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Goshia [24]
3 years ago
10

6p to the 3rd power − 2p to the 2nd power − 8p

Mathematics
1 answer:
bija089 [108]3 years ago
4 0
The expression is
{6p}^{3}  -  {2p}^{2}  - 8p
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Marysya12 [62]

Step-by-step explanation:

-0.45 = 9/20

2.77= 2 7/9

5.55  = 5 5/9

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1. Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial i
Katena32 [7]
Polynomial with real coefficients always has even number of complex roots. We know that one of them is 2 + 5i so the second one will be  2 - 5i and:

f(x)=\big(x-4\big)\big(x-(-8)\big)\big(x-(2+5i)\big)\big(x-(2-5i)\big)=\\\\=(x-4)(x+8)(x-2-5i)(x-2+5i)=\\\\=(x^2-4x+8x-32)(x^2-2x+5ix-2x+4-10i-5ix+10i-25i^2)\\\\=\big(x^2+4x-32\big)\big(x^2-4x+4-25\cdot(-1)\big)=\\\\=(x^2+4x-32)(x^2-4x+29)=\\\\=x^4-4x^3+29x^2+4x^3-16x^2+116x-32x^2+128x-928=\\\\=\boxed{x^4-19x^2+244x-928}

Answer B.
8 0
3 years ago
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Graph the following Line:<br> 6x+3y=6
yulyashka [42]

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Visit geogebra.com

Step-by-step explanation:

none needed <3

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3 years ago
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During the 7th examination of the Offspring cohort in the Framingham Heart Study, there were 1219 participants being treated for
AlexFokin [52]

Answer:

95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

Step-by-step explanation:

We are given that during the 7th examination of the Offspring cohort in the Framing ham Heart Study, there were 1219 participants being treated for hypertension and 2,313 who were not on treatment.

The sample proportion is :  \hat p = x/n = 1219/3532 = 0.345

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

      P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion = 0.345

           n = sample of participants = 3532

           p = population proportion

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

So, 95% confidence interval for the population​ proportion, p is ;

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-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

<u>95% confidence interval for p</u>= [\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }]

    = [ 0.345-1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } } , 0.345+1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } } ]

    = [0.3293 , 0.3607]

Hence, 95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

6 0
3 years ago
Good morning yall &lt;3 say it back free points !!
NeTakaya

Answer:

okay thanks didn't expect this lol

6 0
3 years ago
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