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KATRIN_1 [288]
3 years ago
13

Which represents the polynomial written in standard form? 8x2y2 – StartFraction 3 x cubed y Over 2 EndFraction 4x4 – 7xy3 –7xy3

8x2y2 – StartFraction 3 x cubed y Over 2 EndFraction 4x4 4x4 – 7xy3 – StartFraction 3 x cubed y Over 2 EndFraction 8x2y2 4x4 8x2y2 – StartFraction 3 x cubed y Over 2 EndFraction – 7xy3 –7xy3 – StartFraction 3 x cubed y Over 2 EndFraction 8x2y2 4x4.
Mathematics
1 answer:
olga55 [171]3 years ago
8 0

Polynomial equation is a equation which is formed with coefficients variables and exponents with basic mathematics operation and equality sign. The given option the option A matches correctly with the above polynomial equation which is,

-7xy^3+8x^2y^2-\dfrac{3x^3y}{2}+4x^4

Hence the option A is the correct option.

<h3>Given information-</h3>

The given polynomial equation in the problem is,

8x^2y^2-\dfrac{3x^3y}{2}+4x^4-7xy^3

<h3>Polynomial equation </h3>

Polynomial equation is a equation which is formed with coefficients variables and exponents with basic mathematics operation and equality sign.

In the above polynomial equation the variable are<em> x </em>and <em>y </em>and the highest power of the variable <em>x</em> is four and highest power of the variable <em>y</em> is three.

Arrange the polynomial equation in the power of the variable <em>x. </em>Thus,

4x^4-\dfrac{3x^3y}{2}+8x^2y^2-7xy^3

Arrange the polynomial equation in the power of the variable <em>y. </em>Thus,

-7xy^3+8x^2y^2-\dfrac{3x^3y}{2}+4x^4

From the given option the option A matches correctly with the above polynomial equation which is,

-7xy^3+8x^2y^2-\dfrac{3x^3y}{2}+4x^4

Thus the option A is the correct option.

Learn more about the polynomial equation  here;

brainly.com/question/25958000

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Answer:

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                         P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean weight = \frac{\sum X}{n} = 21.88 ounces

            s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }  = 0.201 ounces

            n = sample of boxes = 12

            \mu = population mean weight

<em>Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.</em>

<u>So, 90% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-1.796 < t_1_1 < 1.796) = 0.90  {As the critical value of t at 11 degrees of

                                                  freedom are -1.796 & 1.796 with P = 5%}  

P(-1.796 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.796) = 0.90

P( -1.796 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 1.796 \times {\frac{s}{\sqrt{n} } } ) = 0.90

P( \bar X-1.796 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+1.796 \times {\frac{s}{\sqrt{n} } } ) = 0.90

<u>90% confidence interval for</u> \mu = [ \bar X-1.796 \times {\frac{s}{\sqrt{n} } } , \bar X+1.796 \times {\frac{s}{\sqrt{n} } } ]

                                        = [ 21.88-1.796 \times {\frac{0.201}{\sqrt{12} } } , 21.88+1.796 \times {\frac{0.201}{\sqrt{12} } } ]

                                        = [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

8 0
3 years ago
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