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Ilya [14]
3 years ago
7

4, 5, 6, 7, 6, 8, 5, 7, 4, 4, 6,5 Find the modes of this data set.

Mathematics
2 answers:
Zepler [3.9K]3 years ago
3 0

Answer:

4,5 and 6

Step-by-step explanation:

because they repeat the most times and they are repeated the exact same amount

IRINA_888 [86]3 years ago
3 0

Answer:

The modes are 4, 5, and 6.

Step-by-step explanation:

4, 5, 6, 7, 6, 8, 5, 7, 4, 4, 6, 5.

Organize them in order to make it easier.

--> 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 8.

There are three 4's, 5's, and 6's.

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1. Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants.
german

Answer:

Step-by-step explanation:

1.

To write the form of the partial fraction decomposition of the rational expression:

We have:

\mathbf{\dfrac{8x-4}{x(x^2+1)^2}= \dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2}}

2.

Using partial fraction decomposition to find the definite integral of:

\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}dx

By using the long division method; we have:

x^2-8x-20 | \dfrac{2x}{2x^3-16x^2-39x+20 }

                  - 2x^3 -16x^2-40x

                 <u>                                         </u>

                                            x+ 20

So;

\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x+\dfrac{x+20}{x^2-8x-20}

By using partial fraction decomposition:

\dfrac{x+20}{(x-10)(x+2)}= \dfrac{A}{x-10}+\dfrac{B}{x+2}

                         = \dfrac{A(x+2)+B(x-10)}{(x-10)(x+2)}

x + 20 = A(x + 2) + B(x - 10)

x + 20 = (A + B)x + (2A - 10B)

Now;  we have to relate like terms on both sides; we have:

A + B = 1   ;   2A - 10 B = 20

By solvong the expressions above; we have:

A = \dfrac{5}{2}     B =  \dfrac{3}{2}

Now;

\dfrac{x+20}{(x-10)(x+2)} = \dfrac{5}{2(x-10)} + \dfrac{3}{2(x+2)}

Thus;

\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)}

Now; the integral is:

\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx =  \int \begin {bmatrix} 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)} \end {bmatrix} \ dx

\mathbf{\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx =  x^2 + \dfrac{5}{2}In | x-10|\dfrac{3}{2} In |x+2|+C}

3. Due to the fact that the maximum words this text box can contain are 5000 words, we decided to write the solution for question 3 and upload it in an image format.

Please check to the attached image below for the solution to question number 3.

4 0
3 years ago
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