

P( $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ( $p_1-p_2$ ) < $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

<u>95% confidence interval for</u> ( $p_1-p_2$ ) =

[ $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ , $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ]

= [ $(0.25-0.275)-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ , $(0.25-0.275)+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ ]

= [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

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3 years ago

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2 years ago

Read 2 more answers

Help please this my second time posting this please help

Nesterboy [21]

Answer:

x = √30

Step-by-step explanation:

From small triangle BDC:

using Pythagorean theorem

CB² = BD² + DC²

x² = BD² + 3²

Fron triangles BDC and ADB.

ΔBDC has long leg BD and short leg DC.

ΔADB has long leg AD and shirt leg BD.

AD : BD = BD : DC

7 : BD = BD : 3

7*3 = BD*BD

BD² = 7*3 = 21

x² = BD² + 3² = 21+9 = 30

x² = 30

x = √30

4 0

3 years ago