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AleksAgata [21]
3 years ago
5

9-8x^2+2x^4 is this a trinomial​

Mathematics
1 answer:
Svetradugi [14.3K]3 years ago
8 0

Yes. A monomial is a term composed by a coefficient and powers of variables multiplied with each other, for example

3xy^3z^2

is a monomial.

In your case, the monomials are

9,\quad 8x^2,\quad 2x^4

As the name suggests, a trinomial is the sum of three monomials, which is exactly your case.

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13%=0.13

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What is the coefficient of the term of degree 4 in this polynomial?<br> x^8+2x^4 - 4x^3 + x^2-1
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2

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the coefficient is the number that multiply the letter

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3 years ago
Radical expression of 4d 3/8
DanielleElmas [232]

Answer:

\boxed{4 \sqrt[8]{ {d}^{3} } }

Step-by-step explanation:

=  > 4 {d}^{ \frac{3}{8} }   \\  \\ =   > 4({d}^{3 \times  \frac{1}{8} }) \\  \\  =  > 4( {d}^{3}  \times   {d}^{ \frac{1}{8} } ) \\  \\  =  > 4( {d}^{3}  \times  \sqrt[8]{d} ) \\  \\  =  > 4  \sqrt[8]{ {d}^{3} }

8 0
2 years ago
In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the fol
Julli [10]

Answer:

0.071,1.928

Step-by-step explanation:

                                                Downtown Store   North Mall Store

Sample size   n                             25                        20

Sample mean \bar{x}                         $9                        $8

Sample standard deviation  s       $2                        $1

n_1=25\\n_2=20

\bar{x_1}=9\\ \bar{x_2}=8

s_1=2\\s_2=1

x_1-x_2=9-8=1

Standard error of difference of means = \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}

Standard error of difference of means = \sqrt{\frac{2^2}{25}+\frac{1^2}{20}}

Standard error of difference of means = 0.458

Degree of freedom = \frac{\sqrt{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1}+\frac{(\frac{s_2^2}{n_2})^2}{n_2-1}}

Degree of freedom = \frac{\sqrt{(\frac{2^2}{25}+\frac{1^2}{20}})^2}{\frac{(\frac{2^2}{25})^2}{25-1}+\frac{(\frac{1^2}{20})^2}{20-1}}

Degree of freedom =36

So, z value at 95% confidence interval and 36 degree of freedom = 2.0280

Confidence interval = (x_1-x_2)-z \times SE(x_1-x_2),(x_1-x_2)+z \times SE(x_1-x_2)

Confidence interval = 1-(2.0280)\times 0.458,1+(2.0280)\times 0.458

Confidence interval = 0.071,1.928

Hence Option A is true

Confidence interval is  0.071,1.928

4 0
3 years ago
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